ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.3

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.3

Question 1.
Fill in the blanks :
(i) If events A and B are mutually exclusive, then P (A ∩ B) is ______________, and if A and B are independent, then P (A ∩ B) is ______________

(ii) If P (A) = 0.5, P (B) = 0.4 and P (A ∩ B) = 0.2, then the statement that events A and B are independent is __________ and value of P (A ∪ B) is

(iii) When a die is rolled twice, the probability of getting a 2 followed by a 4 is ______________

(iv) When two cards are drawn from a pack of cards with replacement, the probability of getting an ace followed by a king is ________, while the probability of getting two hearts is ________

(v) When two cards are drawn from a pack of cards without replacement, the probability of getting an ace followed by a king is ________ while the probability of getting two aces is ________

(vi) If A and B are independent events and P (A) = 0.40, P (B) = 0.30, then P (A and B) is ________ P (A or B) is ________ P (not A) is _________ and P (neither A nor B) is _________
Answer:
(i) P (A ∩ B) = 0
and P (A ∩ B) = P (A) P (B)

(ii) Here P(A) = 0.5 ; P (B) = 0.4
P (A) P (B)= 0.5 x 0.4 = 0.2
= P (A ∩ B)
Thus A and B are independent P(A ∪ B) = P (A) + P (B) – P (A ∩ B)
= P (A) + P (B) – P (A) P (B)
= 0.5 + 0.4 – 0.2
= 0.7

(iii) \(\frac{1}{36}\). since P (getting 2) = \(\frac{1}{6}\)
and P (getting 4) = \(\frac{1}{6}\)
P (getting 2 followed by a 4) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)

(iv) Probability of an ace followed by a king
= \(\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)
Prob. of getting two hearts = \(\frac{13}{52} \times \frac{13}{52}=\frac{1}{16}\)

(v) Prob. of an ace followed by a king
= \(\frac{4}{52} \times \frac{4}{51}\) (with0ut replacement)
Prob. of getting two aces = \(\frac{4}{52} \times \frac{3}{51}\)

(vi) P (A and B) = P (A) P (B)
= 0.40 × 0.3 = 0.12
P (A or B) = P (A) + P (B) – P (A ∩ B)
= 0.4 + 0.3 – 0.12
= 0.58
P (not A) =P(A) = 1 – P(A)= 1 – 0.4 = 0.6
P (neither A nor B) = P (A ∩ B)
= P(A ∪ B)
= 1 – P (A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)]
= 1 – 0.4 – 0.3 + 0.12
= 1.12 – 0.7
= 0-42

Question 2.
What is the probability of obtaining an even prime number on each die when a pair of dice is rolled ?
Answer:
Since probability of getting even prime number on one die = \(\frac{1}{6}\)
[since 2 be the only even prime number]
reqd. probability of getting an even prime number on each die in a throw of two dice
= \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)

Question 3.
(i) Events A and B are such that P(A) = \(\frac{1}{2}\), P (B) = \(\frac{7}{12}\) and P (not A or not B) = \(\frac{1}{4}\). State whether A and B are independent. (NCERT)
(ii) Let E and F be events with P(E) = \(\frac{3}{5}\), P (F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\). Are E and F independent ? (NCERT)
Answer:
(0 Given P (A) = \(\frac{1}{2}\); P (B) = \(\frac{7}{12}\)
P (not A or not B) = P (A ∪ B)
= P\((\overline{A \cap B})\)
\(\frac{1}{4}\) = 1 – P(Ā ∪ B̄)
P(A ∩ B) = \(\frac{3}{4}\)
and P(A).P(B) = \(\frac{1}{2} \times \frac{7}{12}=\frac{7}{24}\)

P(A ∩ B) ≠ P(A).P(B)
Thus, A and B are not independent events.

(ii) P(E) . P (F) = \(\frac{3}{5} \times \frac{3}{10}\)
= \(\frac{9}{50}\) ≠ P(E ∩ F) = \(\frac{1}{5}\)
∴ E and F are not independent events.

Question 4.
(i) If P (A) = \(\frac{3}{5}\) and P (B) = \(\frac{1}{5}\) then find P (A ∩ B) if A and B are independent events. (NCERT)
Answer:
Since A and B are independent events
P(A ∩ B) = P(A) P(B)
= \(\frac{3}{5} \times \frac{1}{5}=\frac{3}{25}\)

(ii) Given two independent events A and B such that P (A) = 0.3 and P (B) = 0.6, find P (A’ ∩ B’).
Answer:
Given P(A) = 0.3; P(B) = 0.6
since A and B are independent events then
so, are A’ and B’
∴ P(A’ ∩ B’) = P(A’) P(B’)
= (1 – P(A)) (1 – P(B))
= (1 – 0.3)(1 – 0.6)
= 0.7 × 0.4
= 0.28

(iii) If A and B are two independent events such that P (not B) = 0.65 and P(A ∪ B) = 0.85, then find P (A).
Answer:
Given P (not B) = p (B) = 0.65
∴ P(B) = 1 – P(B) = 1 – 0.65 = 0.35
Given P (A u B) = 0-85
⇒ 0.85 = P (A) + P (B) – P (A ∩ B)
⇒ 0.85 = P (A) + P (B) – P (A) P (B)
[since A and B are independent events ∴ P (A∩ B) = P (A) P (B)]
⇒ 0.85 = P (A)+ 0.35 – P (A) 0.35
⇒ 0.5 = P (A) (1 – 0.35)
⇒ P(A) = \(\frac{0.5}{0.65}=\frac{10}{13}\)

Question 5.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
Answer:
∴ Sample space = S = {H,T, 1,2, 3, 4, 5,6}
A : event that head appears on coin
B : event that 3 appears on die.
P(A) = \(\frac{1}{2}\); P(B) = \(\frac{1}{6}\)
Clearly P(A ∩B) =\(\frac{1}{2} \times \frac{1}{6}=\frac{1}{12}\) = P(A) . P(B)
A and B are independent events

Question 6.
A die is thrown. If E is the event ‘the number appearing is a multiple of 3’ and F be the event ‘the number appearing is even’, then find whether E and F are independent. (NCERT)
Answer:
Given E : event that the number appearing is a multiple of 3
F : event that the number appearing is even.
∴ E = {3, 6} ; F = {2, 4, 6}
∴ E ∩ F = {6}
∴ P (E ∩ F) = \(\frac{1}{6}\)
P(E) × P(F) = \(\frac{2}{6} \times \frac{3}{6}=\frac{1}{6}\)
∴ P (E ∩ F) = P (E) . P (F)
∴ E and F are independent events

Question 7.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event ‘number is even’ and B be the event’ number is red’. Are A and B independent ? (NCERT)
Answer:
Given A : event that number is even ;
B : event that number is red
∴ A = {2, 4, 6} and B = {1, 2, 3}
∴ A ∩ B = {2}
∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\) and P(B) = \(\frac{3}{6}=\frac{1}{2}\)
and P(A ∩ B) = \(\frac{1}{6}\)
thus P(A ∩ B) ≠ P(A) . P(B)
Thus A and B are not independent events.

Question 8.
If P (A) = 0.2, P (B) =p, P (A ∪ B) = 06 and A and B are given to be independent events, find the value of p.
Answer:
Given P(A) = 0.2,P(B) = p;
P (A ∪ B) = 0.6
Also A and B are independent events.
P(A ∩ B) = P(A) . P(B) …………….(1)

We know that,
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 0.6 = 0.2 + p – 0.2 × p [using (1)]
⇒ 0.4 = 0.8 ⇒ p = \(\frac{0 \cdot 4}{0.8}=\frac{1}{2}\)

Question 9.
A coin is tossed thrice and all the eight outcomes are assumed to be equally likely. In which of the cases are the events E₁ and E₂ independent ? In which cases are they mutually exclusive ?
(i) E₁ : the first throw results in head E₂ : the last throw results in tail
(ii) E₁ : the number of heads is two E₂ : the last throw results in head
(iii) E₁ : the number of heads is odd E₂ : the number of tails is odd.
Answer:
Here S = Sample space
= {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
n (S) = 8

(i) E₁ = {HHH, HHT, HTH, HTT}
n(E₁) = 4

E₂ = {HHT, HTT, THT, TTT}
∴ n (E₂) = 4

E₁ ∩ E₂ = {HHT, HTT}
∴ n (E₁ ∩ E₂) = 2
Thus P (E₁) = \(\frac{4}{8}=\frac{1}{2}\) ;P(E₂) = \(\frac{4}{8}=\frac{1}{2}\)
and P (E₁ ∩ E₂) = \(\frac{2}{8}=\frac{1}{4}=\frac{1}{2} \times \frac{1}{2}\)
= P (E₁) × P (E₂)
Thus E₁ and E₂ are independent events.
Further P(E₁ ∩ E₂) ≠ 0
E₁ and E₂ are not mutually exclusive.

