Practicing ML Aggarwal Class 12 Solutions Chapter 10 Probability Ex 10.1 is the ultimate need for students who intend to score good marks in examinations.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.1

Question 1.

Given P (A) = \(\frac{3}{5}\) and P (B) = \(\frac{1}{5}\) find P(A or B), given that A and B are mutually exclusive events. (NCERT)

Answer:

Given P (A) = \(\frac{3}{5}\) and P (B) = \(\frac{1}{5}\)

We know that P (A or B) = P (A ∪ B)

= P (A) + P (B) – P (A ∩ B) …(1)

Since A and B are mutually exclusive events

∴ A ∩ B = Φ ∴ P(A ∩ B) = 0

∴ from (1); P(A ∪ B) = P (A) + P (B)

= \(\frac{3}{5}+\frac{1}{5}=\frac{4}{5}\)

Question 2.

The probability of an event A occurring is 0.5 and of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.

Answer:

Given P (A) = 0.5 ; P (B) = 0.3

Since A and B are mutually exclusive A ∩ B = Φ

⇒ P (A ∩ B) = 0

∴ P (A’ ∩ B’) = P ((A ∪ B)’)

= 1 – P (A ∪ B)

= 1 – [P(A) + P(B) – P(A ∩ B)]

= 1 – [0.5 + 0.3 – 0] = 1 – 0.8

= 0.2

Question 3.

If A and B are mutually exclusive events, P (A) = 0.35 and P (B) = 0.45, then find

(i) P (A’)

(ii) P (B’)

(iii) P(A ∪ B)

(iv) P(A ∩ B)

(v) P (A ∩ B’)

(vi) P (A’ ∩ B’) (NCERT Exemplar)

Answer:

Given P (A) = 0.35 and P (B) = 0.45

Since A and B are mutually exclusive events

A ∩ B = Φ; P (A ∩ B) = 0

(i) P (A’) = 1 – P (A)

= 1 – 0.35

= 0.65

(ii) P (B’) = 1 – P (B)

= 1 – 0.45

= 0.55

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.35 + 0.45 – 0 = 0.8

(iv) P (A ∩ B) = 0

(v) P (A ∩ B’) = P (A) – P(A ∩ B)

= 0.35 – 0

= 0.35

(vi) P(A’ ∩ B’) = P((A ∪ B)’)

= 1 – P (A ∪ B)

= 1 – {P (A) + P (B) – P (A ∩ B)}

= 1 – {0.35 + 0.45 – 0}

= 1 – 0.8

= 0.2

Question 4.

If E and Fare events such that P(E) = \(\frac{1}{4}\) P(F) = \(\frac{1}{2}\) and P (E and F) = \(\frac{1}{8}\) find

(i) P(E or F)

(ii) P (not E and not F). (NCERT)

Answer:

Given P (E) = \(\frac{1}{4}\), P (F) = \(\frac{1}{2}\);

P (E and F) = P (E ∩ F) = \(\frac{1}{8}\)

(i) P (E or F) = P (E ∪ F)

= P (E) + P (F) – P (E ∩ F)

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

= \(\frac{5}{8}\)

(ii) P (not E and not F) = P (E’ ∩ F’)

= P{(E ∪ F)’}

= 1 – P(E ∪ F)

= 1 – {P (E) + P (F) – P (E ∩ F)}

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

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

= 1 – \(\frac{5}{8}=\frac{3}{8}\)

Question 5.

A and B are events such that P(A) = 0.42, P (B) = 0.48 and P (A and B) = 0.16. Determine

(i) P (not A)

(ii) P (not B)

(iii) P (A or B). (NCERT)

Answer:

Given P (A) = 0.42 ; P (B) = 0.48 ;

P (A and B) = 0.16

(i) P (not A) = P (A’) = 1 – P (A)

= 1 – 0.42

= 0.58

(ii) P (not B) = P (B’) = 1 – P (B)

= 1 – 0.48

= 0.52

(iii) P (A or B) = P (A ∪ B)

= P (A) + P (B) – P (A ∩ B)

= 0.42 + 0.48 – 0.16

= 0.9 – 0.16

= 0.74

Question 6.

