OP Malhotra Class 12 Maths Solutions Chapter 18 Probability Ex 18(d)

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S Chand Class 12 ICSE Maths Solutions Chapter 18 Probability Ex 18(d)

Question 1.
The probability of happening of the event A is V and that of the event B is ‘b\ Given that A and B are independent events; calculate the probability of
(i) happening of both the events A and B
(ii) not happening of both the events A and B
(iii) event A happens and B does not happen
(iv) event B happens and A does not happen
(v) the event B does not happen
Solution:
Given prob. of happening of event A = P (A) = a and P (B) = b
(i) prob. of happening of both the events A and B = P (A ∩ B) = P (A) P (B) = ab [since A and B are independent event]

(ii) required prob. = \(P(\bar{A} \cap \bar{B})=P(\bar{A}) P(\bar{B})\) = (1 – a) (1 – b) [∵ \(\frac { 1 }{ 2 }\) are independent events]

(iii) required prob. = \(P(A \cap \bar{B})=P(A) P(\bar{B})\) = P (A) [ 1 – P (B)] = a (1 – b)

(iv) required probability = \(P(\bar{A} \cap B)=P(\bar{A}) P(B)\) = [1 – P (A)] P (B) = (1 – a) b

(v) required prob. = \(\mathrm{P}(\overline{\mathrm{B}})\) = 1 – P (B) = 1 – b

Question 2.
Given that P (A) = 0.4, P (.B) = 0.7, P (A ∩ B) = 0.2, find
(i) P(A/B)
(ii) P(A’/B’)
(iii) P(A/B’)
(iv) P(A’/B’)
Solution:
Given P (A) = 0.4; P (B) = 0.7 ; P (A ∩ B) = 0.2

Question 3.
Given that P (A) = 0.8, P (B) = 0.7, P(C) = 0.6, P (A/B) = 0.8, P (C/B) = 0.7, P(A ∩ C) = 0.48, determine whether:
(i) A and B are independent, (ii) A and C are independent (ii) B and C are independent.
Solution:
Given P (A) = 0.8 ; P (B) = 0.7 ; P (C) = 0.6, P (A/B) = 0.8, P (C/B) = 0.7, P (A ∩ C) = 0.48 ;
(i) Given P (A/B) = 0.8 ⇒ \(\frac{P(A \cap B)}{P(B)}\) = 0.8 ⇒ P (A ∩ B) = 0.8 x 0.7 = 0.56
Also P (A) . P (B) = 0.8 x 0.7 = 0.56
∴ P (A ∩ B) = P (A) P (B)
Thus A and B are independent events

(ii) Now P (A ∩ C) = 0.48 = 0.8 x 0.6 = P (A) x P (C)
Thus A and C are independent events.

(iii) Now, P (C/B) = 0.7 ⇒ \(\frac{P(B \cap C)}{P(B)}\) = 0.7
⇒ P(B∩C) = 0.7 x 0.7 = 0.49
and P (B) . P (C) = 0.7 x 0.6 = 0.42
∴ P (B ∩ C) ≠ P (B) P (C)
Thus B and C are not independent events.

Question 4.
Given that C and D are independent and that P(C/D) = \(\frac { 2 }{ 3 }\), P(C∩D) = \(\frac { 1 }{ 3 }\), find
(i) P(C)
(ii) P(D)
Solution:

Question 5.
The events A and B are such that P (A’) = \(\frac { 3 }{ 4 }\), P (B) = \(\frac { 1 }{ 3 }\) and P (A ∪ B) = \(\frac { 2 }{ 3 }\). show that A and B are neither mutually exclusive nor independent.
Solution:
Given P (A) = \(\frac { 2 }{ 5 }\), P (B) = \(\frac { 1 }{ 6 }\); P(A∪B) = \(\frac { 13 }{ 30 }\)
We know that, P (A ∩ B) = P (A) + P (B) – P (A ∪ B)
= \(\frac{2}{5}+\frac{1}{6}-\frac{13}{30}=\frac{12+5-13}{30}=\frac{4}{30}=\frac{2}{15}\)
since P (A ∩ B) ≠ 0
Thus A and B are not mutually exclusive events.
Now P(A) . P(B) = \(\frac{2}{5} \times \frac{1}{6}=\frac{1}{15} \neq \mathrm{P}(\mathrm{A} \cap \mathrm{B})\)
Thus A and B are not independent events.
Hence A and B are neither mutually exclusive nor independent events.

Question 6.
The events A and B are such that P (A’) = \(\frac { 3 }{ 4 }\) , P (A/B) = \(\frac { 1 }{ 3 }\), P(A ∪ B)= \(\frac { 2 }{ 3 }\), where A’ denotes the event “X does not occur”. Find
(i) P(A)
(ii) P (A ∩ B)
(iii) P (B)
(iv) P (A/B’)
where B’ denotes the events “5 does not occur”. Deter mine whether A and B are independent.
Solution:

Question 7.
A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.
Solution:
Given a bag contains 4 white, 7 black and 5 red balls.
∴ Total no. of balls in a bag = 4 + 7 + 5 = 16
Since three balls are drawn one by one without replacement.
∴ required probability = \(\frac{4}{16} \times \frac{7}{15} \times \frac{5}{14}=\frac{1}{24}\)

Question 8.
There are three urns A, B and C Urn A contains 4 white balls and 5 blue balls. Urn B contains 4 white balls and 3 blue balls. Urn C contain 3 white balls and 6 blue balls. One ball is drawn from each of the urns. What is the probability that out of these three balls drawn two are white balls and one is blue ball?
Solution:
Given urn A contains 4 white balls and 5 blue balls
urn B contains 4 white and 3 blue balls
urn C contains 3 white and 6 blue balls
Since one ball is drawn from each of the three urns s.t out of there 3 balls, two are white and other is blue drawing of these balls can be done in three ways.
(1) white, white, blue
(2) white, blue, white
(3) blue, white, white
∴ required probability = P (WWB) + P (WBW) + P (BWW)
= \(\frac{4}{9} \times \frac{4}{7} \times \frac{6}{9}+\frac{4}{9} \times \frac{3}{7} \times \frac{3}{9}+\frac{5}{9} \times \frac{4}{7} \times \frac{3}{9}\)
= \(\frac{32}{189}+\frac{12}{189}+\frac{20}{189}=\frac{64}{189}\)

Question 9.
There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is cast. If the face 1 or 3 rums up, a ball is taken from the first bag. and if any other face turns up, a ball is chosen from the second bag. Find the probability of choosing a black ball.
Solution:
Let us define the events are as follows :
E : fare 1 or 3 turns up
A : black ball chosen from first bag
B : black ball chosen from second bag.
Thus P(E) = \(\frac{2}{6}=\frac{1}{3}\) ; P(\(\overline{E}\)) = 1 – P(E) = 1 – \(\frac{1}{3}=\frac{2}{3}\)
required probability = P (E) . P (A) + P (\(\overline{E}\)) P (B) = \(\frac{1}{3} \times \frac{3}{7}+\frac{2}{3} \times \frac{4}{7}=\frac{11}{21}\)

Question 10.
Three bags contain 5 white and 8 red; 7 white and 6 red; 6 white and 5 red balls respectively. One ball is drawn from each bag at random. Find the probability that all the three balls drawn are of the same colour.
Solution:
Given
bag I contains 5W and 8R
bag II contains 7W and 6R
bag III contains 6W and 5R
Since one ball is drawn from each bag at random.
∴ required probability = P (WWW) + P (RRR)
= \(\frac{5}{13} \times \frac{7}{13} \times \frac{6}{11}+\frac{8}{13} \times \frac{6}{13} \times \frac{5}{11}=\frac{210}{1859}+\frac{240}{1859}=\frac{450}{1859}\)

Question 11.
A bag contains 4 white and 2 black balls. Another contains 3 white and 5 black balls. If one ball is drawn from eact bag, find the probability that (i) both are white (ii) both are black (iii) one is white and one is black.
Solution:
bag I contains 4W, 2B
bag II contains 3W, 5B
since one ball is drawn from each bag
(i) ∴ required prob. that both balls are white = P(WW) = P(W) P(W) = \(\frac{4}{6} \times \frac{3}{8}=\frac{1}{4}\)
(ii) required prob. = P (BB) = \(\frac{2}{6} \times \frac{5}{8}=\frac{5}{24}\)
(iii) required probability = P (WB) + P (BW) = \(\frac{4}{6} \times \frac{5}{8}+\frac{2}{6} \times \frac{3}{8}=\frac{26}{48}=\frac{13}{24}\)

Question 12.
(i) The bag A contains 5 red and 3 green balls and bag B contains 3 red and 5 green balls. One ball is drawn from bag A and two from bag B. Find the probability that of the three balls drawn two are red and one is green.
(ii) Bag A contains 3 red and 5 black balls and bag B contains 2 red and 3 black balls. One ball is drawn from bag A and two from bag B. Find the probability that out of 3 balls drawn two are black and one is red.
Solution:
(i) Given bag A contains 5R and 3G and bag B contains 3R and 5G
Since one ball is drawn from bag A and two from bag B s.t out of 3 drawing balls, two are red and one is green.
∴ required probability = P (red ball from bag A and one red and one green from bag B) – P (green ball from bag A, two red from bag B)
= \(\frac{5}{8} \times \frac{{ }^5 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^8 \mathrm{C}_2}+\frac{3}{8} \times \frac{{ }^3 \mathrm{C}_2}{{ }^8 \mathrm{C}_2}=\frac{5}{8} \times \frac{5 \times 3}{\frac{8 \times 7}{2}}+\frac{3}{8} \times \frac{3}{\frac{8 \times 7}{2}}\)
= \(\frac{75+9}{8 \times 4 \times 7}=\frac{84}{32 \times 7}=\frac{12}{32}=\frac{3}{8}\)

