The availability of step-by-step ML Aggarwal Class 12 ISC Solutions Chapter 10 Probability Chapter Test can make challenging problems more manageable.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Chapter Test

Question 1.

Two coins are tossed together. Find P(E | F) if E : no tails appear and F : no heads appear.

Answer:

When two coins are tossed together S = {HH, HT, TH, TT}

E : no tails appear = {HH}

F : no heads appears = {TT}

∴ E ∩ F = Φ ∴ P (E ∩ F) = 0

∴ P (E/F) = \(\frac{P(E \cap F)}{P(F)}\) = 0

Question 2.

Given that the numbers appearing on rolling two dice together are different, what is the probability that the sum of the numbers appearing on the dice is 6 ?

Answer:

When two dice rolled together

Then total no. of outcomes = 6^{2} = 36

E : numbers appearing on rolling two dice together are different

F : Sum of the numbers appearing on dice is 6.

Out of 36 outcomes, the outcomes {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} are leaving out.

∴ n (E) = 36 – 6 = 30

Here E ∩ F = {(1, 5), (5, 1), (2, 4), (4, 2)}

∴ n (E ∩ F) = 4

Since (3, 3) lies in F but not in E.

n(E ∩ F) = 4

Thus P (F/E) = \(\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{E})}=\frac{4}{30}=\frac{2}{15}\)

Question 3.

If A and B are independent events, and P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\) then find P (A^{c} ∩ B^{c}).

Answer:

Given P (A) = \(\frac{1}{2}\) P (B) = \(\frac{1}{3}\)

P (A^{c} ∩ B^{c}) = P ((A ∪ B)^{c})

= l- P(A ∪ B)

= 1 – [P (A) + P (B) – P (A ∩ B)]

= 1 – P (A) – P (B) + P (A) P (B)

[v A and B are independent events Then P (A ∩ B) = P (A) P (B)]

= (1 – P(A))(1 – P(B))

= (1 – \(\frac{1}{2}\)) (1 – \(\frac{1}{3}\))

= \(\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}\)

Question 3(Old).

Choose the correct answer in each of the following :

(i) If A and B are two events such that P (A) ≠ 0 and P (B | A) = 1 then

(a) A ⊂ B

(b) B ⊂ A

(c) B = Φ

(d) A = Φ

Answer:

Since P (B/A) = 1 ⇒ \(\frac{P(A \cap B)}{P(A)}\) = 1

P (A ∩ B) = P (A)

Thus A ⊂ B, ⇒ A ∩ B = A

P (A ∩ B) = P (A)

(ii) If P (A | B) > P (A) then

(a) P (B | A) < P (B)

(b) P (A ∩ B) < P (A) . P(B) (c) P (B | A) > P (B)

(d) P (B | A) = P (B)

Answer:

Given P (A/B) > P (A)

⇒ \(\frac{P(A \cap B)}{P(B)}\) >P(A)

P (A ∩ B) > P (A) P (B)

Now P(B/A) = \(\frac{P(A \cap B)}{P(A)}>\frac{P(A) P(B)}{P(A)}\) = P(B)

[using (1)]

(iii) If A and B are any two events such that P (A) + P (B) – P (A and B) = P (A) then

(a) P (B | A) = 1

(b) P (A | B) = 1

(c) P (B | A) = 0

(d) P (A | B) = 0 (NCERT)

Answer:

Given P (A) + P (B) – P (A ∩ B) = P (A)

⇒ P (B) = P (A ∩ B)

⇒ 1 = \(\frac{P(A \cap B)}{P(B)}\) = P(A/B)

Question 4.

The probabilities that a husband and wife will be alive 20 years from now are 0-7 M and 0-8. Find that in 20 years, husband ‘ will be a widower.

Answer:

Let A: event that husband will alive 20 years from now

B : event that wife will alive 20 years from now.

∴ P (A) = 0.7 ; P (B) = 0.8

P(Ā) = 1 – P (A) = 1 – 0.7 = 0.3

P(B̄) = 1 – P(B) = 1 – 0.8 = 0.2

∴ required probability = P(A ∩ B̄)

= P(A)P(B̄) = 0.7 × 0.2 = 0.14

[∵ A and B are independent events]

Question 5.