A and B are two mutually exclusive events of an experiment. If P (not A) = 0.65, P (A ∪ B) = 0.65 and P (B) = p, find the value of p.

Answer:

Given A and B are mutually exclusive events

∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0

Also, P (not A) = 0.65

⇒ P (A’) = 0.65

⇒ 1 – P (A) = 0.65

P (A) = 1 – 0.65 = 0.35

given P(A ∪ B)= 0.65 ; P (B) = p

We know that,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

⇒ 0.65 = 0.35 + p – 0

⇒ p = 0.65 – 0.35

= 0.30

Question 7.

E and F are two events associated with a random experiment for which P(E) = 0.60, P (E or F) = 0.85, P (E and F) = 0.42. Find P (F).

Answer:

Given P (E) = 0.60 ;

P (E or F) = P (E ∪ F) = 0.85 ;

P (E and F) = P (E ∩ F) = 0.42

We know that

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

⇒ 0.85 = 0.60 + P(F) – 0.42

⇒ P (F) = 0.85 + 0.42 – 0.60

= 0.67

Question 8.

If A, B and C are mutually exclusive and exhaustive events and it is known that P (A ∪ B) = 0.63, calculate P (C).

Answer:

Since A, B and C are mutually exclusive and exhaustive events.

A ∩ B = B ∩ C = C ∩ A = Φ = A ∩ B ∩C and P(A ∪ B ∪ C) = 1

We know that,

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P (B ∩ C) – P (C ∩ A) + P (A ∩ B ∩ C) …(1)

Since P (A ∩ B) = 0 = P (B ∩ C)

= P(C ∩ A) = P(A ∩ B ∩ C)

from (1); 1 = P (A) + P (B) + P (C) …(2)

given P (A ∪ B) = 0.63

⇒ 0.63 = P (A) + P (B) – P (A ∩ B)

P (A) + P (B) = 0.63

∴ from (2); P (C) = 1 – 0.63 = 0.37

Question 9.

A, B, C are three mutually exclusive and exhaustive events associated with a random experiment. Find P (A), it being given that

P(B) = \(\frac{3}{2}\) P(A)and P (C) = \(\frac{1}{2}\)P(B).

Answer:

Since A, B, C are mutually exclusive and exhaustive events

P (A) + P (B) + P (C) = 1 …(1)

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

P(C) = \(\frac{1}{2}\) P(B) = \(\frac{3}{4}\)P(A)

from (1); we get

P (A) + \(\frac{3}{2}\)P(A) + \(\frac{3}{4}\)p(A) = 1

⇒ P (A)[1 + \(\frac{3}{2}+\frac{3}{4}\)] = 1

⇒ P(A) . \(\left(\frac{4+6+3}{4}\right)\) = 1

⇒ P(A) = \(\frac{4}{13}\)

Question 10.

A die is thrown once. Let A be the event that the number obtained is greater than 3 and B be the event that the number obtained is less than 5, then find P(A ∪ B).

Answer:

When a die is thrown once

Total number of possible outcomes = 6, {1,2, 3, 4, 5, 6}

Given A: event that the number obtained is greater than 3

∴ A = {4, 5, 6}

and B : event that number obtained is less than 5

∴ B = {1, 2, 3, 4}

∴ A ∪ B = {1, 2, 3, 4, 5, 6}

Hence required probability = P(A ∪ B)

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

Question 11.

A and B are two candidates seeking admission in an engineering college. The probability that A is selected is 0.5 and the probability that both are selected is atmost 0.3. Is it possible that the probability of B getting selected is 0.7 ? (NCERT Exemplar)

Answer:

Given probability of selection of A = P (A) = 0.5

Since P (A ∪ B) ≤ 1

⇒ P (A) + P (B) – P (A ∩ B) ≤ 1

⇒ 0.5 + P (B) – P (A ∩ B) ≤ 1

⇒ P(B) ≤ 1 – 0.5 + P(A ∩ B)

⇒ P(B) ≤ 0.5 + P(A ∩ B) …(1)

given P(A ∩ B) ≤ 0.3

∴ from (1); P(B) ≤ 0.5 + 0.3

= 0.8

Thus, the probability of B getting selected is 0.7 is logically valid.