(ii) Given bag A contains 3 red and 5 black balls bag B contains 2 red and 3 black balls
Since one ball is drawn from bag A and two from bag B s.t out of 3 drawing balls, one is red and two are black.
So there are two possibilities :
(I) drawing one red ball from bag A and 2 black balls from bag B
(II) drawing one black ball from bag A and 1 red, 1 black ball from bag B
∴ required probability = P (I) + P (II)
= \(\frac{3}{8} \times \frac{{ }^3 \mathrm{C}_2}{{ }^5 \mathrm{C}_2}+\frac{5}{8} \times \frac{{ }^2 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^5 \mathrm{C}_2}=\frac{3 \times 3+5 \times 2 \times 3}{8 \times \frac{5 \times 4}{2}}=\frac{39}{80}\)

Question 13.
Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls one by one without replacement. What is the probability that
(i) both the balls are of same colour?
(ii) at least one ball is red?
Solution:
Given urn contains 2 white, 3 red and 4 black balls
∴ Total no. of balls = 2 + 3 + 4 = 9
(i) required probability = P (WW) + P (RR) + P (BB)
= \(\frac{2}{9} \times \frac{1}{8}+\frac{3}{9} \times \frac{2}{8}+\frac{4}{9} \times \frac{3}{8}=\frac{2+6+12}{72}=\frac{20}{72}=\frac{5}{18}\)
(ii) required probability = P (RW) + P (RB) + P (RR) + P (WR) + P (BR)
= \(\frac{3}{9} \times \frac{2}{8}+\frac{3}{9} \times \frac{4}{8}+\frac{3}{9} \times \frac{2}{8}+\frac{2}{9} \times \frac{3}{8}+\frac{4}{9} \times \frac{3}{8}\)
= \(\frac{6+12+6+6+12}{72}=\frac{42}{72}=\frac{7}{12}\)

Question 14.
(i) Two cards are drawn without replace- ment from a well shuffled pack of 52 cards. Find then probability that one is a spade and the other is a queen of red colour.
(ii) Two cards are drawn without replace- ment from a well shuffled pack of 52 cards. What is the probability that one is a red queen and the other is a king of black colour?
(iii) Two cards are drawn one by one without replacement from a pack of 52 cards. What is the probability that one is the red and the other is a black card?
(iv) Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement Find the probability that one of these is a queen and the other is a king of the opposite shade.
Solution:
(i) prob. of drawing a spade card in first draw = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
and prob. of drawing a queen of red colour in second draw = \(\frac { 2 }{ 51 }\)
prob. of drawing = queen of red colour in first draw = \(\frac { 2 }{ 52 }\)
and prob. of drawing a spade card in 2nd draw = \(\frac { 13 }{ 51 }\)
∴ required probability = \(\frac{1}{4} \times \frac{2}{51}+\frac{2}{52} \times \frac{13}{51}=\frac{1}{102}+\frac{1}{102}=\frac{2}{102}=\frac{1}{51}\)

(ii) prob. of drawing a red queen in first draw = \(\frac { 2 }{ 52 }\)
and prob. of drawing a king of black colour in second draw = \(\frac { 2 }{ 51 }\) [since cards are drawn without replacement]
prob. of drawing a king of black colour in first draw = \(\frac { 2 }{ 52 }\)
and prob. of drawing a red queen in second draw = \(\frac { 2 }{ 51 }\)
∴ required Probability = \(\frac{2}{52} \times \frac{2}{51}+\frac{2}{52} \times \frac{2}{51}=\frac{8}{52 \times 51}=\frac{2}{663}\)

(iii) prob. of drawing a red card in first draw = \(\frac { 26 }{ 52 }\)
and prob. of drawing a black card in second draw = \(\frac { 26 }{ 52 }\)
Also prob. of drawing a black card in first draw = \(\frac { 26 }{ 52 }\)
and prob. of drawing a red card in 2nd draw = \(\frac { 26 }{ 51 }\)
∴ required prob. = \(2 \times \frac{26}{52} \times \frac{26}{51}=\frac{26}{51}\)

(iv) prob. of drawing a queen in first draw = \(\frac { 4 }{ 52 }\)
and prob. of drawing a king of opposite shade = \(\frac { 2 }{ 51 }\)
Since cards are drawn one after other without replacement.
∴ prob. of drawing a king card in first draw = \(\frac { 4 }{ 52 }\)
prob. of drawing a queen of opposite shade in 2nd draw = \(\frac { 2 }{ 51 }\)
∴ required probability = \(\frac{4}{52} \times \frac{2}{51}+\frac{4}{52} \times \frac{2}{51}=\frac{16}{52 \times 51}=\frac{4}{663}\)

Question 15.
Find the probability of drawing one rupee coin from a purse with two compartments one of which contains 3 fifty paisa coins and 2 one-rupee coins and other contains 2 fifty paisa coins and 3 one rupee coins.
Solution:
prob. of drawing a first compartment = prob. of drawing second compartment = \(\frac { 1 }{ 2 }\)
Given
compartment I contains 3 fifty paise coin and 2 one rupee coin
compartment II contains 2 fifty paisa and 3 one rupee coin
∴ required probability = P (choosing one rupee coin from compartment-I) + P (choosing one rupee coin from compartment-II)
= \(\frac{1}{2} \times \frac{2}{5}+\frac{1}{2} \times \frac{3}{5}=\frac{5}{10}=\frac{1}{2}\)

Question 16.
A bag contains 4 yellow and 5 red balls and another bag contains 6 yellow and 3 red balls. A ball is drawn from the first bag and without seeing its colour, it is put into the second bag. Find the probability that if now a ball is drawn from the second bag, it is of yellow colour.
Solution:
Given bag-I contains 4 yellow and 5 red balls and bag-II contains 6 yellow and 3 red balls
Case-I.
When a yellow ball transferred from I to II
Then bag-II contains 7 yellow and 3 red balls
Thus prob. of drawing a yellow ball from second bag P (I) = \(\frac { 4 }{ 9 }\) x \(\frac { 7 }{ 10 }\)

Case-II.
When a red ball transferred from I to II then bag-II contains 6 yellow and 4 red balls

Question 17.
A bag contains 4 white and 3 black bails. Four balls are successively drawn out with replacement. Find the probability that they are alternately of different colours.
Solution:
Given bag contains 4 white and 3 black balls
∴ Prob. of drawing white ball = \(\frac { 4 }{ 3 }\) = P (W)
prob. of drawing black ball = P (B) = \(\frac { 3 }{ 7 }\)
Thus required probability = P (WBWB, BWBW) = P (WBWB) + P (BWBW)
= \(\frac{4}{7} \times \frac{3}{7} \times \frac{4}{7} \times \frac{3}{7}+\frac{3}{7} \times \frac{4}{7} \times \frac{3}{7} \times \frac{4}{7}=2 \times \frac{12 \times 12}{7^4}=\frac{288}{2401}\)

Question 18.
(i) A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.
(ii) A bag contains 2 white and 4 black balls while another bag contains 4 white and 2 black balls. A bag is selected at random and a ball is drawn. Find the probability that the ball drawn is of black colour.
Solution:
(i) Given bag A contains 4 red and 3 black balls and bag B contains 2 red and 4 black balls
∴ prob. of selecting a bag A = prob. of selecting a bag B = \(\frac { 1 }{ 2 }\)
Thus prob. of drawing a red ball from bag A = \(\frac{1}{2} \times \frac{4}{7}=\frac{2}{7}\)
Similarly prob. of drawing a red ball from bag B = \(\frac{1}{2} \times \frac{2}{6}=\frac{1}{6}\)
∴ required prob. of drawing a red ball from either of two bags = \(\frac{2}{7}+\frac{1}{6}=\frac{19}{42}\)

(ii) Given bag A contains 2 white and 4 black balls
and bag B contains 4 white and 2 black balls
∴ prob. of choosing bag A = prob. of choosing bag B = \(\frac { 1 }{ 2 }\)
Thus prob. of drawing a black ball from bag A = \(\frac{4}{6} \times \frac{1}{2}\)
prob. of drawing a black ball from bag B = \(\frac{1}{2} \times \frac{2}{6}\)
∴ required prob. = \(\frac{4}{12}+\frac{2}{12}=\frac{6}{12}=\frac{1}{2}\)

Question 19.
(i) A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases are they likely to contradict each in stating the same fact?
(ii) A problem of statistics is given to three students A, B and C whose chances of solving it are \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\) and \(\frac { 1 }{ 4 }\) respectively, find the probability that only one of them solves the problem.
(iii) A speaks truth in 75% of the cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in starting the same fact ? In what percentage of cases do they (a) contradict, (b) agree with each other?
Solution:
(i) Given P (A speaks truth) = P₁ = 60% = \(\frac { 60 }{ 100 }\)

(ii) Let E₁, E₂ and E₃ be the events that the problem is solved by A, B and C respectively.
Let p₁, p₂ and p₃ be their corresponding probabilities.