An urn contains 4 square pieces, 5 round pieces and 3 triangular pieces. Two pieces are drawn. Find the probability of drawing two different shapes.

Answer:

Total no. of square pieces = 4

Total no. of round pieces = 5

Total no. of triangular pieces = 3

∴ Total no. of pieces = 4 + 5 + 3 = 12

required probability = P (drawing one square and one round piece) + P (drawing one square and one triangular) + P (drawing one round and one triangular piece)

= \(\frac{{ }^4 \mathrm{C}_1 \times{ }^5 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2}+\frac{{ }^4 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2}+\frac{{ }^5 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^{12} \mathrm{C}_2}\)

= \(\frac{4 \times 5+4 \times 3+5 \times 3}{\frac{12 \times 11}{2}}=\frac{47 \times 2}{12 \times 11}=\frac{47}{66}\)

Question 6.

Three faces of a fair die are yellow, two faces red and one blue. The die is tossed A three times. Find the probability that colours yellow, red and blue appear in the first, second and third toss respectively.

Answer:

Given, no. of yellow faces of a fair die = 3

no. of red faces of a fair die = 2

no. of blue faces of a fair die = 1

Total no. of faces in a fair die = 3 + 2 + 1 = 6

P (getting yellow face) = \(\frac{3}{6}=\frac{1}{2}\)

P (getting red face) = \(\frac{2}{6}=\frac{1}{3}\)

P (getting blue face) = \(\frac{1}{6}\)

∴ required probability = \(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{6}=\frac{1}{36}\)

Question 7.

Find the probability that a given three letter word (in English) has all letters repeated.

Answer:

For three letter word, first place can be filled 26 26 26

be 26 alphabets similarly 2nd and 3rd place can be filled by 26 alphabets each.

Thus the total no. of possible 3 letter words

can be formed when letters may repeated = 26 × 26 × 26

Since 3 letters of the word can be filled by any of the 26 English alphabets and all letters are same e.g. aaa, bbb, ………………. zzz.

So total no. of favourable outcomes = 26

∴ required probability = \(\frac{\text { favourable outcomes }}{\text { Total outcomes }}\)

= \(\frac{26}{26 \times 26 \times 26}=\frac{1}{676}\)

Question 8.

Is it possible to construct an unfair die such that probability of getting a six is \(\frac{1}{2}\) that of getting any other number is \(\frac{1}{6}\)?

Answer:

If prob. of getting a six = \(\frac{1}{2}\)

Prob. of getting 1,2,3,4 and 5 each is equal t0 \(\frac{1}{6}\)

Here sum of probabilities of all events

= \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{2}\)

= \(\frac{5}{6}+\frac{1}{2}=\frac{5+3}{6}=\frac{8}{6}=\frac{4}{3}\) ≠ 1

So construction of such on unfair die is not possible.

Question 9.

From each of the four married couples, one of the partners is selected at random. Find the probability that the selected are of the same sex.

Answer:

Since the prob. of getting male from each married couple = \(\frac{1}{2}\)

Prob. of selecting female from each married couple = \(\frac{1}{2}\)

required probability = P (selecting male partner from each married couple) + P (selecting female partner from each married couple)

= \(\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^4=\frac{1}{8}\)

[since there are 4 married couples]

Question 10.

What is the probability that a couple’s ^ second child will be a boy, given that the first child is a girl ?

Answer:

Here sample space with the experiment be {GB, GG}

Total no. of outcomes = 2

no. of favourable outcomes =1 {GB}

required probability = \(\frac{1}{2}\)

Question 11.

Find the probability that when a hand of 7 cards is drawn from a well-shuffled pack of 52 cards, it contains

(i) all kings

(ii) exactly 3 kings

(iii) atleast 3 kings.

Answer:

Total number of ways of selecting 7 cards out of 52 cards = ^{52}C_{7}

(i) required prob. of containing all 4 kings

(ii) required prob. of containing atleast 3 kings

(iii) required prob. of containing atleast 3 kings

Question 12.

Four married couples have gathered in a room. Two persons are selected at i’^-Trandom from amongst them. Find the probability that the selected persons are

(i) a husband and his wife

(ii) a gentleman and a lady but not a couple.