Question 12.

The probability of an event A occurring is \(\frac{2}{3}\) and the probability of event B not occurring is \(\frac{5}{9}\). If the probability of getting success in atleast one of the two events is \(\frac{4}{5}\) what is the probability of success in both the events ?

Answer:

Given P (A) = \(\frac{2}{3}\); P (B’) = \(\frac{5}{9}\)

∴ P(B) = 1 – P(B’) = 1 – \(\frac{5}{9}=\frac{4}{9}\)

also P (atleast one of two events) = \(\frac{4}{5}\)

⇒ P(A ∪ B) = \(\frac{4}{5}\)

∴ required probability in success of both events

= P (A ∩ B)

= P(A) + P(B) – P(A ∪ B)

= \(\frac{2}{3}+\frac{4}{9}-\frac{4}{5}=\frac{90+60-108}{135}\)

= \(\frac{42}{135}=\frac{14}{45}\)

Question 13.

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 multiple of 3 or 7? (ISC 2005)

Answer:

Total number of outcomes = 20 = n (S)

A : events that ticket has a number which is multiple of 3.

B : event that ticket has a number which is multiple of 7.

A = {3, 6, 9, 12, 15, 18} and B = {7, 14}

∴ A ∩ B = Φ; n (A) = 6 ; n (B) = 2

∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{20}=\frac{3}{10}\)

P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{2}{20}=\frac{1}{10}\)

∴ P (A ∩ B) = 0

We know that,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= \(\frac{3}{10}+\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

Question 14.

The probability that a student will pass the final examination in both English and Hindi is 0*5 and the probability of passing neither is 0-1. If the probability of passing the English examination is 0-75 what is the probability of passing the Hindi examination ?

Answer:

Let E: student will pass the final examination in English

H: student will pass the final examination in Hindi.

given, P (E) = 0-75 ; P (E ∩ H) = 0.5 and P (E’ ∩ H’) = P {(E ∪ H)’} = 0.1

⇒ 1 – P (E ∪ H) = 0.1

⇒ P(E ∪ H)= 1 – 0.1 =0.9

⇒ P (E) + P (H) – P (E ∩ H) = 0.9

⇒ 0.75+ P(H ) – 0.5 = 0.9

⇒ P(H) = 0.9 + 0.5 – 0.75

= 0.65

Thus, the required probability of passing the Hindi Examination = 0.65.

Question 15.

In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random find the probability that

(i) the student opted for NCC or NSS.

(ii) the student has opted neither NCC nor NSS.

(iii) the student has opted NSS but not NCC.

Answer:

Let A : event that a student opted for NCC

B : event that a student opted for NSS.

∴ n (A) = 30 ; n (B) = 32 ; n (A ∩ B) = 24 ; n (S) = 60

Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{30}{60}=\frac{1}{2}\)

P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{32}{60}=\frac{8}{15}\)

P (A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{24}{60}=\frac{2}{5}\)

(i) required probability that the student opted for NCC or NSS

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

= \(\frac{1}{2}+\frac{8}{15}-\frac{2}{5}=\frac{15+16-12}{30}=\frac{19}{30}\)

(ii) required probability that the student has opted neither NCC nor NSS = P(A’ ∩ B’) = P((A ∪ B)’)

= 1 – P (A ∪ B)

= 1 – [P (A) + P (B) – P (A ∩ B)]

= 1 – \(\left[\frac{1}{2}+\frac{8}{15}-\frac{2}{5}\right]\)

= 1 – \(\frac{19}{30}=\frac{11}{30}\)

(iii) required probability that the student has opted NSS but not NCC

= P(B ∩ A’) = P(A’ ∩ B)

= P(B) – P(A ∩ B) = \(\frac{8}{15}-\frac{2}{5}=\frac{8-6}{15}=\frac{2}{15}\)

Question 16.

A bag contains 150 nuts and 50 bolts. Half of the bolts and half of the nuts are rusted. One item is drawn at random from the bag, find the probability that it is either rusted or a bolt.