(ii) E₁ = {HHT, HTH, THH} ; E₂ = {HHH, HTH, THH, TTH}
∴ E₁ ∩ E₂ = {THH, HTH}
Thus n(E₁) = 3 ; n(E₂) = 4; n (E₁ ∩ E₂) = 2
P(E₁ ∩ E₂) = \(\frac{n\left(\mathrm{E}_1 \cap \mathrm{E}_2\right)}{n(\mathrm{~S})}=\frac{2}{8}=\frac{1}{4}\)
P(E1) = \(\frac{n\left(\mathrm{E}_1\right)}{n(\mathrm{~S})}=\frac{3}{8}\); P(E₂) = \(\frac{n\left(\mathrm{E}_2\right)}{n(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}\)
Clearly P(E₁ ∩ E₂) ≠ P (E₁) P(E₂)
Hence E₁ and E₂ are not independent events.
Further P(E₁ ∩ E₂) ≠ 0
E₁ and E₂ are not mutually exclusive events.

(iii) E₁ = {HHH, HTT, THT, TTH}
∴ n (E₁) = 4
E₂ = {TTT, THH, HTH, HHT}
∴ n (E₂) = 4
∴ E₁ ∩ E₂ = { } or Φ n(E₁ ∩ E₂) = 0
Here P(E₁ ∩ E₂) = \(\frac{0}{8}\) = 0
P(E₁)P(E₂) = \(\frac{4}{8} \times \frac{4}{8}=\frac{1}{4}\)
∴ P(E₁ ∩ E₂) ≠ P (E₁) P (E₂)
Hence E₁ and E₂ are not independent events but mutually exclusive.

Question 10.
Out of 8 outstanding students of a school, in which there are 3 boys and 5 girls, a team of 4 students is to be selected for a quiz competition. Find the probability that 2 boys and 2 girls are selected.
Answer:
Total no. of ways of selecting 4 students 1 out of 8
= ⁸C₄ = \(\frac{8 !}{4 ! 4 !}=\frac{8 \times 7 \times 6 \times 5}{24}\) = 70 = n(S)
Total no. of ways of selecting 2 boys out of 3 and 2 girls out of 5
= ³C₂ × ⁵C₂
∴ required probability = \(\frac{{ }^3 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2}{70}\)
= \(\frac{3 \times 10}{70}=\frac{3}{7}\)

Question 10 (old).
An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after another without replacement. What is the probability that both balls drawn are black ? (NCERT)
Answer:
Let A and B are the event of getting a black ball in 1st and 2nd draw.
∴ required probability = P (A ∩ B)
= P (A) . P (B/A) …(1)
P (A) = prob. of getting a black ball in 1st draw = \(\frac{10}{15}=\frac{2}{3}\)
P (B/A) = prob. of getting a black ball in 2nd draw when a black ball is already drawn in 1st draw = \(\frac{9}{14}\)
∴ from (1); we have
Required probability = \(\frac{2}{3} \times \frac{9}{14}=\frac{3}{7}\)

Question 11.
Two cards are drawn at random and one by one without replacement from a well- shuffled pack of 52 playing cards. Find the probability that one card is red and the other is black.
Answer:
required probability = P (BR) +P (RB)
= \(\frac{26}{52} \times \frac{26}{51}+\frac{26}{52} \times \frac{26}{51}\)
= \(\frac{2 \times 26 \times 26}{52 \times 51}=\frac{26}{51}\)

Question 12.
(i) A bag contains 4 white, 3 red and 2 blue balls. Four balls are drawn one by one without replacement Find the probability that all balls are white.
Answer:
Total no. of balls in a bag = 4 + 3 + 2
= 9
Required prob. = P (w₁w₂w₃w₄)
= P(w₁)P(w₂)P(w₃)P(w₄)
= \(\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}=\frac{1}{126}\)

(ii) A bag contains 4 white, 3 red and 2 blue balls. Four balls are drawn one by one with replacement. Find the probability that all balls are white.
Answer:
Total no. of balls in a bag = 9 No. of white balls = 4
∴ Required prob. that all ball are white when balls are drawn one by one with replacement
= \(\frac{4}{9} \times \frac{4}{9} \times \frac{4}{9} \times \frac{4}{9}=\left(\frac{4}{9}\right)^4\)

Question 13.
(i) A bag contains 4 white, 3 red and 2 blue balls. Four balls are drawn one by one without replacement Find the probability that no ball is white.
(ii) A bag contains 4 white, 3 red and 2 blue balls. Four balls are drawn one after the other with replacement Find the probability that no ball is white.
Answer:
(i) Total no. of balls in a bag = 4 + 3 + 2
= 9
∴ Total no. of ways of selecting 4 balls out of 9 = ⁹C₄
∴ Total no. of ways of selecting no. white ball means each chosen ball is either red or blue = ⁵C₄
∴ required prob. = \(\frac{{ }^5 C_4}{{ }^9 C_4}\)
= \(\frac{5}{\frac{9 !}{4 ! 5 !}}=\frac{5 \times 4 \times 5 !}{9 \times 8 \times 7 \times 6 \times 5 !}\)
= \(\frac{5 \times 24}{9 \times 8 \times 7 \times 6}=\frac{5}{126}\)

(ii) Total no. of balls in a bag = 4 + 3 + 2 = 9
P (of selecting one ball other than white) = \(\frac{5}{9} \times \frac{5}{9} \times \frac{5}{9} \times \frac{5}{9}=\left(\frac{5}{9}\right)^4\)
∴ required prob. = \(\frac{5}{9} \times \frac{5}{9} \times \frac{5}{9} \times \frac{5}{9}=\left(\frac{5}{9}\right)^4\)

Question 14.
(i) A bag contains 5 red, 6 white and 7 black , balls. Two balls are drawn at random. What is the probability that both balls are red or both are black ?
Sol.
Total number of balls = 5 + 6 + 7= 18
∴ required probability = P(RR) + P(BB)
= \(\frac{{ }^5 C_2}{{ }^{18} C_2}+\frac{{ }^7 C_2}{{ }^{18} C_2}\)
= \(\frac{5 \times 4}{18 \times 17}+\frac{7 \times 6}{18 \times 17}=\frac{62}{18 \times 17}=\frac{31}{153}\)

(ii) There are 2 red and 3 black balls in a bag. Three balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.
Answer:
Since there are 2 red and 3 black balls in a bag
∴ Total no. of balls in a bag = 2 + 3 = 5
P (getting 2 red ball and 1 black ball)
= \(\frac{{ }^2 \mathrm{C}_2 \times{ }^3 \mathrm{C}_1}{{ }^5 \mathrm{C}_3}=\frac{3 \times 2}{5 \times 4}=\frac{6}{20}=\frac{3}{10}\)
P (getting 1 red and 2 black balls)
= \(\frac{{ }^2 C_1 \times{ }^3 C_2}{{ }^5 C_3}=\frac{2 \times 3 \times 2}{5 \times 4}=\frac{3}{5}\)
Thus required probability = \(\frac{3}{10}+\frac{3}{5}=\frac{9}{10}\)

Question 15.
A bag cointains 5 white, 7 red and 4 black ^ balls. Four balls are drawn one by one w5 with replacement. What is the probability that none is white ?
Answer:
Since a bag contains 5 white, 7 red and 4 black balls
∴ Total no. of balls = 5 + 7 + 4 = 16
required prob. of getting no white ball = Prob. of getting a ball other than white i.e. out of (7 + 4) = 11 balls in each drawing
= \(\frac{11}{16} \times \frac{11}{16} \times \frac{11}{16} \times \frac{11}{16}=\left(\frac{11}{16}\right)^4\)

Question 16.
A bag contains 1 white, 5 red and 4 black balls. If three balls are drawn one by one ^ without replacement, then find the probability of drawing a white ball followed by two red balls.
Answer:
Since a bag contains 1 white, 5 red and 4 black balls
∴ Total no. of balls in a bag =1+5 + 4 = 10
∴ required probability = P (WRR)
= P (W) . P (R) . P (R)
= \(\frac{1}{10} \times \frac{5}{9} \times \frac{4}{8}=\frac{1}{36}\)