Answer:

Given no. of married couples = 4

∴ Total no. of persons = 8

(i) Total no. of ways of selecting 2 persons out of 8 persons

required probability = \(\frac{{ }^4 \mathrm{C}_1}{{ }^8 \mathrm{C}_2}=\frac{4 \times 2}{8 \times 7}=\frac{1}{7}\)

[since there are 4 married couples]

(ii) P (of selecting a gentleman and a lady)

= \(\frac{{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^8 \mathrm{C}_2}=\frac{4 \times 4 \times 2}{8 \times 7}=\frac{4}{7}\)

∴ required probability = P (of selecting a gentleman and a lady) – P (selecting a couple)

= \(\frac{4}{7}-\frac{1}{7}=\frac{3}{7}\)

Question 13.

A bag contains 5 red, 4 blue and m green balls. Two balls are drawn at random from the bag. If the probability of both being green is \(\frac{1}{7}\) then find m.

Answer:

Given no. of red balls = 5

no. of blue balls = 4

and no. of green balls = m

∴ Total no. of balls = 5 + 4 + m = 9 + m

∴ prob. of drawing 2 green balls = \(\frac{{ }^m \mathrm{C}_2}{{ }^{9+m} \mathrm{C}_2}\)

also given, prob. of drawing both green balls = \(\frac{1}{7}\)

⇒ \(\frac{{ }^m \mathrm{C}_2}{{ }^{m+9} \mathrm{C}_2}=\frac{1}{7}\)

⇒ \(\frac{m !}{\frac{(m-2) ! 2 !}{\frac{(m+9) !}{(m+7) ! 2 !}}}=\frac{1}{7}\)

⇒ \(\frac{m(m-1)}{(m+9)(m+8)}=\frac{1}{7}\)

⇒ 7m (m – 1) = m^{2} + 17m + 72

⇒ 6m^{2} – 24m – 72 = 0

⇒ m^{2} – 4m – 12 = 0

⇒ (m – 6) (m + 2) = 0

⇒ m = 6, – 2

Since number of balls cannot be negative m = 6

Hence the required no. of green balls = 6

Question 14.

The odds that a book will be reviewed favourably by three independent critics are 5 : 2, 4 : 3 and 3 : 4 respectively. What is the probability that of the three reviews, a majority will be favourable ?

Answer:

The odds that a book will be reviewed favourably by three independent critics be 5 : 2, 4 : 3 and 3 :4.

Question 15.

Four dice are thrown. What is the probability that the sum of the numbers appearing on j^the dice is 23 ?

Answer:

When four dice are thrown.

Total no. of possible outcomes = 6^{4} = 1296

Here favourable outcomes are

{(6, 6, 6, 5), (6, 6, 5, 6), (6. 5, 6, 6). (5. 6, 6. 6)}

∴ no. of favourable outcomes = 4

required probability = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}=\frac{4}{1296}=\frac{1}{324}\)

Question 16.

(i) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4 ?

(ii) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 3 or 4 ?

Answer:

(i) When two dice are thrown together

Total no. of possible outcomes = 6^{2} = 36

Let A : event the sum of numbers on two faces is divisible by 3.

B : event the sum of numbers on two faces is divisible by 4.

A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}

B = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}

A ∩ B = {6, 6}

P(A) = \(\frac{12}{36}=\frac{1}{3}\); P(B) = \(\frac{9}{36}=\frac{1}{4}\)

P(A ∩ B) = \(\frac{1}{36}\)

∴ required probability = P(Ā ∩B̄) = P\((\overline{A \cup B})\)

= 1 – P(A ∪ B)

= 1 – {P(A) + P(B) – P(A ∩B)

= 1 – \(\frac{1}{3}-\frac{1}{4}+\frac{1}{36}\)

= \(\frac{36-12-9+1}{36}=\frac{16}{36}=\frac{4}{9}\)

(ii) required probability = P(A ∪ B)

= P(A) + P(B) – P(A ∩ B)

= \(\frac{1}{3}+\frac{1}{4}-\frac{1}{36}=\frac{12+9-1}{36}\)

= \(\frac{20}{36}=\frac{5}{9}\)

Question 18.

In an essay competition, the odds in favour of the four competitors A, B, C and D are 1 : 2, 1 : 3, 1 : 4 and 1 : 5 respectively. Find the probability that one of them wins the competition.