(iii) Given P (A speaks truth) = 75% = p₁ = \(\frac { 75 }{ 100 }\)
P (A tell a lie) = q₁ = 1 – p₁ = \(\frac { 25 }{ 100 }\)
P (B speaks truth) = p₂ = 80% = \(\frac { 80 }{ 100 }\)
P (B tell a lie) = q₂ = 20% = \(\frac { 20 }{ 100 }\)
∴ required probability that both A and B are likely to contradict each other on same fact
\(p_1 q_2+p_2 q_1=\frac{75}{100} \times \frac{20}{100}+\frac{80}{100} \times \frac{25}{100}=\frac{1500+2000}{10000}=\frac{3500}{10000}=\frac{35}{100}\)
Thus both persons A and B contradict each other in 35% cases.

(b) ∴ required probability = p₁p₂ + q₁q₂\(\frac{75}{100} \times \frac{80}{100}+\frac{25}{100} \times \frac{20}{100}=\frac{6000+500}{10000}=\frac{65}{100}\)
Thus in 65% cases, both A and B agree with each other on same fact.

Question 20.
(i) A husband and wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is \(\frac { 1 }{ 7 }\) and that of wife’s selection is \(\frac { 1 }{ 5 }\). What is the probability that (a) both of them will be selected (b) only one of them will be selected (c) none of them will be selected (d) At least one of them will be selected ?

(ii) A candidate is selected for interview for three posts. For the first post there are 5 candidates, for the second there are 8 and for the third there are 7. What are the chances for his getting at least one post?

(iii) The probability of a student/4 passing an examination is \(\frac { 5 }{ 8 }\) and that of B passing is \(\frac { 2 }{ 3 }\)
Assuming that the two events “A passes” and “B passes” are independent, find the probability of only one of them passing the examination.
Solution:
(i) Given P (husband’s selection) = p₁ = \(\frac { 1 }{ 7 }\)

(ii) Let us define the events are as follows :
A : candidate selected for first post
B : candidate selected for 2nd post
C: candidate selected for 3rd post

(iii) Given E₁ : student A pass the examination.
E₁ : student B pass the examination.

Question 21.
(i) The probability of hitting a target by three marksmen is \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\) and \(\frac { 1 }{ 4 }\) respectively. Find the probability that one and only one of them will hit the target when they fire simultaneously.
(ii) A, B and C shoot to hit a target. If A hits target 4 times in 5 trials, B hits it 3 times in 4 trials and C hits it 2 times in 3 trials, what is the probability that the target is hit by at least 2 persons?
Solution:
(i) p₁ = prob. (that first mark man hit the target) =\(\frac { 1 }{ 2 }\)
p₂ = prob. (that second marks man hit the target) = \(\frac { 1 }{ 3 }\)
p₃ = prob. (that third marks man hit the target) = \(\frac { 1 }{ 4 }\)

(ii) Let E₁ : A hits the target
E₁: B hits the target
E₃ : C hits the target

Question 22.
The odds that a Ph.D. thesis will be favourably reviewed by three inde- pendent examiners are 5 to 2.4 to 3, and 3 to 4 respectively. What is the probability that of the three examiners:
(a) all reject the thesis, (b) all approve the thesis, (c) a majority approve the thesis?
Solution:
Given odds in favour of examiner A are in the ratio 5 : 2
∴ P(A) = \(\frac{5}{5+2}=\frac{5}{7} ; \text { and } \mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A})=1-\frac{5}{7}=\frac{2}{7}\)
also odds in favour of examiner B are in the ratio 4 : 3 4 4
∴ P(B) = \(\frac{4}{4+3}=\frac{4}{7} ; \quad P(\bar{B})=1-\frac{4}{7}=\frac{3}{7}\)
odds in favour of examiner C are in the ratio 3 : 4
∴ P(C) = \(\frac{3}{3+4}=\frac{3}{7} ; \quad P(\bar{C})=1-\frac{3}{7}=\frac{4}{7}\)
(a) Thus required probability that all the three examiners reject the thesis = \(P(\bar{A}) P(\bar{B}) P(\bar{C})\)
= \(\frac{2}{7} \times \frac{3}{7} \times \frac{4}{7}=\frac{24}{343}\)

(b) Thus, required prob. that all the three examiners will approve the thesis = P (A) P (B) P (C)
= \(\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}=\frac{60}{343}\)

(c) ∴ required prob. that majority of examiners will approve the thesis
= P (\(\overline{A}\)) P (B) P (C) + P (A) P (\(\overline{B}\)) P (C) + P (A) P (B) P (\(\overline{C}\)) + P (A) P (B) P (C)
= \(\frac{2}{7} \times \frac{4}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7}=\frac{24+45+80+60}{343}=\frac{209}{343}\)

Question 23.
(i) Two persons A and B throw a die alternately till one of them gets a three and wins the game. Find their respective probabilities of winning, if A begins.
(ii) A, B, C in order cut a pack of cards, replacing them after each cut, on the condition that the first who cut a spade shall win a prize; find their respective chances of winning, assuming that the game may continue indefinitely.
Solution:
(i) Let p = prob. of success i.e. prob. of getting 3 in single toss of a die = \(\frac { 1 }{ 6 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Since A starts the game so A can win in first throw, 3rd throw, 5th throw and so on
∴ probability of A’s winning in first throw = \(\frac { 1 }{ 6 }\) = p
probability of A’s winning in 3rd throw = qqp = (\(\frac { 5 }{ 6 }\))²\(\frac { 1 }{ 6 }\)
probability of A’s winning in 5th throw = qqqqp = \(\left(\frac{5}{6}\right)^4 \frac{1}{6}\) and so on
Since all thesecases are mutually exclusive
∴ required prob. of A’s winning =p + qqp + qqqqp + …

Since A and B are mutually exclusive and exhaustive events.
∴ required probability of B’s winning the game first = 1 – P (A) = 1 – \(\frac { 6 }{ 11 }\) = \(\frac { 5 }{ 11 }\)

(ii) Let p = probability of success = prob. of getting a spade card = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)
When A starts the game so he can win in 1st throw, 4th throw, 7th throw and so on.
Thus prob. of A’s winning in first throw = p = \(\frac { 1 }{ 4 }\)
prob. of A’s winning in 4th throw = qqqp = (\(\frac { 3 }{ 4 }\))³\(\frac { 1 }{ 4 }\)
prob. of A’s winning in 7th throw = qqqqqqp = (\(\frac { 3 }{ 4 }\))⁶\(\frac { 1 }{ 4 }\) and so on
Therefore prob. of A’s winning the game first =p + qqqp + qqqqqqp +

Clearly B will win the game in 2 nd throw, 5 th throw and 8 th throw and so on.
∴ prob. of B’s winning the game first = qp + qqqqp + qqqqqqqp + ……… ∞

Question 24.
(i) Three persons A, B, C throw a die in succession in the same order till one of them gets a ‘six’ and wins the game. If A starts the game, find their respective probabilities of winning.
(ii) A and B take turn in throwing two dice, the first to throw 9 being awarded. Show that if A has the first throw, their chances of winning are in the ratio 9 : 8.
Solution:
(i) Let p = prob. of success = prob. of getting a six = \(\frac { 1 }{ 6 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Given A starts the game so A can win in first throw, 4th throw, 7th throw and so on.
∴ prob. of A’s winning the game first =p + qqqp + qqqqqqp + ….. ∞

B wins the game if he throws a six in 2nd, 5th and 8th throw.
i.e. B wins the game first in 2nd, 5th and 8th throw

Since A, B and C are mutually exclusive and exhaustive events
∴ P (C wins) = 1 – P (A wins) – P (B wins) = \(1-\frac{36}{91}-\frac{30}{91}=\frac{91-66}{91}=\frac{25}{91}\)

(ii) Let p = prob. of success = prob. of getting 9 in a single throw of two dice
⇒ p = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 9 }\) = \(\frac { 8 }{ 9 }\)
[Here favourable cases = {(3, 6), (4, 5), (5, 4), (6, 3)} and Total no. of outcomes = 6² = 36]
given A has the first throw. So A can wins the game in first throw, 3rd throw, 5th throw and so on.
∴ required prob. of A’s winning = p + qqp + qqqqp + …. ∞

Since A and B are mutually exclusive and exhaustive events.
∴ P (B wins) = 1 – P (A’s wins) = 1 – \(\frac { 9 }{ 17 }\) = \(\frac { 8 }{ 17 }\)
Thus, the required ratio of chance of winning A and B
= P (A wins): P (B wins) = \(\frac { 9 }{ 17 }\) : \(\frac { 8 }{ 17 }\) i.e. 9 : 8

Question 25.
(i) A man alternately tosses a coin and throws a die beginning with the coin. Find the probability he gets a head in the coin before he gets a 5 or 6 in the die.
(ii) A and B throw with a pair of dice. A wins if he throws 7 before B throws 8 and B wins if he throws 8 before A throws 7. If A begins, show that his chance of winning is \(\frac { 36 }{ 61 }\). What is B’s chance of winning?
(iii) A and B throw a pair of dice alternately. A wins the game, if he gets a total of 1 and B
wins the game, if he gets a total of 10. If A starts the game, then find the probability that B wins.
(iv) In a hockey match, both tear is A and B scored same number of goals up to the end of the game, so to decide the winner, the referee asked both the captains to throw die alternately and decided that the team, whose captain gets a six first, will be declared the winner, if the captain of team A was asked to start, then find heir respective probabilities of winning the match and state whether the decision of the referee was fair or not.
Solution:
(i) Let us define the events are as follows :
A : Man gets head and B : Man gets 5 or 6

(ii) Let us define the events are as follows :
A : A gets 7 with a pair of dice = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B : B gets 8 = {(2, 6), (3, 5), (4, 4), (5,3), (6,2)}

Since A starts the game. So A can wins in first throw, 3rd throw, 5th throw and so on.
Thus P (prob. of A’s winning the game first) = p₁ + q₁q₂P₁ + q₁q₂q₁q₂p₁ + … ∞

Since A and B are mutually exclusive and exhausitive events.
∴ P(B) = 1 – P(A) = 1 – \(\frac { 36 }{ 61 }\) = \(\frac { 25 }{ 61 }\)

(iii) Let us define the events are as follows :
A : A gets a total of 7 with pair a dice B : B gets a total of 10
∴ A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B = {(4, 6), (5, 5), (6, 4)}
Since A starts the game. Thus A can wins the game first in first throw, 3rd throw, 5th throw and so on.