Question 17.
From a bag containing 20 tickets, numbered from 1 to 20, two tickets are drawn at random. Find the probability that
(i) both the tickets have prime numbers on them
(ii) on one there is a prime number and on the other there is a multiple of 4.
Answer:
Total no. of outcomes = 20
Let E : event that ticket have prime number
∴ E = {2,3, 5, 7, 11, 13, 17, 19}
Thus required probability = P (E)
= \(\frac{{ }^8 \mathrm{C}_2}{{ }^{20} \mathrm{C}_2}=\frac{8 \times 7}{20 \times 19}=\frac{14}{95}\)

(ii) The total no. of ways of drawing 2 tickets out of 20 = ²⁰C₂
Total no. of ways of drawing a prime no. on one ticket = ⁸C₁
Number of ways of drawing (getting) a multiple of 4 on other ticket = ⁵C₁
{4, 8, 12, 16, 20}
∴ required probability
= \(\frac{{ }^8 \mathrm{C}_1 \times{ }^5 \mathrm{C}_1}{{ }^{20} \mathrm{C}_2}=\frac{8 \times 5 \times 2}{20 \times 19}=\frac{4}{19}\)

Question 18.
An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting .
(i) 2 red balls
(ii) 2 blue balls
(iii) one red and one blue ball.
Answer:
Since given urn contains 4 red and 7 blue balls.
∴ Total no. of balls in a urn = 4 + 7 = 11.
(i) required prob. = P (RR) = P (R) . P (R)
= \(\frac{4}{11} \times \frac{4}{11}=\frac{16}{121}\)

(ii) required prob. = P (BB) = P (B) . P (B)
= \(\frac{7}{11} \times \frac{7}{11}=\frac{49}{121}\)

(iii) required prob. = P (BR) + P (RB)
= P (B) P (R) + P (R) P (B)
= 2 × \(\frac{7}{11} \times \frac{4}{11}=\frac{56}{121}\)

Question 18 (old).
A box contains 100 tickets numbered 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.
Answer:
Total no. of outcomes = 100.
A : event that the ticket bearing a number divisible by 10
∴ A = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100}
∴ P(A) = \(\frac{10}{100}=\frac{1}{10}\)
Thus required probability
= P (AAAAA)
= P (A) P (A) P (A) P (A) P (A)
= \(\left(\frac{1}{10}\right)^5\)
= 0.00001

Question 19.
Three balls are drawn one by one without replacement from an urn containing 3 white, 3 red and 3 black balls. What is the probability that atleast one ball is red?
Answer:
Given urn contains 3 white, 3 red and 3 black balls.
∴ Total no. of balls in given urn =3 + 3 + 3 = 9
P (atleast one red ball) = 1 – P (no red ball)
Let A be the event of drawing a non-red ball in first draw and B be the event of drawing a non red ball in second draw.
and C : event of drawing a-non-red ball in 3rd draw
∴ P(A) = \(\frac{6}{9}=\frac{2}{3}\)
P(A) after drawing a non red ball in first draw, so there are 5 non-red balls out of 8 balls.
P (B/A) = \(\frac{6}{9}=\frac{2}{3}\)
after drawing a non-red balls in first two draws.
So, there are 4 non-red balls out of 7 balls
P (C/A ∩ B) = \(\frac{4}{7}\)
Thus P (no red ball) = P (A ∩ B ∩ C)
= P (A) . P (B/A) . P (C/A ∩ B)
= \(\frac{2}{3} \times \frac{5}{8} \times \frac{4}{7}=\frac{5}{21}\)
required prob. = 1 – P (no red ball)
= 1 – \(\frac{5}{21}=\frac{16}{21}\)

Question 20.
Three distinct numbers are chosen randomly from the first 50 natural numbers. Find the probability that all the three numbers are divisible by 2 and 3 both.
Answer:
Here total no. of ways of selecting 3 natural numbers out of 50 = ⁵⁰C₃
= n(S)
E: event of that numbers are divisible by 2 and 3 i.e. by 6 = {6,12, 18,24,30,36,42, 48}
∴ n (E) = No. of ways of selecting 3 numbers out of 8 = ⁸C₃
Thus required prob. = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\)
= \(\frac{{ }^8 C_3}{{ }^{50} C_3}=\frac{8 \times 7 \times 6}{50 \times 49 \times 48}=\frac{1}{350}\)

Question 20 (old).
In bag A, there are 5 white and 8 red ^ balls, in bag B, 7 white and 6 red balls, Cr and in bag C, 6 white and 5 red balls. One ball is taken out at random from each bag. Find the probability that all the three balls are of the same colour.
Answer:
Given bag A contains 5 white and 8 red balls bag B contains 7 white and 6 red balls and bag C contains 6 white and 5 red balls
P (W₁W₂W₃) = Prob. (of getting white ball from bag A, B and C)
= P (W₁) P (W₂) P (W₃)
= \(\frac{5}{13} \times \frac{7}{13} \times \frac{6}{11}=\frac{210}{1859}\)
P (R₁R₂R₃) = prob. (of getting red ball from bag A, B and C)
= P(R₁)P(R₂)P(R₃)
= \(\frac{8}{13} \times \frac{6}{13} \times \frac{5}{11}\)
= \(\frac{240}{1859}\)
∴ required probability
= P(W₁W₂W₃) + P(R₁R₂R₃)
= \(\frac{210+240}{1859}=\frac{450}{1859}\)

Question 21.
The probability of finding a green signal on a busy crossing X is 3Q%. What is the probability of finding a green signal on two consecutive days out of three ?
Answer:
Given P (green signal or busy day)
= \(\frac{30}{100}=\frac{3}{10}\)
P (not getting a green signal) = 1 – \(\frac{3}{10}=\frac{7}{10}\)
Required prob. = P (getting a green signal on first two days)
+ P (getting a green signal on last two days)
= \(\frac{3}{10} \times \frac{3}{10} \times \frac{7}{10}+\frac{7}{10} \times \frac{3}{10} \times \frac{3}{10}\)
= \(\frac{63}{1000}+\frac{63}{1000}=\frac{126}{1000}=\frac{63}{500}\)

Question 22.
One bag contains 3 white balls, 7 red balls and 15 black balls. Another bag contains 10 white balls, 6 red balls and 9 black balls. One ball is taken from each bag. Find the probability that both balls will be of the same colour. (ISC 2005)
Answer:
bag I contains 3 white, 7 red and 15 black balls and bag II contains 10 white, 6 red and 9 black balls
P (of getting one white from I bag and one white from bag II)
= P (W₁, W₂) = P (W₁) P (W₂)
= \(\frac{3}{25} \times \frac{10}{25}=\frac{30}{625}=\frac{6}{125}\)
P(R₁R₂) = P(R₁)P(R₂)
= \(\frac{7}{25} \times \frac{6}{25}=\frac{42}{625}\)
P(B₁B₂) = P(B₁)P(B₂)
= \(\frac{15}{25} \times \frac{9}{25}=\frac{27}{125}\)
required probability = P (W₁W₂) + P (R₁R₂) + P (B₁B₂)
= \(\frac{6}{125}+\frac{42}{625}+\frac{27}{125}\)
= \(\frac{30+42+135}{625}=\frac{207}{625}\)

Question 22 (old).
X is taking up subjects Mathematics, Physics and Chemistry in the examination. His ^y^probabilities of getting grade A in these subjects are 0-2, 0-3 and 0-5 respectively. Find the probability that he gets
(i) Grade A in all subjects
(ii) Grade A is no subject
(iii) Grade A in two subjects.
Answer:
Consider the following events
M = event that X got grade A in mathematics
P = event that X got grade A in physics
C = event that X got grade A in chemistry’
Then P (M) = 0.2 ;P(p) = 0.3; P(C) = 0.5
(i) required probability = P (M ∩ p ∩ C) = P(M) P(p) . P (C)
= 0.2 × 0.3 × 0.5 = 0-03

(ii) required probability = P(M ∩P ∩ C) = P(M) P(p) P(C)
= [1 – 0.2] [1 – 0.3][1 – 0.5]
= (0.8) (0.7) (0-5) = 0.28

(iii) required probability
= P(M ∩ p ∩ C) + P(M ∩ p ∩ C) + P(M ∩ P ∩ C)
= P(M) P(p) P(C)+ P(M) – P(p)P(C) + P(M) P (p) P (C)
= (0.2) (0.3) (1 – 0.5) + (0.2) (1 – 0.3) (0.5) + (1 – 0.2) (0.3) (0.5)
= 0.03 + 0.07 + 0.12
= 0.22