Answer:

Given odds in favour of four competitors A, B, C and D are 1 : 2, 1 : 3, 1 :4 and 1 : 5 respectively.

∴ P(A) = \(\frac{1}{3}\)

P(B) = \(\frac{1}{4}\)

P(C) = \(\frac{1}{5}\)

P(D) = \(\frac{1}{6}\)

P(Ā) = 1 – P(A)

= 1 – \(\frac{1}{3}=\frac{2}{3}\)

P(B̄) = 1 – P(B)

= 1 – \(\frac{1}{4}=\frac{3}{4}\)

P(C̄) = 1 – P(C)

= 1 – \(\frac{1}{5}=\frac{4}{5}\)

P(D̄) = 1 – P(D)

= 1 – \(\frac{1}{6}=\frac{5}{6}\)

required probability that one of them will win the competition

= P (A) + P (B) + P (C) + P (D)

= \(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{20+15+12+10}{60}\)

= \(\frac{57}{60}=\frac{19}{20}\)

Question 19.

A bag contains 3 white, 3 black and 2 red bails. Three balls are drawn one by one without replacement. Find the probability that the third ball is red.

Answer:

Given bag contains 3 white, 3 black and 2 red balls Total no. of balls = 3 + 3 + 2 = 8

∴ required probability = P (WBR) + P (BWR) + P (WWR) + P (BBR) + P (BRR) + P (RBR) + P (WRR) + P (RWR)

Question 20.

There are three urns A, B and C. A contains 4 white balls and 5 blue balls. Urn B contains 4 white balls and 3 blue balls. Urn C contains 3 w hite balls and 6 blue balls. One ball is drawn from each of these urns. What is the probability that out of these three balls drawn, two are white and one is blue ?

Answer:

urn A contains 4 white and 5 blue balls

urn B contains 4 white and 3 blue balls

urn C contains 3 white and 6 blue balls

required probability = P (W_{1}W_{2}B_{3}) + P (W_{1}B_{2}W_{3}) + P (B_{1}W_{2}W_{3})

= P (W_{1}) P (W_{2}) P (B_{3}) + P (W_{1}) P (B_{2}) P (W_{3}) + P (B_{1}) P (W_{2}) P (W_{3})

= \(\frac{4}{9} \times \frac{4}{7} \times \frac{6}{9}+\frac{4}{9} \times \frac{3}{7} \times \frac{3}{9}+\frac{5}{9} \times \frac{4}{7} \times \frac{3}{9}\)

= \(\frac{96+36+60}{9 \times 7 \times 9}\)

= \(\frac{64}{189}\)

Question 21

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive ? (NCERT)

Answer:

Let the second order determinant A = \(\left|\begin{array}{ll}

a & b \\

c & d

\end{array}\right|\)

Where each entry a, b, c or d is either 0 or 1.

Thus for each entry we have 2 possibilities

Total no. of possible cases = 24 = 16

Here favourable outcomes are those determinants whose value is positive and these are

\(\left|\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right| ;\left|\begin{array}{ll}

1 & 1 \\

0 & 1

\end{array}\right| ;\left|\begin{array}{ll}

1 & 0 \\

1 & 1

\end{array}\right|\)

In each case the value of Δ = 1 > 0.

required probability = \(\frac{3}{16}\)

Question 22.

A bag contains (n + 1) coins. It is known that one of these coins shows head on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that the toss results in head is \(\frac{7}{12}\) then find the value of n.

Answer:

Total no. of coins in bag = n + 1

Here total no. of fair coins = n

and no. of biased coin = 1

Let us define the events E_{1}, E_{2} and A as follows :

E_{1} : drawing a fair coin from bag

E_{2} : drawing an unfair coin from bag

A : toss results in head.

Then P (E_{1}) = \(\frac{n}{n+1}\) ; P (E_{2}) = \(\frac{1}{n+1}\)

Here E_{1} and E_{2} both mutually exclusive and exhaustive events.

Then P(A/E_{1}) = \(\frac{1}{2}\); P(A/E_{2}) = \(\frac{2}{2}\)

Thus by law of total probability; we have

P(A) = P(E_{1}) P(A/E_{1}) + P(E_{2})P(A/E_{2})

⇒ \(\frac{7}{12}=\frac{n}{n+1} \cdot \frac{1}{2}+\frac{1}{n+1}\) . 1

⇒ \(\frac{7}{12}=\frac{n+2}{2(n+1)}\)

⇒ 7(n + 1) = 6(n + 2)

⇒ 7n – 6n = 12 – 7

⇒ n = 5

Question 23.