(iv) Let p = prob. of success = prob. of getting a six = \(\frac { 1 }{ 6 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)
Since team A’s captain starts the game.
So, A can wins in first throw, 3rd throw, 5th throw and so on.
∴ P (A’s winning of the game first) = p + qqp + qqqqp + …. ∞
= \(\frac{1}{6}+\left(\frac{5}{6}\right)^2 \frac{1}{6}+\left(\frac{5}{6}\right)^4 \frac{1}{6}+\ldots . \infty\)
= \(\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{6}{11}\)
Since A and B are mutually exclusive and exhaustive events.
Thus P (B’s winning) = 1 – P (A’s winning) = 1 – \(\frac { 6 }{ 11 }\) = \(\frac { 5 }{ 11 }\)
Since P (A) ≠ P (B). Thus the decision of Refree is not fair.

Question 26.
A bag contains 6 red 5 blue balls and another bag contains 5 red and 8 blue balls. A ball is drawn from the first bag and without noticing colour is put in the second bag. A ball is then drawn from the second bag. Find the probability that the bail drawn is blue in colour.
Solution:
Given
bag I contains 6 red and 5 blue balls
bag II contains 5 red and 8 blue balls
Case-I. When a red ball transferred from bag I to bag II
Then bag II contains 6 red and 8 blue balls.
∴ Prob. that the ball drawn from second bag be blue in colour = \(\frac { 6 }{ 11 }\) x \(\frac { 8 }{ 14 }\)

Case-II. When a blue ball transferred from bag I to bag II.
Then bag II contains 5 red ball and 9 blue balls.
Thus the probability that the ball drawn from second bag be blue in colour = \(\frac { 5 }{ 11 }\) x \(\frac { 9 }{ 14 }\)
Thus required probability = \(\frac{6}{11} \times \frac{8}{14}+\frac{5}{11} \times \frac{9}{14}=\frac{93}{154}\)

Question 27.
Box A contains 3 red balls and 2 black balls and box B contains 2 red balls and 3 black balls. One ball is drawn from box A and placed in box B. Then one bail is drawn at random from box B and placed in box A. Find the probability that the composition of the balls in the two boxes remains unaltered.
Solution:
Let us define the events are as follows :
E₁ : Red ball is transferred from bag A to bag B
E₂ : black ball is transferred from bag A to bag B
E₃ : Red ball is transferred from bag B to bag A
E₄ : black ball is transferred from bag B to bag A
Thus, \(P\left(E_1\right)=\frac{3}{5} ; P\left(E_2\right)=\frac{2}{5} ; P\left(E_3\right)=\frac{3}{6} ; P\left(E_4\right)=\frac{4}{6}\)
∴ required probability = \(P\left(E_1\right) P\left(E_3\right)+P\left(E_2\right) P\left(E_4\right)=\frac{3}{5} \times \frac{3}{6}+\frac{2}{5} \times \frac{4}{6}=\frac{17}{30}\)

Question 28.
Three peopled, B and C have probabilities \(\frac { 3 }{ 5 }\), \(\frac { 2 }{ 5 }\) and \(\frac { 3 }{ 4 }\) respectively of hitting a target with a rifle shot. Calculate the probability of there being exactly 2 hits if each A, B and C fires once at the target. A, B and C decide to participate in a contest in which each fires once and the first to hit the target receive a prize of ₹ 94. If they agree to fire in the order A, B and then C, calculate how much of ₹ 94 prize money each should contribute so that this game is fair.
Solution:
Given P (A) = prob. of hitting the target by A = \(\frac { 3 }{ 5 }\) ; P (B) = prob. of hitting the target by B = \(\frac { 2 }{ 5 }\)
P (C) = prob. of hitting the target by C = \(\frac { 3 }{ 4 }\)

(ii) Probability of A’s success
∴ prob. of B’s success = prob. of a’s failure x prob. of B’s success = \(\frac { 2 }{ 5 }\) x \(\frac { 2 }{ 5 }\) = \(\frac { 4 }{ 25 }\)
Since C takes a chance only if A and B both fails
∴ prob. of C s success = \(\frac{2}{5} \times \frac{3}{5} \times \frac{3}{4}=\frac{18}{100}=\frac{9}{50}\)
Thus A, B and C contribute in the ratio \(\frac{3}{5}: \frac{4}{25}: \frac{9}{20} \text { i.e. } \frac{3}{5} \times 50: \frac{4}{25} \times 50: \frac{9}{50} \times 50\) i.e. 30 : 8 : 9
∴ A’s contribution = ₹\(\left(\frac{30}{47} \times 94\right)\) = Rs. 60
B’s contribution = ₹\(\left(\frac{8}{47} \times 94\right)\) = Rs. 16
C’s contribution = ₹\(\left(\frac{9}{47} \times 94\right)\) = Rs. 18

Example

Question 1.
A bag contains 20 balls marked 1 to 20. One ball is drawn at random from the bag. What is the probability that the ball drawn is marked with a number which is a multiple of 5 or 7?
Solution:
Here S = {1,2, 3, 4, ….,20}
Let A : event that the ball is marked with multiple of 5 = {5, 10, 15, 20}
B : event that the ball is marked with multiple of 7 = {7, 14}

Question 2.
A problem in mathematics is given to four students A, B, C and D. Their chances of solving the problem are \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\), \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 5 }\) respectively. What is the probability that the problem will be Solved?
Solution:

Question 3.
The probability that a contractor will get a plumbing contract is \(\frac { 2 }{ 3 }\) and electric contract is \(\frac { 4 }{ 9 }\). If the probability of getting at least one contract is \(\frac { 4 }{ 5 }\), find the probability that he will get both the contracts.
Solution:
Let us define the events are as follows:
A : Contractor will get a plumbing contract
B : Contractor will get an electric contract

Question 4.
A bag has 4 red and 5 black, a second bag has 3 red and 7 black balls. One ball is drawn. Find the probability that two balls are black and one is red.
Solution:
Given bag-I contains 4 and 5 black balls
bag-II contains 3 red and 7 black balls
required probability = P (one black ball from bag-I and one black, one red ball from bag II) + P (one red ball from bag-I and black balls from bag
= \(\frac{5}{9} \times \frac{3 \times 7}{{ }^{10} \mathrm{C}_2}+\frac{4}{9} \times \frac{{ }^7 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}=\frac{5 \times 3 \times 7+2 \times 7 \times 6}{9 \times \frac{10 \times 9}{2}}=\frac{105+84}{810}=\frac{189}{405}=\frac{7}{15}\)

Question 5.
Tickets numbered from 1 to 20 are mixed up together and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 7 ?
Solution:
Let us define the events are as follows :
A : event that the ticket has a number which is multiple of 3 = {3, 6, 9, 12, 15, 18}
B : event that the ticket has a number which is multiple of 7 = {7, 14}

Question 6.
In a certain city, the probability of not reading the morning newspaper by the residents is \(\frac { 1 }{ 2 }\) and that of not reading the evening newspaper is \(\frac { 2 }{ 5 }\). The probability of reading both the newspapers is \(\frac { 1 }{ 5 }\). Find the probability that a resident reads either the morning or evening or both the papers.
Solution:
Let us define the events as follows :
A : reading morning newspaper by the residents
B : reading evening newspaper by the residents

Question 7.
A candidate is selected for interview of management trainees in 3 companies. For the first company, there are 12 candidates for the second there are 15 candidates and for the third, there are 10 candidates. Find the probability that he is selected by at least one of the companies.
Solution:
Let us define the events are as follows :
A : candidate selected by first company
B : candidate selected by second company
C : candidate selected by third company

Question 8.
The probability of A, B and C solving a problem are \(\frac { 1 }{ 3 }\), \(\frac { 2 }{ 7 }\) and \(\frac { 3 }{ 8 }\) respectively. If all try and solve the problem simultaneously, find the probability that only one of them will solve it.
Solution:
Let us define the events are as follows :
E₁ : problem is solved by A
E₂ : problem is solved by B
E₃ : problem is solved by C