Question 23.
An urn contains 7 white, 5 black and 3 red balls. Two balls are drawn at random.
Find the probability that
(i) both the balls are red
(ii) one ball is red, the other ¡s black
(iii) one ball is white.
Answer:
Given urn contains 7 white, 5 black and 3 red balls
Total no. of balls in urn = 7 + 5 + 3 = 15
Total no. of ways of drawing 2 balls out of 15 = ¹⁵C₂

(i) ∴ required prob. (that both balls are red)
= \(\frac{{ }^3 \mathrm{C}_2}{{ }^{15} \mathrm{C}_2}=\frac{3 \times 2}{15 \times 14}=\frac{1}{5 \times 7}=\frac{1}{35}\)

(ii) required prob. = \(\frac{{ }^3 C_1 \times{ }^5 C_1}{{ }^{15} C_2}=\frac{3 \times 5 \times 2}{15 \times 14}=\frac{1}{7}\)

(iii) required prob. = P(WB) + P(WR)
= \(\frac{{ }^7 \mathrm{C}_1 \times{ }^5 \mathrm{C}_1+{ }^7 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{15} \mathrm{C}_2}\)
= \(\frac{7 \times 5+7 \times 3}{\frac{15 \times 14}{2}}=\frac{56}{105}=\frac{8}{15}\)

Question 24.
In a group of sudents, there are 3 boys and 3 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl, or 3 girls and 1 boy are selected.
Answer:
Given, there are 3 boys and 3 girls in a group.
∴ Total no. of ways of selecting four students out of 6 = ⁶C₄
Total no. of ways of selecting either 3 boys and 1 girl or 3 girls and 1 boy = ³C₃ × ³C₁ + ³C₃ × ³C₁
required probability
= \(\frac{{ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_4}=\frac{1 \times 3+1 \times 3}{\frac{6 \times 5}{2}}=\frac{6}{15}=\frac{2}{5}\)

Question 25.
Three cards are drawn from a pack of well-shuffled 52 cards. Find the probability that
(i) all the three cards are of the same suit
Answer:
Since there are 4 suits in a pack of 52 cards
Total no. of outcomes = ⁵²C₃ [i.e. selecting 3 cards out of 52 cards]
Each suit contains 13 cards.
So selection of 3 cards from each suit can be done in ¹³C₃ ways each.
∴ reqd. probability = 4 × \(\frac{{ }^{13} C_3}{{ }^{52} C_3}=\frac{4 \times 13 \times 12 \times 11}{52 \times 51 \times 50}=\frac{22}{425}\)

(ii) one is a king, the other is a queen and the third is a jack.
Answer:
reqd. prob. = \(\frac{{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_3}=\frac{4 \times 4 \times 4 \times 6}{52 \times 51 \times 50}=\frac{16}{5525}\)

Question 26.
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 ? (ISC 2014)
Answer:
bag I contains 5 white balls and 4 black balls
bag II contains 7 white balls and 9 black balls
∴ required probability P (white ball from bag I and two black balls from bag II) + P (black ball from bag-I and one white and one black ball from bag II)
= \(\frac{5}{9} \times \frac{{ }^9 \mathrm{C}_2}{{ }^{16} \mathrm{C}_2}+\frac{4}{9} \cdot \frac{{ }^7 \mathrm{C}_1 \times{ }^9 \mathrm{C}_1}{{ }^{16} \mathrm{C}_2}\)
= \(\frac{5}{9} \times \frac{9 \times 8}{16 \times 15}+\frac{4}{9} \times \frac{7 \times 9 \times 2}{16 \times 15}\)
= \(\frac{1}{6}+\frac{7}{30}=\frac{12}{30}=\frac{2}{5}\)

Question 27.
Two dice are rolled once. Find the probability that
(i) the numbers on two dice are different
Answer:
Total no. of outcomes = 6² = 36
Required probability = 1 – P (of getting same numbers on both dice)
= 1 –\(\frac{6}{36}\) = 1- \(\frac{1}{6}=\frac{5}{6}\)
[since favourable cases are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]

(ii) the total of numbers on the two dice is atleast 4.
Answer:
required probability P (atleast 4)
= 1 – P (atmost 3)
Here favourable cases are {(1, 1), (1, 2), (2, 1)}
required probability = 1 – \(\frac{1}{12}=\frac{11}{12}\) = 1 –\(\frac{1}{12}=\frac{11}{12}\)

(iii) the product of the numbers on two dice Is odd.
Answer:
When two dice are rolled then n(S) = 6² = 36
Let E : Event that product of the numbers on two dice is odd
= {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5))
∴ n(E) = 9
Thus required prob. \(\frac{n(E)}{n(S)}=\frac{9}{36}=\frac{1}{4}\)

Question 28.
In a single throw of three dice, find the probability of getting
(i) same number on the three dice
(ii) a total of 5
(iii) a total of atmost 5
(iv) a total of atleast 5.
Answer:
Total exhaustive cases = 6 × 6 × 6 = 216
(i) favourable cases are {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6))
Thus required probability = \(\frac{6}{216}=\frac{1}{36}\)

(ii) Favourable cases = {(1, 1, 3), (1, 3, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1), (3, 1, 1)}
∴ required probability = \(\frac{6}{216}=\frac{1}{36}\)

(iii) Favourable cases {(1, 1, 1), (1, 2, 1), (1, 1, 2). (2, 1, 1), (1, 1, 3), (1, 3, 1), (1,2, 2), (2, 1,2), (2, 2, 1), (3, 1, 1))
∴ required probability = \(\frac{10}{216}=\frac{5}{108}\)

(iv) P (a total of atleat 5) = 1 – P (atmost 4)
= 1 – \(\frac{4}{216}\) = 1 – \(\frac{4}{216}\)
[∵ Favourable cases are {(1, 1, 1), (1, 2, 1), (1, 1, 2), (2, 1, 1)}]

Question 28 (old).
The probabilities of two students A and B coming to school in time are \(\frac{3}{7}\) and \(\frac{5}{7}\) respectively. Assuming that the events, ‘A coming in time’ and ‘B coming in time” are independent, find the probability of only one of them coming to school in time.
Write atleast one advantage of coming to school in time.
Answer:
Let A: event that student A coming to school in time
B : event that student B coming to school in time
∴ P(A) = \(\frac{3}{7}\); P(B) = \(\frac{5}{7}\)
required probability
= P(A n B̄) + P(Ā n B)
= P (A) P (B̄) + P (Ā) P (B)
[Since A and B̄; Ā and B are independent events]
= \(\frac{3}{7} \times\left(1-\frac{5}{7}\right)+\left(1-\frac{3}{7}\right) \times \frac{5}{7}\)
= \(\frac{3}{7} \times \frac{2}{7}+\frac{4}{7} \times \frac{5}{7}=\frac{26}{49}\)
Since coming to school is a good habit, since students donot miss any topic of their subject. Also later on habit of being in time is very helpful throughout the life.

Question 29.
The probability of two students A and B coming to school on time are \(\frac{2}{7}\) and \(\frac{4}{7}\) respectively. Assuming that the events ‘A coming on time’ and ‘B coming on time are independent’, find the probability of only one of them coming to school on time.
Answer:
Let E₁ : event that student A coming to school on time
E₂: event that student B coming to school on time
ThenP(E₁) = \(\frac{2}{7}\); P(E₂) = \(\frac{4}{7}\)
given E₁ and E₂ are independent events.
∴ required probability that only one of them coming to school on time
= P(E₁ ∩ Ē₂) + P(Ē₁ ∩ E₂)
= P(E₁)P(Ē₂) + P(Ē₁)P(E₂)
[since E₁ and E₂ are independent events then so are Ē₁ and E₂ ; E₁ and Ē₂]
= P(E₁) [1 – P(E₂)] + [1 – P(E₁)]P(E₂)
= \(\frac{2}{7}\left[1-\frac{4}{7}\right]+\left[1-\frac{2}{7}\right] \frac{4}{7}\)
= \(\frac{2}{7} \times \frac{3}{7}+\frac{5}{7} \times \frac{4}{7}=\frac{26}{49}\)