A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. (NCERT)

Answer:

Let us define the events E_{1}, E_{2} and A as follows:

E_{1}: bag I is chosen

E_{2}: bag I is chosen

A: drawn ball will be red

Then P(E_{1}) = P(E_{2}) = \(\frac{1}{2}\)

∴ both events E_{1} and E_{2} are mutually exclusive and exhaustive events.

P (A/E_{1}) = prob. of drawing red ball from bag I

= \(\frac{4}{4+4}=\frac{1}{2}\)

P (A/E_{2}) = probability of drawing red ball from bag II

=\(\frac{2}{2+6}=\frac{2}{8}=\frac{1}{4}\)

we want to find P (E_{1}/A)

Then by Baye’s Theorem, we have

Question 24.

A company launches a new product and estimated that a person who comes across its advertisement will buy the product with a probability of 0.7, and who does ’ not see the advertisement will buy the product with a probability of 0.3. If 70% of the people come across the advertisement, then what is the probability that a person who buys the product had come across the advertisement ?

Answer:

Let us define the events E_{1}, E_{2} and A are as follows:

E_{1} : people came across the advertisement

E_{2} : people does not came across the advertisement

A : person will buy the product

Then P (E_{1}) = 70% = \(\frac{70}{100}\) ;

P(E_{2}) = 30% = \(\frac{30}{100}\)

E_{1} and E_{2} are mutually exclusive and exhaustive events.

P(A/E_{1}) = probability of that person will buy the product and he does not came across the advertisement = 0.7

P(A/E_{2}) = probability that person will buy the product and he already came across the advertisement = 1 – 0.7 = 0.3

We want to find P (E_{1}/A)

Then by Baye’s Theorem, we have

Question 25.

Probability that A speaks truth is \(\frac{4}{5}\). A coin is tossed. A reports that head appears. What is the probability that actually it was head ? (NCERT)

Answer:

Let us define the events E_{1}, E_{2} and A are as follows :

E_{1} : Coin shows head

E_{2} : Coin does not shows head

A : A reports that head appears

ThenP(E_{1}) = \(\frac{1}{2}\) = P (E_{2})

E_{1} and E_{2} are mutually exclusive and exhaustive events.

P (A/E_{1}) = probability that A reports that head appears and given that head has occured.

probability that man speaks truth = \(\frac{4}{5}\)

P (A/E_{2}) = prob. that A reports that head appears and given that head has not occured.

= probability that A does not speaks truth

= 1 –\(\frac{4}{5}=\frac{1}{5}\)

Then by Baye’s Theorem, we have

P(E_{1}/A) = \(\frac{P\left(A / E_1\right) P\left(E_1\right)}{P\left(A / E_1\right) P\left(E_1\right)+P\left(A / E_2\right) P\left(E_2\right)}\)

= \(\frac{\frac{4}{5} \times \frac{1}{2}}{\frac{4}{5} \times \frac{1}{2}+\frac{1}{5} \times \frac{1}{2}}=\frac{\frac{4}{5}}{\frac{4}{5}+\frac{1}{5}}\)

= \(\frac{4}{5}\)

Question 26.

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ?

Answer:

Let E_{1}, E_{2}, E_{3} and A are the events defined as follows :

E_{1} : A scooter driver is insured

E_{2} : A car driver is insured

E_{3} : A truck driver is insured

Let A : event that insured person met an accident

P (A/E_{1}) = probability that insured scooter driver will meet an accident = 0.01

P (A/E_{2}) = probability that insured car driver will meet an accident = 0.03

P (A/E_{3}) = probability that the person will meet an accident given that he is a truck driver = 0.15

We want to find P (E_{1}/A).

Then by Baye’s Theorem, we have

Question 27.

Three bags contain balls as shown in the table below :

A bag is chosen at random and two balls are drawn from it. They happened to be white and red. What is the probability that they came from bag III ?