Question 9.
A and B throw two dice each. If A gets a sum of 9 on his two dice, then find the probability of 3 gening a higher sum.
Solution:
When two dice are thrown then n (S) = 36
If A getting a sum of 9 with two dice.
Then B should get a sum of 10, 11 or 12 i.e. B = {(4,6), (5, 5), (6, 4), (5, 6), (6, 5), (6,6)}
∴ n (B) = 6
Thus required probability = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{6}{36}=\frac{1}{6}\)

Question 10.
Kamal and Monica appear for an inter- view for two vacancies. The probability of Kamal’s selection is \(\frac { 1 }{ 3 }\) and that oPMonica’s selection is \(\frac { 1 }{ 5 }\). Find the probability that only one of them will be selected.
Solution:
Let us define the events are as follows :
A: Komal’s selection
B : Monica’s selection
∴ P(A) = \(\frac { 1 }{ 3 }\) ; P(B) = \(\frac { 1 }{ 5 }\)
Thus required Probability = \(P(A \cap \bar{B})+P(\bar{A} \cap B)=P(A) P(\bar{B})+P(\bar{A}) P(B)\)
= \(\frac{1}{3}\left(1-\frac{1}{5}\right)+\left(1-\frac{1}{3}\right) \frac{1}{5}=\frac{1}{3} \times \frac{4}{5}+\frac{2}{3} \times \frac{1}{5}=\frac{6}{15}=\frac{2}{5}\)

Question 11.
The bag ‘A’ contains 3 white and 2 black balls while the bag ‘B’ contains 2 white and 5 black balls. One of the bags is chosen at random and a ball is drawn from it. What is the probability that the ball is white.
Solution:
Given bag A contains 3 white and 2 black balls and bag B contains 2 white and 5 black bails
Also, probability of chosing bag A = prob. of chosing bag B = \(\frac { 1 }{ 2 }\)
∴ prob. of drawing a white ball from bag A = \(\frac{1}{2} \times \frac{3}{5}=\frac{3}{10}\)
prob. of drawing a white ball from bag B = \(\frac{1}{2} \times \frac{2}{7}=\frac{1}{7}\)
∴ required prob. that white ball is drawn from either of the bag = \(\frac{3}{10}+\frac{1}{7}=\frac{21+10}{70}=\frac{31}{70}\)

Question 12.
In a single throw of two dice find the probability of getting a total of at most 9.
Solution:
When two dice are thrown
Then total no. of exhausitve cases = 6² = 36
E : event of getting a total of 10, 11 or 12 = {(4,6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}
∴ P(E) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
Thus required probability’ of getting a total of atmost 9 = 1 – P(E) = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\)

Question 13.
Three are 3 urns A, B and C. Urn A contains 4 red balls and 3 black balls. Urn B contains 5 red balls and 4 black balls. Urn C contains 4 red balls and 4 black balls. One ball is drawn from each of these urns. What is the probability that the 3 balls drawn consist of 2 red balls and 1 black ball?
Solution:
Given urn A contains 4 red balls and 3 black balls urn B contains 5 red balls and 4 black balls and urn C contains 4 red balls and 4 black balls So there are following possibilities :
(I) red ball from urn A, red ball from B, black ball from C
(II) red ball from urn A, black ball from B, red ball from C
(III) black ball from urn A, red ball from B and from urn C
∴ required probability = P (I) + P (II) + P (III)
= \(\frac{4}{7} \times \frac{5}{9} \times \frac{4}{8}+\frac{4}{7} \times \frac{4}{9} \times \frac{4}{8}+\frac{3}{7} \times \frac{5}{9} \times \frac{4}{8}=\frac{80+64+60}{504}=\frac{204}{504}=\frac{17}{42}\)

Question 14.
The probability that a teacher will give an unannounced test during any class meeting is \(\frac { 1 }{ 5 }\). If a student is absent twice, find the probability that the student will miss at least one test.
Solution:
Let us define the events as follows :
E₁ : event that the student misses the first test and E₂ : the student misses the second test
∴ E₁ and E₂ are independent events.
Thus P (E₁) = P (E₂) = \(\frac { 1 }{ 5 }\)
∴ required probability = 1 – \(P\left(\overline{\mathrm{E}}_1\right) P\left(\overline{\mathrm{E}}_2\right)=1-\left(1-\frac{1}{5}\right)\left(1-\frac{1}{5}\right)=1-\frac{4}{5} \times \frac{4}{5}=\frac{9}{25}\)

Question 15.
Bag A contains 5 white and 4 black balls, any bag B contains 7 white and 6 black balls. One rail is drawn from the bag A and without noticing its colour, is put in the bag B. If the ball is drawn from bag B, find the probability that it is black in colour.
Solution:
Case-I: When a white ball is transferred from bag A to bag B then bag B contains 8 white balls and 6 black balls.
∴ probability of drawing a black ball from bag B = \(\frac { 5 }{ 9 }\) x \(\frac { 6 }{ 14 }\)

Case-II: When a black ball is transferred from bag A to bag B then bag B contains 7 white and 7 black balls.
∴ prob. of drawing a black ball from bag B = \(\frac { 4 }{ 9 }\) x \(\frac { 7 }{ 14 }\)
Hence the required probability = \(\frac{5}{9} \times \frac{6}{14}+\frac{4}{9} \times \frac{7}{14}=\frac{58}{126}=\frac{29}{63}\)

Question 16.
An article manufactured by a company consists of two parts A and B. In the process of manufacture of part/4, 9 out of 104 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part B. Calculate the probability that the article manufactured will not be defective.
Solution:
Let us define the events are as follows :
A : part A is defective and B : part B is defective

Question 17.
Two horses are considered for a race. The probability of selection of the first horse is \(\frac { 1 }{ 4 }\) and that of the second is \(\frac { 1 }{ 3 }\). What is the probability that:
(i) both of them will be selected?
(iii) none of them will be selected?
Solution:
Let E₁ : event that first horse is selected for race;
E₂ : event that second horse is selected for race
∴ P(E₁) = \(\frac { 1 }{ 4 }\) ; P(E₂) = \(\frac { 1 }{ 3 }\)
sines E₁ and E₂ are independent events.

Question 18.
Akhii and Vijay appear for an interview for two vacancies. The probability of Akhil’s selection \(\frac { 1 }{ 4 }\) and Vijay’s selection is \(\frac { 2 }{ 3 }\). Find the probability that only one of them will be selected.
Solution:
Let us define the events as follows :
E₁ : event that Akhii is selected for job; E₂ : event that vijay is selected for Job

Question 19.
There are two bags. One bag contains six green and three red balls. The second bag contains five green and four red balls. One ball is transferred from the first bag to the second bag. Then one ball is drawn form the second bag. Find the probability that it is a red bail.
Solution:
Case-I: When a green ball transferred from bag A to bag B. Then bag B contains 6 green balls and four red balls.
Thus the probability of drawing a red ball from bag B = \(\frac { 6 }{ 9 }\) x \(\frac { 4 }{ 10 }\)

Case-II: When a red ball transferred from bag A to bag B. Then bag B contains 5 green and 5 red balls.
Thus the prob. of drawing a red ball from bag B = \(\frac{3}{9} \times \frac{5}{10}\)
∴ required probability = \(\frac{6}{9} \times \frac{4}{10}+\frac{3}{9} \times \frac{5}{10}=\frac{24+15}{90}=\frac{39}{90}=\frac{13}{30}\)

Question 20.
A word consists of 9 different alphabets, in which there are 4 consonants and 5 vowels. Three alphabets are chosen at random. What is the probability that more than one vowel will be selected?
Solution:
Given no. of consonants = 4 and no. of vowels = 5
∴ Total no. of alphabets = 4 + 5 = 9
There are two possibilities.
(I) 2 vowels, 1 consonant (II) 3 vowels
∴ required probability = \(\frac{{ }^5 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1}{{ }^9 \mathrm{C}_3}+\frac{{ }^5 \mathrm{C}_3}{{ }^9 \mathrm{C}_3}=\frac{\frac{5 \times 4}{2} \times 4+\frac{5 \times 4}{2}}{\frac{9 \times 8 \times 7}{6}}=\frac{40+10}{84}=\frac{50}{84}=\frac{25}{42}\)

Question 21.
A purse contains 4 silver and 5 copper coins. A second purse contains 3 silver and 7 copper coins. If a coin is taken out at random from one of the purses, what is the probability that it is a copper coin ?
Solution:
Given purse-I contains 4 silver and 5 copper coins
purse-II contains 3 silver and 7 copper coins
∴ prob. of selection of purse-I = prob. of selection of purse-II = \(\frac { 1 }{ 2 }\)
Case-I. When a coin is taken out from purse-I
∴ prob. of drawing a copper com from purse-II = \(\frac { 1 }{ 2 }\) x \(\frac { 5 }{ 9 }\)

Case-II. When a coin is taken out from purse-II
Then probability of drawing a copper coin from purse-II = \(\frac { 1 }{ 2 }\) x \(\frac { 7 }{ 10 }\)
∴ required Protability = \(\frac{5}{18}+\frac{7}{20}=\frac{50+63}{180}=\frac{113}{180}\)

Question 22.
Aman and Bhuwan throw a pair of dice alternately. In order to win, they have to get a sum of 8. Find their respective probabilities of Winning if Aman starts the game.
Solution:
When a pair of dice thrown,
Total no. of cases = 6² = 36
E : getting a sum of 8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