Question 29(old).
(i) Kamal and Monika appear for an interview for two vacancies. The probability of Kamal’s selection is \(\frac{1}{3}\) and that of Monika’s selection is \(\frac{1}{5}\). Find the probability that only one of them will be selected. (ISC 2007)
(ii) Akhil and Vijay appear for an interview for two vacancies. The probability of Akhil’s selection is \(\frac{1}{4}\) and Vijay’s selection is \(\frac{2}{3}\). Find the probability that only one of them will be selected. (ISC 2010)
Answer:
Probability of Kamal’s selection = P(K) = \(\frac{1}{3}\)
probability of Monika’s selection = P (M) = \(\frac{1}{5}\)
P (K̄) = 1 – \(\frac{1}{3}=\frac{2}{3}\)
P(M̄) = 1 – P(M) = 1 – \(\frac{1}{5}=\frac{4}{5}\)

(i) required probability
= P(M ∩ K̄) + P(M̄ ∩ K)
= P (M) + P (K) – 2 P (M ∩ K)
= \(\frac{1}{5}+\frac{1}{3}-2 \times \frac{1}{3} \times \frac{1}{5}\)
= \(\frac{3+5-2}{15}=\frac{6}{15}=\frac{2}{5}\)

(ii) Given prob. of Akhil selection = \(\frac{1}{4}\) = P(A)
Prob. of Vijay selection = \(\frac{2}{3}\) = P(V)
Since A and V are independent events Thus reqd. prob. that only one of them will be selected
= P(A ∩ V̄) + P(Ā ∩ V)
= P(A) [1 – P(V)] + [1 – P(A)] P(V)
= \(\frac{1}{4}\left(1-\frac{2}{3}\right)+\left(1-\frac{1}{4}\right) \frac{2}{3}\)
= \(\frac{1}{4} \times \frac{1}{3}+\frac{3}{4} \times \frac{2}{3}=\frac{7}{12}\)

Question 30.
(i)A problem is given to three students whose chances of solving it are \(\frac{1}{4}, \frac{1}{5}\) and \(\frac{1}{3}\) respectively. Find the probability that the problem is solved.
Answer:
Let A, B and C be the events that the three students respectively solving the problem
P(A) = \(\frac{1}{4}\); P(B) = \(\frac{1}{5}\); P(C) = \(\frac{1}{3}\)
∴ P(Ā) = probability that problem is not solved by student
A = 1 – P(A) = 1 – \(\frac{1}{4}=\frac{3}{4}\)
P(B̄) = 1 – P(B) = 1 – \(\frac{1}{5}=\frac{4}{5}\)
and P(C̄) = 1 – P(C) = 1 – \(\frac{1}{3}=\frac{2}{3}\)
∴ Probability that problem is not solved = P (Ā) P (B̄) P (C̄) = \(\frac{3}{4} \times \frac{4}{5} \times \frac{2}{3}=\frac{2}{5}\)
∴ required probability that problem is solved = 1 – probability that problem is not solved
= 1 – \(\frac{2}{5}=\frac{3}{5}\)

(ii) The probability of hitting a target at the first, second and third shots are \(\frac{3}{5}, \frac{2}{5}\) and \(\frac{3}{4}\). Calculate the probability that the target gets hit. (ISC 2004)
Answer:
Let A : be the event that target being hit at first shot.
B : event that target being hit at second shot.
C : event that target being hit at third shot. .
∴ required probability that target being hit
= 1 – P(target is not being hit by any shot) …(1)
Given P (A) = \(\frac{3}{5}\); P (B) = \(\frac{2}{5}\); P (C) = \(\frac{3}{4}\)
∴ P(Ā) = 1 – P(A) = 1 – \(\frac{3}{5}=\frac{2}{5}\) ; P (B̄) = 1 – P (B) = 1 – \(\frac{2}{5}=\frac{3}{5}\)
and P(C̄) = 1 – P (C) = 1 – \(\frac{3}{4}=\frac{1}{4}\)
∴ required probability = 1 – P(ĀB̄C̄) = 1 – P(Ā)P(B̄)P(C̄)
=1 – \(\frac{2}{5} \times \frac{3}{5} \times \frac{1}{4}\)
= 1 – \(\frac{3}{50}=\frac{47}{50}\)

(iii) A problem in Mathematics is given to three students whose chances of solving it are \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{4}\) respectively. What is the probability that exactly one of them solves the problem.
Answer:
Let A, B and C be the .events that the three students respectively solving the problem. Then A, B and C are independent events.
Also given, P (A) = \(\frac{1}{2}\); P (B) = \(\frac{1}{3}\)
and P (C) = \(\frac{1}{4}\)
∴ required probability = P(A ∩ B ∩ C) + P(A ∩ B ∩ C) + P(A ∩ B ∩ C)
= P (A) P (B) P (C) + P (A) P (B) P (C) + P (A) P (B) P (C)
= P (A) (1 – P (B)) (1 – P (C)) + (1 – P (A)) P (B) (1 – P (C)) + P (C) (1 – P (A)) (1 – P (B))
= \(\frac{1}{2}\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{2}\right) \frac{1}{3}\left(1-\frac{1}{4}\right)+\frac{1}{4}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\)
= \(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}+\frac{1}{2} \times \frac{1}{3} \times \frac{3}{4}+\frac{1}{4} \times \frac{1}{2} \times \frac{2}{3}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{6+3+2}{24}=\frac{11}{24}\)

Question 31.
Three persons A, B and C shoot to hit a target Their probabilities of hitting the target are \(\frac{5}{6}, \frac{4}{5}\) and \(\frac{3}{4}\) respectively. Find the probability that
(i) exactly two persons hit the target
(ii) atleast one person hits the target (ISC 2020)
Answer:
P(A) = \(\); P(B) = \(\); P(C) = \(\)
P(Ā) = 1 – P(A) = 1 – \(\frac{5}{6}=\frac{1}{6}\)
P(B̄) = 1 – P(B) = 1 – \(\frac{4}{5}=\frac{1}{5}\)
and P(C̄) = 1 – P(C) = 1 – \(\frac{3}{4}=\frac{1}{4}\)

(i) P (exactly two persons hit the target)
= P(ABC̄) + P(AB̄C) + P(ĀBC)
= P (A) P (B) P(C̄) + P (A) P(B̄) P (C) + P(Ā) P (B) P (C)
= \(\frac{5}{6} \times \frac{4}{5} \times \frac{1}{4}+\frac{5}{6} \times \frac{1}{5} \times \frac{3}{4}+\frac{1}{6} \times \frac{4}{5} \times \frac{3}{4}\)
= \(\frac{20+15+12}{120}=\frac{47}{120}\)

(ii) P (atleast one person hits the target)
= 1 – P(Ā B̄ C̄)
= 1 – P(Ā) P(B̄) P(C̄)
= 1 – \(\frac{1}{6} \times \frac{1}{5} \times \frac{1}{4}=\frac{119}{120}\)

Question 31 (Old).
The probability that a boy will not pass M.B.A. examination is \(\frac{3}{5}\) and that a girl will not pass is \(\frac{4}{5}\). Calculate the probability that atleast one of them passes the examination.
What ideal conditions a student should keep in mind while appearing in an examination ? (Value Based)
Answer:
Let A : event that a boy will pass MBA examination
B : event that a girl will pass MBA examination.
Given P(A) = 1 – P(Ā) = 1 – \(\frac{3}{5}=\frac{2}{5}\) [∵ given P(Ā) = \(\frac{3}{5}\)]
and P(B̄) = \(\frac{4}{5}\)
required prob. = 1 – P (that neither boy nor girl will pass MBA exam)
= 1 – P(Ā n B̄) =1 – P(Ā) P(B̄) [Since Ā and B̄ are independent events]
= 1 – \(\frac{3}{5} \times \frac{4}{5}=\frac{25-12}{25}=\frac{13}{25}\)

While appearing in an examination, a student should have no intention of copying or cheating. Since cheating or copying will help the students to pass the exam but develop the habit of dishonesty, which leads to corruption.