Answer:

Given total no. of nuts in a bag = 150

and total no. of bolts in a bag = 50

Total no. of items in a bag = 150 + 50

= 200 = n (S)

No. of rusted items in a bag

= \(\frac{1}{2}\) × 150 + \(\frac{1}{2}\) × 50

= 75 + 25

= 100

Let A : event of drawing a rusted item

B : event of drawing a bolt

∴ n (A) = No. of rusted items = 100

∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{100}{200}\)

n(B) = No. of drawing a bolt = 50

∴ P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{50}{200}\)

A ∩ B : event of drawing a rusted bolt

∴ P(A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{25}{200}\)

Thus required prob. = P(A ∪ B)

= P(A) + P(B) – P(A ∩ B)

= \(\frac{100}{200}+\frac{50}{200}-\frac{25}{200}=\frac{125}{200}=\frac{5}{8}\)

Question 16 (odd).

For a post, three persons A, B and C appear in an interview. The probability of A being selected in twice that of B and the probability of B being selected is thrice that of C. If the post is filled, what are the probabilities of A, B and C being selected ?

Answer:

Given P (A) = 2 P (B) …(1)

and P (B) = 3 P (C)

∴ from(1); P(A) = 6P(C)

Since A, B and C are mutually exclusive and exhaustive events.

P (A) + P (B) + P (C) = 1

⇒ 6P (C) + 3P (C) + P (C) = 1

⇒ 10P (C) = 1

⇒ P(C) = \(\frac{1}{10}\)

∴ P (A) = 6 P (C) = \(\frac{6}{10}\)

and P(B) = 3P(C) = \(\frac{3}{10}\)

Question 17.

A card is drawn from a well shuffled pack of playing cards. What is the probability that it is either a spade or an ace or both ?

Answer:

Let A : event that the card drawn is spade B : event that the card drawn is an ace.

∴ required probability = P (A ∪ B)

= P (A) + P (B) – P (A ∩ B)

= \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}=\frac{4}{13}\)

Question 18.

A card is drawn at random from a pack of 52 playing cards. What is the probability that the card drawn is neither a spade nor a queen ? (ISC 2006)

Answer:

Let A : event of drawing a spade card

B : event of drawing a queen card

P (A) = \(\frac{13}{52}\) [since there are 13 spade cards]

and P (B) = \(\frac{4}{52}\) [since there are 4 queens]

P (A ∩ B) = P [drawing a queen of spade] = \(\frac{1}{52}\)

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}=\frac{16}{52}\)

Thus, required probability of drawing neither a spade nor a queen = P (A’ ∩ B’) = P((A ∪ B)’)

= 1 – P(A ∪ B) = 1 – \(\frac{16}{52}=\frac{36}{52}=\frac{9}{13}\)

Question 19.

A card is drawn at random from well shuffled pack of 52 playing cards. Find the probability that it is neither a king nor a red card.

Answer:

Let A : event of drawing a red card

B : event of drawing a king card

∴ P (A) = \(\frac{26}{52}=\frac{1}{2}\)

[since there are 26 red cards]

P(B) = \(\frac{4}{52}=\frac{1}{13}\) [since there are 4 kings]

∴ P (A ∩ B) = P (drawing a red king) = \(\frac{2}{52}\)

[one king of heart and other of diamond] We know that,

P(A ∪ B) = P (A) + P(B) – P(A ∩ B)

= \(\frac{26}{52}+\frac{4}{52}-\frac{2}{52}=\frac{28}{52}=\frac{7}{13}\)

Thus required probability of getting neither a king nor a red card = P (A’ n B’)

= P((A ∪ B)’) = 1 – P(A ∪ B)

= 1 – \(\frac{7}{13}=\frac{6}{13}\)

Question 20.

Find the probability of getting an odd number on the first die or a total of 8 in a single throw of two dice.

Answer:

In a single throw of two dice.

Total exhaustive cases = n (S) = 36

Let A : event of getting an odd number on first die

B : getting a total of 8.