Since P (Aman’s wins) + P (bhuman’s wins) = 1
As both events are mutually exclusive and exhaustive.
∴ P (bhuw’an s wins) = 1 – \(\frac { 36 }{ 67 }\) = \(\frac { 31 }{ 67 }\)

Question 23.
Three persons ^4, B and C shoot to hit a target. If in trials A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 trials. Find the probability that
(i) Exactly two persons hit the target
(ii) At least two persons hit the target
Solution:
(i) Let E₁ : A hits the targets; E₂ : B hits the target; E₃ : C hits the target

(ii) Let E₁ : A hits the targets; E₂ : B hits the target; E₃ : C hits the target

Question 24.
A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random from the box, what is the probability that either both are rusted or both are bolts?
Solution:
Total no. of nuts = 40;
Total no. of bolts = 30
∴ Total no. of items = 70
Total no. of rusted items = 15 + 20 = 35
∴ Total no. of ways of drawing 2 items = \({ }^{70} \mathrm{C}_2\)
Let E₁ : both items drawn are rusted items
E₂ : both item drawn are bolts

Question 25.
A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and the second draw 3 red balls.
Solution:
Let us define the events are as follows :
E₁ : 3 balls drawn in the first draw are all white
E₂ : 3 balls drawn in the second draw are all red.
Total no.of ways of drawing 3 balls out of 13 balls = \({ }^{13} \mathrm{C}_3\)
No. of favourable cases = Total no. of ways of drawing 3 white balls out of 5 = \({ }^5 \mathrm{C}_3\)
∴ P(E₁) = \(\frac{{ }^5 \mathrm{C}_3}{{ }^{13} \mathrm{C}_3}=\frac{\frac{5 \times 4}{2}}{\frac{13 \times 12 \times 11}{6}}=\frac{10 \times 6}{13 \times 12 \times 11}=\frac{5}{143}\)
Since balls are drawn without replacement.
Thus prob. of drawing 3 red balls when 3 white balls have already been drawn in first draw
= P(E₂/E₁) = \(\frac{{ }^8 \mathrm{C}_3}{{ }^{10} \mathrm{C}_3}=\frac{\frac{8 \times 7 \times 6}{6}}{\frac{10 \times 9 \times 8}{6}}=\frac{8 \times 7 \times 6}{10 \times 9 \times 8}=\frac{14}{30}=\frac{7}{15}\)

Question 26.
Bag A contains three red and four white bails; bag B contains two red and three white balls. If one ball is drawn from bag A and two balls from bag B, find the probability that
(i) One ball is red and two balls are white;
(ii) All the three balls are of the same colour.
Solution:
Given bag A contains 3 red balls and 4 white balls
bag B contains 2 red balls and 3 white balls
(i) The possible ways of selecting the balls from given bags is given as under.
(I) One white ball from bag A and one red, one white ball from bag B
(II) One red ball from bag A and 2 white balls from bag B
∴ required prob. of drawing one red and 2 white balls = \(\frac{4}{7} \times \frac{{ }^2 C_1 \times{ }^3 C_1}{{ }^5 C_2}+\frac{3}{7} \times \frac{{ }^3 C_2}{{ }^5 C_2}\)
= \(\frac{4 \times 2 \times 3+3 \times 3}{7 \times \frac{5 \times 4}{2}}=\frac{33}{70}\)

(ii) possible ways of selecting the balls are given as under
(I) one red ball from bag A and two red balls from bag B
(II) one white ball from bag A and two white balls from bag B
∴ required probability = \(\frac{{ }^3 \mathrm{C}_1}{{ }^7 \mathrm{C}_1} \times \frac{{ }^2 \mathrm{C}_2}{{ }^5 \mathrm{C}_2}+\frac{{ }^4 \mathrm{C}_1}{{ }^7 \mathrm{C}_1} \times \frac{{ }^3 \mathrm{C}_2}{{ }^5 \mathrm{C}_2}=\frac{3 \times 1+4 \times 3}{7 \times \frac{5 \times 4}{2}}=\frac{15}{70}=\frac{3}{14}\)

Question 27.
Three persons, Aman, Bipin and Mohan attempt a Mathematics problem independently. The odds in favour of Aman and Mohan solving the problem are 3 : 2 and 4 :1 respectively and the odds against Bipin solving the problem are 2 : 1. Find:
(i) The probability that all the three will solve the problem.
(ii) The probability that problem will be solved.
Solution:
Let us define the events are as follows :
A : Aman solved the problem; B : Bipin solved the problem; M : Mohan solved the problem

Question 28.
In a college, 70% students pass in Physics, 75% pass in Mathematics and 10% students fail in both. One student is chosen at random. What is the probability that:
(i) He passes in Physics and Mathematics?
(ii) He passes in Mathematics given that he passes in Physics?
(ii) He passes in Physics given that he passes in Mathematics? (ISC 2014)
Solution:
Let us define the events are as follows :
E : students pass in Physics; M : students pass in Mathematics

Question 29.
A bag contains 5 white and 4 black balls and another bag contains 7 white and 9 black balls. A ball Is drawn from the first bag and two balls drawn from the second bag. What is the probability of drawing one white and two black balls?
Solution:
Given bag A contains 5 white ball 4 black balls and bag B contains 7 white balls and 9 black balls.
Here drawing of balls from both balls can be done in folowing ways.
(I) One white ball from bag A and 2 black ball from B.
(II) One black ball from bag A arid one white and one black ball from B.
Thus required probability = P(I) + P(II) = \(\frac{{ }^5 \mathrm{C}_1}{{ }^9 \mathrm{C}_1} \times \frac{{ }^9 \mathrm{C}_2}{{ }^{16} \mathrm{C}_2}+\frac{{ }^4 \mathrm{C}_1}{{ }^9 \mathrm{C}_1} \times \frac{{ }^7 \mathrm{C}_1 \times{ }^9 \mathrm{C}_1}{{ }^{16} \mathrm{C}_2}\)
= \(\frac{5 \times \frac{9 \times 8}{2}+4 \times 7 \times 9}{9 \times \frac{16 \times 15}{2}}=\frac{180+252}{1080}=\frac{432}{1080}=\frac{2}{5}\)

Question 30.
An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into the urn Otherwise, it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball shown is black.
Solution:
The probability of drawing a white ball in first draw = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)
The probability that the first drawing to be white ball,
and the second drawing is also white and third is black = P (W₁ W₂ B₃)
= \(\mathrm{P}\left(\mathrm{W}_1\right) \mathrm{P}\left(\mathrm{W}_2 / \mathrm{W}_1\right) \mathrm{P}\left(\mathrm{B}_3 / \mathrm{W}_1 \cap \mathrm{W}_2\right)=\frac{2}{4} \times \frac{1}{3} \times \frac{2}{2}=\frac{1}{6}\)
prob. of drawing first white ball, then black and third is black = P (W₁ B₂ B₃) = \(\frac{2}{4} \times \frac{2}{3} \times \frac{2}{3}=\frac{2}{9}\)
prob. of first drawing a black ball, white ball and third a black ball = P (B₁ W₂ B₃) = \(\frac{2}{4} \times \frac{2}{4} \times \frac{2}{3}=\frac{1}{6}\)
prob. of first drawing a black ball, black ball and third also a black ball = P (B₁ B₂ B₃) = \(\frac{2}{4} \times \frac{2}{4} \times \frac{2}{4}=\frac{1}{8}\)
∴ required probability = \(\frac{1}{6}+\frac{2}{9}+\frac{1}{6}+\frac{1}{8}=\frac{12+16+12+9}{72}=\frac{49}{72}\)

Question 31.
Three persons A, B and C shoot to hit a target. If A hits the target four times in five trials, B hits it three times in four trials and C hits it two times in three trials, find the probability that (i) Exactly two persons hit the target (ii) At least two persons hit the target (iii) None nit the target.
Solution:
(i) Let E₁ : A hits the targets; E₂ : B hits the target; E₃ : C hits the target

Question 32.
A committee of 4 persons has to be chosen from 8 boys and 6 girls, consisting of at least one girl. Find the probability that the committee consists of more girls than boys.
Solution:
Given total no. of boys = 8 and total no. of girls = 6 Let us define the events are as follows :
E₁ : Number of girls more than no. of boys in 4 member committee
E₂ : 4 member committee consists of atleast one girl

Question 33.
An urn contains 10 white and 3 black balls, while another urn contains 3 white and 5 black balls. Two balls are drawn from the first urn and put into the second win and then a ball is drawn from the second urn Find the urn probability that the ball drawn from the second urn a Whitehall.
Solution:
Given urn A contains 10 white and 3 black balls and urn B contains 3 white and 5 black balls
Case-I. When two white balls are transferred from urn A to urn B and prob.of this event be \(\frac{{ }^{10} \mathrm{C}_2}{{ }^{13} \mathrm{C}_2}\) and then urn B contains 5 white and 5 black balls.
∴ probability of getting a white ball = \(\frac{{ }^{10} \mathrm{C}_2}{{ }^{13} \mathrm{C}_2} \times \frac{5}{10}=\frac{\frac{10 \times 9}{2}}{\frac{13 \times 12}{2}} \times \frac{1}{2}=\frac{90}{13 \times 24}=\frac{15}{52}\)