Question 32.
A husband and a 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
(i) both of them will be selected ?
(ii) none of them will be selected ?
(iii) only one of them will be selected ?
(iv) atleast one of them will be selected ?
Answer:
Let H denotes the event that husband is selected and W denotes the event that wife is selected.
P(H) = \(\frac{1}{7}\); P(W) = \(\frac{1}{5}\)
Thus P(H̄) = 1 – \(\frac{1}{7}=\frac{6}{7}\); P(W̄) = 1 – \(\frac{1}{5}=\frac{4}{5}\)

(i) Thus required probability that both of them selected = P (W ∩ H) = P(W) P (H) = \(\frac{1}{5} \times \frac{1}{7}=\frac{1}{35}\)

(ii) required probability = P(W̄ ∩ H̄) = P\((\overline{\mathrm{W} \cup \mathrm{H}})\) = 1 – P(W ∪ H)
= 1 – P (W) – P (H) + P (W ∩ H)
= [1 – P (W)] [1 – P (H)] [∵ W and H are indepenent events.]
= P (W) . P (H) = \(\frac{4}{5} \times \frac{6}{7}=\frac{24}{35}\)

(iii) required probability = P (W n H̄) + P (H n W̄)
= P (W) P (H̄) + P (H) P (W̄)
= \(\frac{1}{5} \times \frac{6}{7}+\frac{1}{7} \times \frac{4}{5}=\frac{10}{35}=\frac{2}{7}\)

(iv) Required probability that atleast one of them is selected
= 1 – P [none of them is selected]
= 1 – \(\frac{24}{35}=\frac{11}{35}\) [by using part (ii)]

Question 33.
A machine operates if all of its three components function. The probability that the first component fails during the year is 014, the probability that the second component fails is 010 and the probability that the third component fails is 0 05. What is the probability that the machine will fail during the year ?
Answer:
Let A : event that the first component fails
B : event that the second component fails
C : event that the third components fails
∴ P(A) = 0.14 ; P (B) = 0.10; P(C) = 0.05
P (Ā) = prob. when first component operate good
P(Ā) = 1 – 0.14 = 0.86
P(B̄) = 1 – 0.10 = 0.90
P(C̄) = 1 – 0.05 = 0.95
∴ required prob. = 1 – P(ĀB̄C̄) = 1 – P(Ā)P(B̄)P(C̄) = 1 – 0.86 × 0.90 × 0.95 = 0.2647

Question 34.
A speaks truth in 60% of the cases, while B in 40% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? Do you think that statement of B is true ?
Answer:
Given P (A speaks truth) = \(\frac{60}{100}=\frac{3}{5}\) = P (A)
p (A tell a lie) = \(\frac{60}{100}=\frac{40}{100}=\frac{2}{5}\) = P(Ā)
P(B speaks truth) = 40% = \(\frac{40}{100}=\frac{4}{10}\) =P(B)
∴ P (B tell a lie) = 1 – \(\frac{4}{10}=\frac{6}{10}\) = P(B̄)
∴ required probability = P (A speaks truth and
B(tell a lie) + P (A tell a lie and B speaks truth)
= P (AB̄) + P (ĀB)
= P(A)P(B̄) + P(Ā)P(B)
= \(\frac{3}{5} \times \frac{6}{10}+\frac{2}{5} \times \frac{4}{10}\)
= \(\frac{18}{50}+\frac{8}{50}=\frac{26}{50}\)
= 52%
Yes, the statement of B is true as it is more likely to be true.

Question 34 (old).
A speaks truth in 75% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact ? Do you think that statement of B is true ?
Answer:
Given P (A speaks truth) = \(\frac{75}{100}=\frac{3}{4}\) = P (A)
P(A, tell a lie) = 1 – \(\frac{75}{100}=\frac{25}{100}=\frac{1}{4}\) = P(A)
P (B speaks truth) = 90% = \(\frac{90}{100}=\frac{9}{10}\) = P (B)
P(B tell a lie) = 1 – \(\frac{9}{10}=\frac{1}{10}\)= P (B)
required probability = P (A speaks truth and B tell a lie) + P (A tell a lie and B speaks truth)
= P (AB̄) + P (ĀB) = P (A) P (B̄) + P (Ā) P (B)
= \(\frac{3}{4} \times \frac{1}{10}+\frac{1}{4} \times \frac{9}{10}=\frac{3}{40}+\frac{9}{40}=\frac{12}{40}=\frac{3}{10}\) = 30%
Yes, the statement of B is true as it is more likely to be true.

Question 35.
Ten Tickets are numbered 1 to 10. Two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of 5 and the other a multiple of 4 when the tickets are drawn
(i) without replacement
(ii) with replacement.
Answer:
Sample space
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = number on tickets is multiple of 5 ={5, 10}
and B = event that number on tickets is multiple of 4 = {4, 8}

(i) ∴ required probability
= prob. of getting multiple of 5 on 1st draw and multiple of 4 on 2nd draw + prob. of getting multiple of 4 on 1st draw and multiple of 5 on 2nd draw
= P (A n B) + P (B n A) = P (A) P (B/A) + P (B) . P (A/B)
= \(\frac{2}{10} \times \frac{2}{9}+\frac{2}{10} \times \frac{2}{9}=\frac{8}{90}=\frac{4}{45}\)

(ii) Required probability = P(A n B) + P(B n A) = P(A) P(B/A) + P(B)P(A/B)
= \(\frac{2}{10} \times \frac{2}{10} \times \frac{2}{10} \times \frac{2}{10}=\frac{8}{100}=\frac{2}{25}\)

Question 36.
A bag contains 5 red discs and 4 blue discs. If 3 discs are drawn from the bag without replacement, then find the probability that
(i) all three are blue
(ii) there are 2 red and 1 blue
(iii) the second disc is red.
Answer:
Since a bag contains 5 red discs and 4 blue discs.
Total no. of discs in a bag = 5 + 4 = 9
(i) required prob. that all three discs are blue = P(B₁B₂B₃) =

(ii) required prob. = P (R₁R₂B₃) + P (R₁B₂R₃) + P (B₁R₂R₃)
= P (R₁) P (R₂) P (B₃) + P (R₁) P (B₂)P (R₃) + P (B₁) P (R₂) P (R₃)
= \(\frac{5}{9} \times \frac{4}{8} \times \frac{4}{7}+\frac{5}{9} \times \frac{4}{8} \times \frac{4}{7}+\frac{4}{9} \times \frac{5}{8} \times \frac{4}{7}\)
= \(\frac{80}{9 \times 8 \times 7}\) × 3
= \(\frac{240}{9 \times 7 \times 7}=\frac{30}{63}=\frac{10}{21}\)

(iii) required probability = P(R₁R₂R₃) + P (R₁R₂B₃) + P (B₁R₂R₃) + P (B₁R₂B₃)
= P (R₁) P (R₂) P (R₃) + P (R₁) P (R₂) P (B₃) + P (B₁) P (R₂) P (R₃) + P (B₁) P (R₂) P (B₃)
= \(\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7}+\frac{5}{9} \times \frac{4}{8} \times \frac{4}{7}+\frac{4}{9} \times \frac{5}{8} \times \frac{4}{7}+\frac{4}{9} \times \frac{5}{8} \times \frac{3}{7}\)
= \(\frac{60}{9 \times 8 \times 7}+\frac{80}{9 \times 8 \times 7}+\frac{80}{9 \times 8 \times 7}+\frac{60}{9 \times 8 \times 7}\)
= \(\frac{280}{9 \times 8 \times 7}=\frac{40}{9 \times 8}=\frac{5}{9}\)

Question 37.
A bag contains 5 white and 3 black balls. Four balls are successively drawn out without replacement. What is the probability that they are alternately of different colours. What would be the probability of this event if balls are drawn With replacement ?
Answer:
Since a bag contains 5 white and 3 black balls
Total no. of balls = 5 + 3 = 8
When four balls are successively drawn out without replacement,
required prob. = P (BWBW) + P (WBWB)
= P (B) P (W) P (B) P (W) + P (W) P (B) P (W) P (B)

When four balls are drawn successively out with replacement
∴ required probability = P (B) P (W) P (B) P (W) + P (W) P (B) P (W) P (B)
= \(\frac{3}{8} \times \frac{5}{8} \times \frac{3}{8} \times \frac{5}{8}+\frac{5}{8} \times \frac{3}{8} \times \frac{5}{8} \times \frac{3}{8}\)
= \(\frac{225}{4096}+\frac{225}{4096}=\frac{450}{4096}=\frac{225}{2048}\)

Question 38.
A bag contains one black and two white balls. A drawing from the bag consists of taking a ball from the bag and keeping it out if it is white but putting it back if it is black. Calculate the probabilities that
(i) the first drawing is a white ball
(ii) the second drawing is a white ball
(iii) the third drawing is a white ball.
Answer:
Since a bag contains one black and 2 white balls
∴ Total no. of balls = 1 + 2 = 3
(i) prob. that first drawing a white ball = \(\frac{2}{3}\)

(ii) required prob. = P (BW) + P (WW) = P (B) P (W) + P (W) P (W)
= \(\frac{1}{3} \times \frac{2}{3}+\frac{2}{3} \times \frac{1}{2}=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\)