Answer:

Let E_{1}, E_{2}, E_{3} and A are the events as follows :

E_{1} : bag-I is chosen;

E_{2} : bag-11 is chosen ;

E_{3} : bag III is chosen

A : a white and red ball is drawn

Then P(E_{1}) = P(E_{2}) = P(E_{3}) = \(\frac{1}{3}\)

Here, E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive events.

P(A/E_{1}) = Prob. of drawing one white and one red ball from bag I

= \(\frac{{ }^1 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{3 \times 2}{6 \times 5}=\frac{1}{5}\)

P(A/E_{2}) = Prob. of drawing one white and one red ball from bag II

= \(\frac{{ }^2 \mathrm{C}_1 \times{ }^1 \mathrm{C}_1}{{ }^4 \mathrm{C}_2}=\frac{2}{\frac{4 \times 3}{2}}=\frac{4}{4 \times 3}=\frac{1}{3}\)

P(A/E_{3}) = Prob. of drawing one white and one red ball from bag III

= \(\frac{{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^9 \mathrm{C}_2}=\frac{4 \times 2 \times 2}{9 \times 8}=\frac{2}{9}\)

We want to find P (E_{3}/A)

Then by Baye’s theorem ; we have

Question 27(Old).

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A ? (NCERT)

Answer:

Let E_{1}, E_{2}, E_{3} and E are the events defined as follows :

E_{1} : Machine A is producing items

E_{2} : Machine B is producing items

E_{3} : Machine C is producing items

E : A defective item is produced

Then P(E_{1}) = \(\frac{50}{100}=\frac{1}{2}\)

P(E_{2}) = \(\frac{30}{100}=\frac{3}{10}\)

P(E_{3}) = \(\frac{20}{100}=\frac{1}{5}\)

E_{1}, E_{2} and E_{3} are mutually exclusive and exhaustive events.

P (E/E_{1}) = P (getting a defective item produced by machine A)

= 1% = \(\frac{1}{100}\)

P (E/E_{2}) = P (defective item is produced by machine B)

= 5% = \(\frac{5}{100}\)

P (E/E_{3}) = P (defective item is produced by machine C)

= 7% = \(\frac{7}{100}\)

We want to find P (E_{1}/E)

Then by Baye’s Theorem, we have

Question 28.

The probability function of a random variable X is given by

where p is a constant. Find the value of p. Calculate P (0 ≤ X < 3) and P (X > 1).

Answer:

Given

Since Σ P (x) = 1 ⇒ P (1) + P (2) + P (3) + …. = 1

⇒ 2p + p + 4p + 0 …. + 0 …….. = 1

⇒ 7p = 1

⇒p = \(\frac{1}{7}\)

P (0 ≤ X < 3) = P (X = 0) + P (X = 1) + P (X = 2) = 0 + 2p + p = 3p = \(\frac{3}{7}\) P(X > 1) = 1 – P(X ≤ 1)= 1 – P(X = 0) – P(X = 1)

= 1 – 0 – 2p

= 1 – \(\frac{2}{7}=\frac{5}{7}\)

Question 29.

Find the probability distribution of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.

Answer:

When two dice thrown, all the 36 outcomes are equally likely so it is case of binomial distribution.

Let p = probability of getting a total of 9 = \(\)

Since favourable outcomes are {(4, 5), (5, 4), (3, 6), (6, 3)}

q = 1 – p = 1 – \(\frac{1}{9}=\frac{8}{9}\) Here n = 2

Given random variable Y represents the number of times a total of 9 appears. Y takes values 0, 1,2.

Now P (X = r) = ^{n}C_{r} p^{r} q^{n – r}

Thus the probability of distribution of Y is given below:

Question 30.

Two cards were drawn without replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of cards of honour (i.e. Jack, Queen, King and Ace).

Answer:

Since there are 16 honour cards i.e. 4 kings, 4 queens, 4 jacks and 4 aces.

Clearly the cards are drawn without replacement so trials or events are not independent,

it is not a problem of binomial distribution.

Let X be the random variable denotes the number of honour cards and X takes values 0, 1,2.