Case-II. When two black balls are transferred from urn A to urn B and then urn B contains 3 white and 7 black balls.
∴ Required probability = \(\frac{{ }^3 \mathrm{C}_2}{{ }^{13} \mathrm{C}_2} \times \frac{3}{10}=\frac{3 \times 3}{\frac{13 \times 12}{2} \times 10}=\frac{3}{260}\)

Case-III. When one black ball and one white ball are transferred from bag A to bag B. Then urn B contains 4 white and 6 black ball.
∴ Required probability = \(\frac{{ }^{10} \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{13} \mathrm{C}_2} \times \frac{4}{10}=\frac{10 \times 3 \times 4}{\frac{13 \times 12}{2} \times 10}=\frac{2}{13}\)
Thus one combining all three cases, we have
Required probability = \(\frac{15}{52}+\frac{2}{260}+\frac{2}{13}=\frac{75+3+40}{260}=\frac{108}{260}=0.45\)

Question 34.
If P(A) = \(\frac { 6 }{ 11 }\), P(B) = \(\frac { 5 }{ 11}\), P(A ∪B) = \(\frac { 7 }{ 11}\), then
(i) P (A ∩ B) = …………..
(ii) P(\(\frac { A }{ B }\)) = …………..
(iii) P(\(\frac { B }{ A }\)) = …………..
Solution:

Question 35.
If P (A) = 0.8 and P(\(\frac { B }{ A }\)) = 0.4, then (i) P (A ∩ B) = ………… (ii) P(A∪B) = …………
Solution:

Question 36.
If A and B are independent events and P(A) = \(\frac { 3 }{ 5 }\), P(B) = \(\frac { 1 }{ 5 }\), then P (A ∩ B) = ………….
Solution:
Since, P (A ∩ B) = P (A) P (B) [∵ A and B are independent events]
= \(\frac { 3 }{ 5 }\) x \(\frac { 1 }{ 5 }\) = \(\frac { 3 }{ 25 }\)

Question 37.
If A and B are independent events with P (A) = 0.3, P (B) = 0.4, then
(i) P(\(\frac { A }{ B }\)) = ……………..
(ii) P(\(\frac { B }{ A }\)) = …………….
Solution:

Question 38.
If P (A) = 0.4, P (B) = 0.8 and P(\(\frac { B }{ A }\)) = 0.6, then
(i) P(A ∪ B) = …………….
(ii) P(\(\left(\frac{\overline{\mathbf{B}}}{{\overline{\mathbf{A}}}\right)\)) = ……………..
Solution:

Question 39.
A pair of dice is thrown 243 times. If getting a sum of 9 is considered a success, then the mean and variance of the number of successes respectively are mean = ……………., variance = …………..
Solution:
Given n = 243
p = prob. of getting a sum of 9 = \(\frac { 4 }{ 36 }\) = \(\frac { 1 }{ 9 }\)
[Here total no. of cases = 6² = 36
and no. of favourable cases = 4, {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ Mean = np = 243 x \(\frac { 1 }{ 9 }\) = 27
and Variance = npq = 243 x \(\frac { 1 }{ 9 }\) x (1 – \(\frac { 1 }{ 9 }\))
= 243 x \(\frac { 1 }{ 9 }\) x \(\frac { 8 }{ 9 }\) = 24

Question 40.
If a die is thrown 5 times, then the probability that an odd number will come up exactly three times is …………….
Solution:
p = prob. of getting a head in single throw of die = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
∴ q = 1 – p = 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) and n = 5
Then by binomial distribution, we have
P (X = r) = \({ }^n \mathrm{C}_r p^r q^{n-r}={ }^5 \mathrm{C}_r\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{5-r}\)
= \({ }^5 C_r\left(\frac{1}{2}\right)^5\)
∴ Required prob. = P (X = 3)
= \({ }^5 C_3\left(\frac{1}{2}\right)^5=\frac{5 !}{2 ! 3 !} \times \frac{1}{2^5}=\frac{10}{32}=\frac{5}{16}\)

Question 41.
The probability distribution of number of heads, when two coins are tossed at a time is given as ……………
Solution:
p = prob. of getting a head in single toss of coin = \(\frac { 1 }{ 2 }\)

Question 42.
If the sum of mean and variance of a binomial distribution for five trials is 4.8, then the probability distribution is ……………
Solution:
According to given condition, we have np + npq = 4.8

Thus required probability distribution is given by (q + p)ⁿ i.e. (\(\frac { 1 }{ 5 }\) + \(\frac { 4 }{ 5 }\))⁵

Question 43.
If X be a random variable taking values x₁, x₂, x₃,…, x_n will probabilities p₁, p₂, p₃ …, p_n respectively, then variance (x) is equal to …………..
Solution:
Required variance
= \(\sum_{i=1}^n p_i x_i^2-\left[\sum_{i=1}^n p_i x_i\right]^2\)

Question 44.
A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find P(A ∩ B).
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ 4 }\)
(c) \(\frac { 1 }{ 6 }\)
(d) \(\frac { 1 }{ 3 }\)
Solution:
Thus, A ∩ B be the event that, number obtained is even on red face
and there is only one even no. i.e. 2 on red face.
∴ n (A ∩ B) = 1 ; n (S) = 6
Thus P (A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{1}{6}\)

Question 45.
If A and B are two events such that A ⊂ B and P (B) ≠ 0, then which of the following is true :
(a) P(\(\frac { A }{ B }\)) = \(\frac { P(B) }{ P(A) }\)
(b) P(\(\frac { A }{ B }\)) < P(A)
(c) \(A\left(\frac{A}{B}\right) \geq P(A)\)
(d) None of these
Solution:
\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)} \geq P(A)\) [∵A ⊂ B ⇒ A ∩ B = A]
[Since A ⊂ B ⇒P (A) ≤ P (B)]

Question 46.
Two events A and B will be independent, if
(a) A and B are mutually exclusive
(b) P(A’∩B’) = [1 – P(A)] [1 – P(B)]
(c) P (A) = P (B)
(d) None of these
Solution:
P (A’ ∩ B’) = P (A’) P (B’)
= [1 – P(A)][1 – P(B)]
[since A and B are independent events then so are A’ and B’ P (A’ ∩ B’) = P (A’) P (B’)]

Question 47.
If A and B are two events such that
P(B) = \(\frac { 3 }{ 5 }\), P (\(\frac { A }{ B }\)) = \(\frac { 1 }{ 2 }\) and P(A∪B)= \(\frac { 4 }{ 5 }\), then P(A) is equal to
Solution:

Question 48.
If P (A) = \(\frac { 1 }{ 2 }\), P (B) = 0, then P(\(\frac { A }{ B }\)) is
(a) 0
(b) \(\frac { 1 }{ 2 }\)
(c) not defined
(d) 1
Solution:
\(P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\) → ∞ [ ∵P(B) = 0]

Question 49.
If A and B are events such that \(\mathbf{P}\left(\frac{\mathbf{A}}{\mathbf{B}}\right)=\mathbf{P}\left(\frac{\mathbf{B}}{\mathbf{A}}\right)\), then
(a) A ⊂ B
(b) B = A
(c) A ∩B = Φ
(d) P (A) = P (B)
Solution:
Given
\(P\left(\frac{A}{B}\right)=P\left(\frac{B}{A}\right)\)
⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{P(B \cap A)}{P(A)}\)
⇒ P (B) = P (A)

Question 50.
If P (A) = \(\frac { 2 }{ 5 }\), P(B) = \(\frac { 3 }{ 10 }\) and P (A ∩ B) = \(\frac { 1 }{ 5 }\), then P(\(\frac { A’ }{ B’ }\)) is equal to
Solution:
We know that

Question 51.
The probability distribution of a discrete random variable X is given below :

X	0	1	2	3
P(X)	\(\frac { 4 }{ k }\)	\(\frac { 6 }{ k }\)	\(\frac { 10 }{ k }\)	\(\frac { 12 }{ k }\)

The value of k is
(a) 8
(b) 16
(c) 32
(d) 0.96
Solution:
Since,
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
⇒ \(\frac{4}{k}+\frac{6}{k}+\frac{10}{k}+\frac{12}{k}\) = 1 ⇒ \(\frac { 32 }{ k }\) = 1
⇒ k = 32

Question 52.
Let A and B be independent events with P(A) = \(\frac { 1 }{ 4 }\) and P (A ∪ B) = 2P(B) – P(A). Find P(B).
(a) \(\frac { 1 }{ 4 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { 2 }{ 5 }\)
Solution:
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
Since, A and B are independent events.
Then P(A ∩ B) = P(A) P(B)
We know that,
P (A ∪ B) = P(A) + P(B) – P(A) P(B)
⇒ 2P(B) – P(A) = P(A) + P(B) – P(A) P(B)
⇒ P(B) = 2P(A) – P(A) P(B)
⇒ P (B) = 2 x \(\frac { 1 }{ 4 }\) – \(\frac { 1 }{ 4 }\) P(B)
⇒ \(\frac { 5 }{ 4 }\) P (B) = \(\frac { 1 }{ 2 }\) p (B) = \(\frac{1}{2} \times \frac{4}{5}=\frac{2}{5}\)

Question 53.
If eight coins are tossed together, then the probability of getting exactly 3 heads is
(a) \(\frac { 1 }{ 256 }\)
(b) \(\frac { 7 }{ 32 }\)
(c) \(\frac { 5 }{ 32 }\)
(d) \(\frac { 3 }{ 32 }\)
Solution:
p = prob. of getting a head in single toss of coin = \(\frac { 1 }{ 2 }\)

Question 54.
A card is picked at random from a pack of cards. Given that the picked card is a queen, what is the probability that it is a spade?
(a) \(\frac { 1 }{ 3 }\)
(b) \(\frac { 4 }{ 13 }\)
(c) \(\frac { 1 }{ 4 }\)
(d) \(\frac { 1 }{ 2 }\)
Solution:
Let A : event that picked card is queen
B : event that picked card is spade.
∴ A ∩ B : event that picked card is queen of spade.