(iii) required prob. = P (BBW) + P (BWW) + P (WBW) + P (WWW)
= \(\frac{1}{3} \times \frac{1}{3} \times \frac{2}{3}+\frac{1}{3} \times \frac{2}{3} \times \frac{1}{2}+\frac{2}{3} \times \frac{1}{2} \times \frac{1}{2}+\frac{2}{3} \times \frac{1}{2} \times \frac{0}{1}\)
= \(\frac{2}{27}+\frac{1}{9}+\frac{1}{6}=\frac{4+6+9}{54}=\frac{19}{54}\)

Question 39.
A bag contains 4 white and 2 black balls. Another contains 3 white and 5 black balls. If one ball is draw from each bag, then find the probability that
(i) both are white
(ii) both are black
(iii) one is white and one is black.
Answer:
Given bag I contains 4 white and 2 black balls and bag II contains 3 white and5 black balls
(i) required prob. = P (W₁W₂)
= \(\frac{4}{6} \times \frac{3}{8}=\frac{12}{48}=\frac{1}{4}\)

(ii) required prob. = P (B₁B₂)
= \(\frac{2}{6} \times \frac{5}{8}=\frac{10}{48}=\frac{5}{24}\)

(in) required prob. = P (B₁W₂) + P (W₁B₂)
= \(\frac{2}{6} \times \frac{3}{8}+\frac{4}{6} \times \frac{5}{8}=\frac{26}{48}=\frac{13}{24}\)

Question 40.
A bag contains 2 black, 4 white and 3 red balls. One ball is drawn at random and kept aside. Then another ball is drawn and kept aside. This process is continued till all the balls are drawn. Find the probability that the ball drawn are in the sequence 2 black, 4 white and 3 red.
Answer:
Given bag contains 2 black, 4 white and 3 red balls
Total no. of balls = 2 + 4 + 3 = 9
∴ required probability = P (B₁B₂W₃W₄W₅W₆R₇R₈R₉)
= \(\left(\frac{2}{9} \times \frac{1}{8}\right)\left(\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4}\right),\left(\frac{3}{3} \cdot \frac{2}{2} \cdot \frac{1}{1}\right)\)
= \(\frac{1}{36} \times \frac{1}{35}=\frac{1}{1260}\)

Question 41.
A policeman fires four bullets on a dacoit. The probability that the dacoit will be killed by one bullet is 0 6. What is the probability that the dacoit is still alive ?
Answer:
Let A : event that dacoit will be killed by one bullet.
∴ P (A) = 0.6
Thus P (dacoit will not killed by one bullet i.e. dacoit will alive)
= P(Ā) = 1 – P(A)
= 1 – 0.6
= 0.4
∴ required probability = p (Ā) P (Ā) P (Ā) P (Ā)
= (0.4) (0.4) (0.4) (0.4)
= \(\frac{4 \times 4 \times 4 \times 4}{10 \times 10 \times 10 \times 10}=\frac{16}{625}\)

Question 42.
A and B toss a coin alternately till one of them gets a head and wins the game. If A starts first, then find the probability that B will win the game.
Answer:
Let E be the event of getting a head in single toss of coin , ∴ P(E) = \(\frac{1}{2}\)
P(Ē) = 1 – \(\frac{1}{2}=\frac{1}{2}\)
Thus B will wins if he gets a head in 2nd throw, 4th throw of 6th throw …………..
His probability of getting a head in 2nd throw = P(Ē ∩ E) = P(Ē).P(E) = \(\frac{1}{2} \cdot \frac{1}{2}=\left(\frac{1}{2}\right)^2\)
Probability of getting a head in 4th throw = P(Ē ∩ Ē ∩ Ē ∩ E)= [P(Ē)]³P(E) = \(\left(\frac{1}{2}\right)^3 \frac{1}{2}=\left(\frac{1}{2}\right)^4\)
and so on.
Thus, the prob. of winning of B
= P(Ē ∩ E) + P(Ē ∩ Ē ∩ Ē ∩ E) + P(Ē ∩ Ē ∩ Ē ∩ Ē ∩ Ē ∩ E) …………………………….∞
= P (Ē) P(E) + P(Ē) P(Ē) P(Ē)P(E) + P(Ē) P(Ē) P(Ē) P(Ē) P(Ē) P(E) ……………………………..∞
= \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^6\) ……………….. ∞
= \(\frac{\left(\frac{1}{2}\right)^2}{1-\left(\frac{1}{2}\right)^2}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\)

Question 43.
Two aeroplanes I and II bomb a target in succession. The probabilities of aeroplanes I and II scoring a hit correctly are 0.3 and 0.2 respectively. The second plane will bomb only if the first plane misses the target. Find the probability that the target is hit by the second plane.
Answer:
Let A : event that the aeroplane I hit the target
B : event that the aeroplane II hit the target
Then P (A) = 0.3 and P (B) = 0.2
∴ P(Ā) = 1 – P (A) = 1 – 0.3 = 0.7
and P(B̄) = 1 – P (B) == 1 – 0.2 = 0.8
Since it is given that, second plane will bomb only if the first plane misses the target.
∴ probability of second plane hitting the target in first round (i.e. when plane I fails to hit the target)
= P(A)P(B) = 0.7 × 0.2
Now second plane will get second chance to hit the target if plane-I has fail in first and third chance and second plane fails to hit the target in first chance.
∴ prob. of second plane to hit the target in 2nd round = p (Ā) P (B̄) P (Ā) P (B)
= 0.7 × 0.8 × 0.7 × 0.2
Similarly probability of second plane of hitting the target in 3rd round
= P (A) P (B) P (A) P (B) P (A) P (B)
= 0.7 × 0.8 × 0.7 × 0.8 × 0.7 × 0.2
and so on.
Thus required probability that the target is hit by second plane
= 0.7 × 0.2 + 0.7 × 0.8 × o.7 × 0.2 + 0.7 × 0.8 + 0.7 × 0.8 × 0.7 × 0.2 + …. ∞.
= 0.7 × 0.2 + (0.7)² × 0.8 × 0.2 + (0.7)³ × (0.8)² × 0.2 + … ∞
= 0.7 × 0.2 [1 + 0.7 × 0.8 + (0.7)² × (0.8)² + …. ∞]
= 0.14[latex]\frac{1}{1-0.7 \times 0.8}[/latex] [S = \(\frac{a}{1-r}\)]
= \(\frac{0.14}{1-0.56}=\frac{0.14}{0.44}=\frac{14}{44}=\frac{7}{22}\)

Question 44.
In a hockey match, both teams 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 a 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, fUjd their respective probabilities of winning the match and state whether the decision of the refree was fair or not.
Answer:
we know that in throwing of single die
probability of getting six= \(\frac{1}{6}\)
Since it is given that captains to throw a die alternatively.
Thus P (A) = prob. of getting six= P(B) = \(\frac{1}{6}\)
P (A) = 1 – \(\) = P(B)
Since A starts the game hence A wins in 1st throw, 3rd throw, 5th throw, and B wins in 2nd throw, 4th throw, 6th throw ….
P(A’s winning) ;
= P(A) + P(Ā B̄ A) + P(Ā B̄ Ā B̄ A) + ………….
= P(A) + P(Ā) P(B̄) P(A) + P(Ā) P(B̄) P(Ā) P(B̄) P(A) + ……………..
= \(\frac{1}{6}+\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}+\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\) + ………….
= \(\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{\frac{1}{6}}{\frac{11}{36}}=\frac{6}{11}\)

and P (B’s wihning)
= P(ĀB) + P(ĀB̄ĀB) +P(ĀB̄ĀB̄ĀB) …… ∞
= P(Ā)P(B) + P(Ā) P (B̄) P(Ā)P'(B) +P(Ā)P(B̄)P(Ā)P(B̄) P(Ā)P(B) ….
= \(\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}\) ………….
= \(\frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{5}{11}\)
The decision was not fair as the probabilities are not equal since the probability of team A winning is more than that Ot team B winning. In fact, their probabilities of winning are in the ratio \(\frac{6}{11}: \frac{5}{11}\) i.e. 6:5.