P(X = 0) = Prob. of drawing no honour card

= Prob. of getting two 2 cards other than honour cards

= \(\frac{{ }^{36} \mathrm{C}_2}{{ }^{52} \mathrm{C}_2}=\frac{36 \times 35}{52 \times 51}=\frac{105}{221}\)

P (X = 1) = prob. of drawing one honour card and one non-honour card

= \(\frac{{ }^{16} \mathrm{C}_1 \times{ }^{36} \mathrm{C}_1}{{ }^{52} \mathrm{C}_2}=\frac{16 \times 36 \times 2}{52 \times 51}=\frac{96}{221}\)

P (X = 2) = prob. of drawing both honour cards

= \(\frac{{ }^{16} \mathrm{C}_2}{{ }^{52} \mathrm{C}_2}=\frac{16 \times 15}{52 \times 51}=\frac{20}{221}\)

Thus the required probability distribution of X is given below :

Question 31.

Find the probability distribution of the number of boys in families with 3 children, assuming ^ equal probabilities for boys and girls.

Answer:

Since probabilities of both boys and girls are same,

i.e. \(\frac{1}{2}\) i.e. trials are independent and hence it is a problem of binomial distribution.

p = probability of getting boy = \(\frac{1}{2}\)

q = 1 -p = 1 – \(\frac{1}{2}=\frac{1}{2}\) and here n = 3

Let X be the binomial variate and be the number of boys in families and X can takes values 0, 1, 2, 3.

Thus, the required probability distribution of X is given by

Question 32.

Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls. Also find the mean and variance of the distribution.

Answer:

Since balls are drawn one by one without replacement so it is not a case of binomial distribution.

Let X be the random variable denotes the no. of white balls drawn in a random sample of 3 draws.

So X can takes values 0, 1, 2, 3.

P(X = 0) = prob. (of drawing no white ball)

= P (drawing 3 red ball one by one without replacement)

= \(\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}=\frac{120}{720}=\frac{1}{6}\)

P (X = 1) = P (of drawing one white ball)

= P (drawing one white and 2 red balls)

= P (WRR, RWR, RRW)

= P (WRR) + P (RWR) + P (RRW)

= \(\frac{4}{10} \times \frac{6}{9} \times \frac{5}{8}+\frac{6}{10} \times \frac{4}{9} \times \frac{5}{8}+\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}\)

= \(\frac{360}{720}=\frac{1}{2}\)

P (X = 2) = P (drawing two white balls and one red ball)

= P (WWR, WRW, RWW)

= P (WWR) + P (WRW) + P (RWW)

= \(\frac{4}{10} \times \frac{3}{9} \times \frac{6}{8}+\frac{4}{10} \times \frac{6}{9} \times \frac{3}{8}+\frac{6}{10} \times \frac{4}{9} \times \frac{3}{8}\)

= \(\frac{216}{720}=\frac{3}{10}\)

P (X = 3) = P (drawing 3 white balls one by one without replacement)

= \(\frac{4}{10} \times \frac{3}{9} \times \frac{2}{8}=\frac{24}{720}=\frac{1}{30}\)

The probability distribution of X is given below ;

Question 36.

The probability distribution of a discrete random variable X is given as under

Calculate :

(i) the value of A if E (X) = 2 -94

(ii) variance of X.

Answer:

Given probability distribution of discrete random variable given below

Mean = E (X) = ΣX P (X) = 1 × \(\frac{1}{2}\) + 2 × \(\frac{1}{5}+\frac{12}{25}+\frac{2 \mathrm{~A}}{10}+\frac{3 \mathrm{~A}}{25}+\frac{5 \mathrm{~A}}{25}\)

⇒ 2.94 = \(\frac{1}{2}+\frac{2}{5}+\frac{12}{25}+\frac{\mathrm{A}}{5}+\frac{3 \mathrm{~A}}{25}+\frac{\mathrm{A}}{5}\)

⇒ 2.94 = \(\frac{25+20+24+10 \mathrm{~A}+6 \mathrm{~A}+10 \mathrm{~A}}{50}\)

⇒ 147 = 69 + 26A

⇒ 78 = 26A

⇒ A = \(\frac{78}{26}\) = 3

probability distribution of X reduces to :

∴ Var (X) = ΣX^{2} P(X) – µ^{2}

Question 34.

A and B play a game in which A’s chance of w inning the game is \(\frac{3}{5}\). In a series of 6 games, find the probability that A will win atleast 4 games.

Answer:

Let p = probability of A’s winning the game = \(\frac{3}{5}\)

q = 1 – p = 1 – \(\frac{3}{5}=\frac{2}{5}\)

Here n = 6

It is case of binomial distribution

Question 35.