Question 55.
A six-faced unbiased die is thrown twice and the sum of the numbers appearing on the upper face is observed to be 7. The probability that the number 3 has appeared at least once is
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 1 }{ 4 }\)
(d) \(\frac { 1 }{ 5 }\)
Solution:
Let A: sum of numbers appearing on upper face is 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2)), (6, 1)}
B : number 3 has appeared atleast once = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}
∴ A ∩ B = {(3,4), (4, 3)}
∴ n (A ∩ B) = 2 ; n (A) = 6
∴ required prob. = P (B/A)
= \(\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=\frac{n(\mathrm{~B} \cap \mathrm{A})}{n(\mathrm{~A})}=\frac{2}{6}=\frac{1}{3}\)

Question 56.
For a married couple, the probability that a husband will vote in an election is 0.5 and the probability that his wife will vote, is 0.4. The probability that the husband votes, given that his wife also votes is 0.7. Then the probability that husband and wife both will vote is
(a) 0.28
(b) 0.20
(c) 0.35
(d) 0.15
Solution:
Let A : event that husband will vote in election
B : event that his wife will vote in election
Then P (A) = 0.5 ; P (B) = 0.4 ;
P (A | B) = 0.7
⇒ \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\)
⇒ P (A ∩ B) = 0.7 x 0.4 = 0.28
∴ required prob. that both will vote = 0.28

Question 57.
If P (A) = \(\frac { 1 }{ 12 }\), P(B) = \(\frac { 5 }{ 12 }\) and P (B/A) = \(\frac { 1 }{ 15 }\), then P(A ∩B) is equal to
(a) \(\frac { 89 }{ 180 }\)
(b) \(\frac { 90 }{ 180 }\)
(c) \(\frac { 91 }{ 180 }\)
(d) \(\frac { 92 }{ 180 }\)
Solution:
Given P (B | A) = \(\frac { 1 }{ 15 }\)
⇒ \(\frac{P(B \cap A)}{P(A)}=\frac{1}{15}\)
⇒ P(A ∩B) = \(\frac { 1 }{ 15 }\)P(A)
= \(\frac{1}{15} \times \frac{1}{12}=\frac{1}{180}\)

Question 58.
A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number of the die and a spade card is
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ 4 }\)
(c) \(\frac { 1 }{ 8 }\)
(d) \(\frac { 3 }{ 4 }\)
Solution:
Consider the following events :
A : getting an even number on die = {2, 4, 6}
B : getting a spade card
Then P(A) = \(\frac { 3 }{ 2 }\) = \(\frac { 1 }{ 2 }\) ; P(B) = \(\frac{13}{52}=\frac{1}{4}\)
Since A and B are independent events.
∴ required probability = P (A ∩ B) = P (A)
P(B) = \(\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}\)

Question 59.
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then the probability of getting exactly one red ball is
(a) \(\frac { 15 }{ 196 }\)
(b) \(\frac { 135 }{ 392 }\)
(c) \(\frac { 15 }{ 56 }\)
(d) \(\frac { 15 }{ 29 }\)
Solution:
Total no. of balls in a bag = 5 + 3 = 8
given no. of red balls = 5
and no. of blue balls = 3
Since three balls are drawn without replacement
∴ required probability of getting exactly one red ball = P (one red ball in 1st draw and one each in blue balls in 2nd and 3rd draw) + P (one blue ball in 1st draw and one red ball in 2nd draw and one blue in 3rd draw) + P (one blue ball in each 1st and 2nd draw and red ball in 3rd draw)
= \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6}+\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6}\)
= \(\frac{90}{8 \times 7 \times 6}=\frac{15}{56}\)

Question 60.
In a college 30% students fail in Physics, 25% fail in Mathematics and 10% fail in both. One student is chosen at random. The probability that she fails in Physics if she has failed in Mathematics is
(a) \(\frac { 1 }{ 10 }\)
(b) \(\frac { 1 }{ 3 }\)
(c) \(\frac { 2 }{ 5 }\)
(d) \(\frac { 9 }{ 20 }\)
Solution:
Consider the following events
A : students fail in Physics
B : students fail in Maths
Then P (A) = 30% = \(\frac { 3 }{ 100 }\) ;
and P(B) = 25% = \(\frac { 25 }{ 100 }\)
and P (A ∩ B) = 10% = \(\frac { 10 }{ 100 }\)
Then required probability = P (A/B)
= \(\frac{P(A \cap B)}{P(B)}=\frac{10 / 100}{25 / 100}=\frac{2}{5}\)

Question 61.
The probability distribution of a discrete random variable X is given below.

X	2	3	4	5
P(X)	\(\frac { 5 }{ k }\)	\(\frac { 7 }{ k }\)	\(\frac { 9 }{ k }\)	\(\frac { 11 }{ k }\)

The value of k is
(a) 8
(b) 16
(c) 32
(d) 48
Answer:
Since sum of probabilities of all events be equal to 1.
∴ P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) = 1
⇒ \(\frac{5}{k}+\frac{7}{k}+\frac{9}{k}+\frac{11}{k}\) = 1
⇒ \(\frac { 32 }{ k }\) ⇒ k = 32

Question 62.
Given two independent events A and B such that P (A) = 0.3 and P (B) = 0.6, find P (A’ ∩ B’).
Solution:
Given P (A) = 0.3 ; P (B) = 0.6
∴ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= P (A) + P (B) – P (A) P (B)
[∵ A and B are independent events.
Then P (A ∩ B) = P (A) P (B)]
= 0.3 + 0.6 – 0.3 x 0.6
= 0.9 – 0.18
= 0.72
∴ P (A’ ∩ B’) = P {(A ∪ B)’}
= 1 – P(A ∪ B)
= 1 – 0.72
= 0.28

Question 63.
Find [P (B/A) + P (A/B)], if P (A) = \(\frac { 3 }{ 10 }\), P(B) = \(\frac { 2 }{ 5 }\) and P(A ∪B) = \(\frac { 3 }{ 5 }\).
Solution:

Question 64.
A die whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find if A and B are independent events.
Solution:
Given A : number obtained is even
B : number obtained is red
Then P (A) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\) ; P (B) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
A ∩ B : number obtained is even on red face = {2}
∴ P (A ∩ B) = \(\frac { 1 }{ 6 }\) and
P(A) P(B) = \(\frac { 1 }{ 2 }\) x \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 4 }\)
Thus P(A∩B) # P(A) P (B)
∴ A and B are not independent events.

Question 65.
The random variable X has a probability distribution P (X) of the form
P(X = x) = \(\left\{\begin{array}{cc}
k, & \text { if } x=0 \\
2 k, & \text { if } x=1 \\
3 k, & \text { if } x=2 \\
0, & \text { otherwise }
\end{array}\right.\)
Determine the value of k.
Solution:
Since ∑P (X = x) = 1
⇒ P (X = 0) + P (X = 1) + P (X = 2) + P (X = x > 3) = 1
⇒ k + 2k + 3k + 0 = 1
⇒ 6k = 1
⇒ k = \(\frac { 1 }{ 6 }\)

Question 66.
From the set of integers {1, 2, 3, 4, 5} two numbers a and b {a ≠ b) are chosen at random. Find the probability that \(\frac { a }{ b }\) is an integer.
Solution:
Total no. of cases = \({ }^5 P_2=\frac{5 !}{3 !}\)
Total no. of favourable cases i.e. \(\frac { a }{ b }\) be an integers
= {(2,1), (3,1), (4, 1), (5, 1), (4, 2)}
= 5
∴ Required prob. = \(\frac { 5 }{ 20 }\) = \(\frac { 1 }{ 4 }\)

Question 67.
A can hit a target 4 times in 5 slots, B can hit 3 times in 4 slots and C can hit 2 times in 3 slots. Calculate the probability that (i) A, B, C all may hit (ii) None of these will hit the target.
Solution:
Given P (A can hit a target) = P (A) = \(\frac { 4 }{ 5 }\)
P (B can hit a target) = P (B) = \(\frac { 3 }{ 4 }\)
P (C can hit a target) = P (C) = \(\frac { 2 }{ 3 }\)
(i) ∴ Required probability that A, B and C all may hit = P (A) P (B) P (C)
= \(\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}=\frac{2}{5}\)

(ii) ∴ Required probability
= \(\mathrm{P}(\overline{\mathrm{A}}) \mathrm{P}(\overline{\mathrm{B}}) \mathrm{P}(\overline{\mathrm{C}})\)
= [1 – P(A)][1 – P(B)][1 – P(C)]
= \(\left[1-\frac{4}{5}\right]\left[1-\frac{3}{4}\right]\left[1-\frac{2}{3}\right]\)
= \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3}=\frac{1}{60}\)