Question 44 (Old).
Aman and Bhuvan 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. (ISC 2011)
Answer:
When a pair of dice is thrown
Then total no. of outcomes = 6² = 36 and all outcomes are equally likely.
Let E be the event of throwing a sum of 8
∴ E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
∴ P(E) = \(\frac{5}{36}\)
∴ Prob. of not throwing a sum of 8 = 1 – \(\frac{5}{36}=\frac{31}{36}\)
As Aman starts the game, prob. of winning in first throw = \(\frac{5}{36}\)
Aman gets the second chance only if Aman and Bhuvan fail in the first round.
prob. of Aman’s winning in 3rd throw = \(\frac{31}{36} \times \frac{31}{36} \times \frac{5}{36}=\left(\frac{31}{36}\right)^2 \frac{5}{36}\)

prob. of Aman’s winning in 5th throw = \(\frac{31}{36} \times \frac{31}{36} \times \frac{31}{36} \times \frac{31}{36} \times \frac{5}{36}=\left(\frac{31}{36}\right)^4 \frac{5}{36}\)
and so on.
Thus, the prob. of Aman’s winning the game
= \(\frac{5}{36}+\left(\frac{31}{36}\right)^2 \frac{5}{36}+\left(\frac{31}{36}\right)^4 \times \frac{5}{36}\) + ………………..
= \(\frac{\frac{5}{36}}{1-\left(\frac{31}{36}\right)^2}=\frac{\frac{5}{36}}{\frac{1296-961}{1296}}\)
= \(\frac{5}{36} \times \frac{1296}{335}=\frac{36}{67}\)

Thus, prob. of Bhuvan’s winning the game = 1 – \(\frac{36}{67}=\frac{31}{67}\)

Question 45.
A and B throw a pair of dice alternately till one of them gets the sum of numbers as multiples of 6 and wins the game. If A starts first, find the probability of B winning the game.
Answer:
When two dice are thrown, then total no. of outcomes = 6² = 36
E : sum of numbers as multiples of 6 = {(1,5), (2,4), (3,3), (4,2), (5,1), (6,6)}
∴ P(E) = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{6}{36}=\frac{1}{6}\) = p
q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\)

Since A starts first, then B can win the game in 2nd throw, 4th throw, 6th throw and so on.
P (B’s winning)
= qp + qqqp + qqqqqp + ……………
= \(\frac{5}{6} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^3 \frac{1}{6}+\left(\frac{5}{6}\right)^5 \frac{1}{6}\) + ………………
= \(\frac{5}{36}\left[1+\frac{25}{36}+\left(\frac{5}{6}\right)^4+\ldots \ldots\right]\)
= \(\frac{5}{36}\left[\frac{1}{1-\frac{25}{36}}\right]\)
= \(\frac{5}{36} \times \frac{36}{11}=\frac{5}{11}\)

Question 46.
A and B cut a pack of 52 playing cards alternately and every time the card is put back. The pack is well-shuffled after each cut. A starts the game and the one who cuts a heart first wins. What are their respective probabilities of winning ?
Answer:
Total no. of outcomes = 52
Let E : event of getting a heart card
∴ P (E) =\(\frac{1}{4}\)
∴ P(E) = prob. of not getting a heart card = 1 – \(\frac{1}{4}=\frac{3}{4}\)
As A shorts the game, prob. of winning in first cut = \(\frac{1}{4}\)
A gets the second chance only if A and B fail in the first round.
∴ prob.of A’s winning in 3rd throw = P(ĒĒE) = P(Ē)P(Ē)P(E) = \(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}=\left(\frac{3}{4}\right)^2 \frac{1}{4}\)
prob. of A’s winning in 5th throw = P(Ē Ē Ē Ē E) = P(Ē) P(Ē) P(Ē) P(Ē) P(E) = \(\left(\frac{3}{4}\right)^4 \frac{1}{4}\)
and so on.

Thus, the required prob. of A’s winning the game
= \(\frac{1}{4}+\left(\frac{3}{4}\right)^2 \frac{1}{4}+\left(\frac{3}{4}\right)^4 \frac{1}{4}\) ………
= \(\frac{\frac{1}{4}}{1-\left(\frac{3}{4}\right)^2}=\frac{\frac{1}{4}}{1-\frac{9}{16}}=\frac{\frac{1}{4}}{\frac{7}{16}}=\frac{1}{4} \times \frac{16}{7}=\frac{4}{7}\)

∴ required prob. of B’s winning = 1 – P (A’s winning)
= 1 – \(\frac{4}{7}=\frac{3}{7}\)

Question 47.
A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. A starts the game, find the probability that B wins the game.
Answer:
When a pair of dice are thrown, total no. of outcomes = 36
A : For getting a sum 7 ={(1, 6)(2, 5), (3, 4),(4, 3), (5, 2), (6, 1)}
B : getting a sum 10 = {(4, 6), (5, 5), (6, 4)}
∴ P(A) = \(\frac{6}{36}=\frac{1}{6}\) ∴ P(A) = 1 – \(\frac{1}{6}=\frac{5}{6}\); P(B) = \(\frac{3}{36}=\frac{1}{12}\) , ∴ P(B) = 1 – \(\frac{1}{12}=\frac{11}{12}\)
Since A starts the game hence B wins in 2nd throw, 4th trow, 6th throw, ….
∴ P(B’s winning) = P(Ā B) + P(Ā B̄ Ā B) + P (Ā B̄ Ā B̄ Ā B) + ….
= P (Ā) P (B) + P (Ā) P (B̄) P (Ā) P (B) + P (Ā) P (B̄) P (Ā) P (B̄) P (Ā) P (B).
= \(\frac{5}{6} \times \frac{1}{12}+\frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{1}{12}+\frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{11}{12} \times \frac{5}{6} \times \frac{1}{12}\) + ………
= \(\frac{5}{6} \times \frac{1}{12}+\left(\frac{5}{6}\right)^2 \times \frac{11}{12} \times \frac{1}{12}+\left(\frac{5}{6}\right)^3\left(\frac{11}{12}\right)^2 \frac{1}{12}\) + ……… ∞
which form an infinite G.P. with a = first term = \(\frac{5}{6} \times \frac{1}{12}\) and r = \(\frac{5}{6} \times \frac{11}{12}\)
P(B’s winning) = \(\frac{\frac{5}{6} \times \frac{1}{12}}{1-\frac{5}{6} \times \frac{11}{12}}=\frac{\frac{5}{72}}{\frac{72-55}{72}}=\frac{5}{17}\)

Question 48.
A, B and C in order toss a fair coin. The first one to throw a head wins. What are their respective chances of winning ? Assume that the game may continue indefinitely.
Answer:
p = P (getting a head) = \(\frac{1}{2}\)
q = p(getting a tail) = \(\frac{1}{2}\)
P(A winning)
= P(A wins in 1st, 4th, 7th throw and soon)
= p + qqqp + qqqqqqp + ………. ∞
= \(\frac{1}{2}+\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^7\) + ………..∞
= \(\frac{\frac{1}{2}}{1-\frac{1}{8}}=\frac{\frac{1}{2}}{\frac{7}{8}}=\frac{4}{7}\)

P (B winning) = P (B wins in 2nd throw or 5th throw or 8th throw and so on)
= qp + q⁴p + q⁷p + …………… ∞
= \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^5+\left(\frac{1}{2}\right)^8\) + ……………..
= \(\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac{\frac{1}{4}}{\frac{7}{8}}=\frac{2}{7}\)

∴ P (C’s winning) = 1 – P (A) – P (B)
= 1 – \(\frac{4}{7}-\frac{2}{7}=\frac{1}{7}\)

Question 49.
A, B and C throw a die one after the other in the same order till one of them gets a ‘6’ and wins the game. Find their respective probability of winning, if A starts the game.
Answer:
p = P (probability of getting a six) = \(\frac{1}{6}\)
q = P(prob. of not getting a six)
= 1 – \(\frac{1}{6}=\frac{5}{6}\)

Now A starts the game so A wins the game in 1st throw or 4th throw or 7th throw and soon.
P (A’s winning) =p + q³p + q⁶p + …. ∞
= \(\frac{1}{6}+\left(\frac{5}{6}\right)^3 \frac{1}{6}+\left(\frac{5}{6}\right)^6 \frac{1}{6}\) + ……….. ∞
= \(\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^3}=\frac{\frac{1}{6}}{\frac{216-125}{216}}\)
= \(\frac{36}{91}\)

Now B wins in 2nd throw or 5th throw or 8th throw and so on.
P (C’s winning) = qp + q⁴p + q⁷p + ……….∞
= \(\frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^4 \frac{1}{6}+\left(\frac{5}{6}\right)^7 \frac{1}{6}\) + ……. ∞
= \(\frac{\frac{5}{36}}{1-\frac{125}{216}}=\frac{30}{91}\)

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