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

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

Also A ∩ B = {(3, 5), (5, 3)}

∴ n (A) = 18 ; n (B) = 5 ; n (A ∩ B) = 2

Thus P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{18}{36}\)

P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{5}{36}\)

P(A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{2}{36}\)

We know that,

P(A ∪ B) = P (A) + P (B) – P (A ∩ B)

= \(\frac{18}{36}+\frac{5}{36}-\frac{2}{36}=\frac{21}{36}=\frac{7}{12}\)

Question 21.

If two dice are thrown simultaneously, find the probability of getting

(i) a sum of 7 or 11

(ii) a doublet or a total of 6.

Answer:

When two dice are thrown simultaneously Total number of outcomes = n (S) = 6^{2} = 36

(i) Let A : event of getting a sum of 7.

B : event of getting a total of 11.

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

B = {(5, 6), (6, 5)}

A ∩ B = Φ P (A ∩ B) = 0 ; n (A) = 6 ;

n (B) = 2

∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{36}=\frac{1}{6}\)

P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{2}{36}=\frac{1}{18}\)

We know that,

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= \(\frac{6}{36}+\frac{2}{36}=\frac{8}{36}=\frac{2}{9}\)

(ii) Let A : event of getting a doublet B : event of getting a total of 6.

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

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

A ∩ B = {(3, 2)} n (A) = 6 ;

n (B) = 5 ; n (A ∩ B) = 1

Thus P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{36}\)

P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{5}{36}\)

and P(A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{1}{36}\)

We know that,

P(A ∪ B) = P (A) + P (B) – P(A ∩ B)

= \(\frac{6}{36}+\frac{5}{36}-\frac{1}{36}=\frac{10}{36}=\frac{5}{18}\)

Question 22.

The probabilities that a student will get A, B, C or D grade are 0.4, 0.35, 0.15 and 0.1 respectively. Find the probability that she will get

(i) B or C grade

(ii) atmost C grade.

Answer:

Given P (A) = P (getting grade A) = 0.4

P (B) = P (getting grade B) = 0.35

P (C) = P (getting grade C) = 0.15

P (D) = P (getting grade D) = 0.1

(i) P (getting B or C grade) = P (B) + P (C)

= 0.35 + 0.15 = 0.5

[∵ B ∩ C = Φ ∴ P (B ∩ C) = 0]

(ii) P (getting atmost C grade)

= P (getting D grade) + P (getting C grade)

= 0.1 + 0.15 = 0.25

Question 23.

Two cards are drawn at random from a pack of 52 cards. What is probability that the cards are either both aces or both black cards ?

Answer:

Out of 52 cards, two cards can be drawn in ^{52}C_{2} ways.

Total no. of outcomes = ^{52}C_{2
}Let A : event of two black cards

B : event of drawing two aces

A ∩ B : event of drawing two black cards and both are aces.

∴ P(A) = \(\frac{{ }^{26} C_2}{{ }^{25} C_2}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)

[since there are 26 black cards and 2 black cards can be drawn in ^{26}C_{2} ways]

Question 24.

A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are defective. If a person takes out 2 fruits at random, find the probability that either both are apples or both are good.

Answer:

So total number of fruits in a basket

= 20 + 10 = 30

out of 30 fruits, 2 fruits can be drawn in ^{30}C_{2} ways

Total no. of outcomes = ^{30}C_{2
}Given Total no. of good apples = (20 – 5) = 15

and total no. of goods arranges = 10 – 3 = 7

and Total no. of good fruits = 15 + 7 = 22

So out of 22 good fruits, 2 good fruits can be drawn in ^{22}C_{2} ways.

Let A : event of drawing two apples

B : event of drawing two good fruits

A ∩ B : event of drawing two good apples.

∴ P (A) = \(\frac{{ }^{20} C_2}{{ }^{30} C_2}\) [since there are 20 apples]

P(B) = \(=\frac{{ }^{22} C_2}{{ }^{30} C_2}\) and P(A ∩ B) = \(\frac{{ }^{15} C_2}{{ }^{30} C_2}\)

[since there are 15 good apples]

required probability = P(A ∪ B)

= P (A) + P (B) – P (A ∩ B)