Suppose that 90% of the people are right handed. What is the probability that atmost 6 people of a random sample of 10 people are right handed ? (NCERT)

Answer:

Let p probability that people are right handed = 90% = \(\frac{90}{100}=\frac{9}{10}\)

q = 1 – p = 1 – \(\frac{9}{10}=\frac{1}{10}\), Here n = 10

Then by binomial distribution, we have

P (X = r) = ^{n}C_{r} p^{r} q^{n – r} = ^{10}C_{r} \(\left(\frac{19}{10}\right)^r\left(\frac{1}{10}\right)^{10-r}\)

required probability = P (X ≤ 6) = 1 – P (X ≥ 7)

= 1 – [P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)]

= 1 – \(\sum_{r=7}^{10}{ }^{10} \mathrm{C}_r\left(\frac{9}{10}\right)^{\mathrm{r}}\left(\frac{1}{10}\right)^{10-r}\)

Question 35(Old).

Let X denote the num ber of colleges w here you w iUapply after your results and P (X = x) denotes your probability of getting admission in x number of colleges. It is given that

where A is a positive constant

(i) Find the value of k.

(ii) What is the probability that you will get admission in exactly two colleges ?

(iii) Find the mean and variance of the probability distribution.

Answer:

(i) Given P (X = x)

Since P (X = x) denotes your probability of getting admission in x number of colleges.

∴ Σp_{i} = 1

⇒ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒ k . 0 + k . 1 + 2k (2) + k (5 – 3) + k (5 – 4) = 1

⇒ 0 + k + 4k + 2k + k = 1

⇒ 8k = 1

⇒ k = \(\frac{1}{8}\)

(ii) required probability = P (X = 2) = 2k × 2 = 4k = \(\frac{4}{8}=\frac{1}{2}\)

(iii)

∴ Mean = μ = ΣX P (X)

= 0^{2} × 0 + 1 ^{2} × k + 2^{2} × 4k + 3^{2} × 2k + 4^{2} × k – \(\)

= k + 8k + 6k + 4k – \(\left(\frac{19}{8}\right)^2\)

= 51k – \(\frac{361}{64}=\frac{51}{8}-\frac{361}{64}\)

= \(\frac{408-361}{54}=\frac{47}{64}\)

Question 36.

If 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random

(i) 1

(ii) 0

(iii) less than 2 bolts will be defective.

Answer:

Let p probability of getting a detective bolt produced by machine

= 20% = \(\frac{20}{100}=\frac{1}{5}\)

q = 1 – p = 1 – \(\frac{1}{5}=\frac{4}{5}\) and here n = 4

Then by binomial distribution, we have

Question 37.

A die is thrown 6 times. What is the probability that there will be (i) no ace (ii) not more than one ace (iii) not more than 4 aces ? (Note that ace means a number 1, or one dot, on , the upper face of the die.)

Answer:

Here p = probability of getting ace in single throw of dice = \(\frac{1}{6}\)

q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\) and here n = 6

Since all the outcomes are equally likely. So it is a problem of binomial distribution.

Then by binomial distribution, we have

P(X = r) = ^{n}C_{r} p^{r} q^{n – r} = ^{6}C_{r}

(i) Prob. of getting no ace = P (X = 0)

= ^{6}C_{0}\(\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^6=\left(\frac{5}{6}\right)^6\)

(ii) Probability of getting not more than one ace

= P (X < 1) = P (X = 0) + P (X = 1)

= ^{6}C_{0}\(\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^6\) + ^{6}C_{1}\(\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^5\)

= \(\left(\frac{5}{6}\right)^5\left[\frac{5}{6}+\frac{6}{6}\right]=\frac{11}{6}\left(\frac{5}{6}\right)^5\)

(iii) Probability of getting not more than 4 aces

= P (X < 4) = 1 – P (X > 4) = 1 – P (X = 5) – P (X = 6)

= 1 – ^{6}C_{5}\(\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)\) – ^{6}C_{5}\(\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^0\)

= 1 – \(\frac{6 \times 5}{6^6}-\frac{1}{6^6}\)

= 1 – \(\frac{31}{6^6}\)

= 1 – \(\left(\frac{1}{6}\right)^6\)