12th NCERT Probability Exercise 13.2 Questions 17

Do or do not

There is no try

There is no try

Question (1)

If P(A) = 3/5 and P(B) = 1/5, find P(A ∩ B) if A and B are independent events.Solution

Since A and B are independent, \[P\left( {A \cap B} \right) = P\left( A \right) \cdot P\left( B \right)\] \[ = P\left( A \right) \cdot P\left( B \right)\] \[ = \frac{3}{5} \times \frac{1}{5}\] \[ = \frac{3}{{25}}\]Question (2)

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.Solution

Let A be the event first card drawn is black.\[P\left( A \right) = \frac{{26{C_1}}}{{52{C_1}}}\] \[ = \frac{{26}}{{52}} = \frac{1}{2}\] Let B be the event that second card drawn is also black given that first is also black.

The next card is drawn without replacement , so there are 51 cards in which 25 are black.

\[P\left( {B/A} \right) = \frac{{25{C_1}}}{{51{C_1}}}\] \[ = \frac{{25}}{{51}}\] The probability both the cards are black

\[P\left( {A \cap B} \right) = P\left( A \right)P\left( {B/A} \right)\] \[ = \frac{1}{2} \times \frac{{25}}{{51}}\] \[ = \frac{{25}}{{102}}\]

Question (3)

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.Solution

A box will be approved for sale if three selected oranges are good. Totally there are 15 oranges , we select 3 oranges one by one .Let A be the event that First orange selected is good.

P (A) = 12/15 = 4/5.

B be the event that second orange is good given that first is also good.

P(B/A) = 11/ 14.

Let C be the event that third orange is good given that first two are good.

P(C /(A ∪B))= 10/13.

So probability of box approved for sale

\[ = \frac{4}{5} \times \frac{{11}}{{14}} \times \frac{{10}}{{13}}\] \[ = \frac{{44}}{{91}}\]

Question (4)

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.Solution

A fair coin and an unbiased die are tossed..The number of elements of sample space n = 12.

Let A be the event head appers on the coin.

∴ P(A) = 6/12 = 1/2.

B is the event number 3 is on die.

∴ P(B) = 2/12 = 1/6.

A ∩ B = {H,3}.

∴ P(A ∩B) = 1/12.

\[P\left( A \right) \times P\left( B \right) = \frac{1}{2} \times \frac{1}{6}\] \[ = \frac{1}{{12}}\] \[ = P\left( {A \cap B} \right)\] A and B are independent events.

Question (5)

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?Solution

A die is thrown so sample space will contain 6 elements. n = 6Let A be the event that the number is even.

A = {2, 4, 6}, P(A) = 3/6 = 1/2.

Let B be the event the number is red.

B = { 1, 2, 3}, P(B) = 3/6 = 1/2.

A ∩ B = { 2}, P( A ∩ B) = 1/6

\[P\left( A \right) \times P\left( B \right) = \frac{1}{2} \times \frac{1}{2}\] \[ = \frac{1}{4}\] \[ \ne P\left( {A \cap B} \right)\] So A and B are not independent events.

Question (6)

Let E and F be events with P(E) = 3/5, P(F) = 3/10 and P(E ∩ F) = 1/5. Are E and F independent?Solution

\[P\left( E \right) \times P\left( F \right)\] \[ = \frac{3}{5} \times \frac{3}{{10}}\] \[\frac{9}{{50}}\] \[ \ne P\left( {E \cap F} \right)\] ∴ E and F are not independent events.Question (7)

Given that the events A and B are such that P(A) = 1/2, P(A ∪ B) = 3/5 and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.Solution

P(A) = 1/2, P(A ∪ B) = 3/5 and P(B) = p.(i) Mutually exclusive.

∴ P( A ∩ B) = 0.

\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] \[\frac{3}{5} = \frac{1}{2} + p - 0\] \[p = \frac{3}{5} - \frac{1}{2}\] \[p = \frac{{6 - 5}}{\begin{array}{l}10\\p = \frac{1}{{10}}\end{array}}\] (ii) A and B are independent.

P (A ∩ B ) = P(A) ×P(B)

= 1/2(p)= p/2

\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] \[\frac{3}{5} = \frac{1}{2} + p - \frac{p}{2}\] \[\frac{p}{2} = \frac{3}{5} - \frac{1}{2}\] \[\frac{p}{2} = \frac{{6 - 5}}{{10}}\] \[p = \frac{{1 \times 2}}{{10}} = \frac{1}{5}\]

Question (8)

. Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find (i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A|B) (iv) P(B|A)Solution

Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4.(i) \[P\left( {A \cap B} \right) = P\left( A \right) \cdot P\left( B \right)\] \[ = 0.3 \times 0.4\] \[ = 0.12\] (ii) \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] \[ = 0.3 + 0.4 - 0.12\] \[ = 0.58\] (iii) \[P\left( {A/B} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\] \[ = \frac{{0.12}}{{0.4}}\] \[ = 0.3\] (iv) \[P\left( {B/A} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\] \[ = \frac{{0.12}}{{0.3}}\] \[ = 0.4\]

Question (9)

If A and B are two events such that P(A) = 1/4, P(B) = 1/2 and P(A ∩ B) = 1/8, find P (not A and not B).Solution

If A and B are two events such that P(A) = 1/4, P(B) = 1/2 and P(A ∩ B) = 1/8, \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] \[ = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}\] \[ = \frac{{2 + 4 - 1}}{8}\] \[ = \frac{5}{8}\] \[{\rm{P (not A and not B) = P}}\left( {A' \cap B'} \right)\] \[P\left( {A \cup B} \right)'\] \[ = 1 - P\left( {A \cup B} \right)\] \[ = 1 - \frac{5}{8}\] \[ = \frac{3}{8}\]Question (10)

Events A and B are such that P(A) = 1/2, P(B) = 7/12 ,and P(not A or not B) = 1/4. State whether A and B are independent?Solution

Events A and B are such that P(A) = 1/2, P(B) = 7/12 ,and P(not A or not B) = 1/4. \[{\rm{P (not A or not B) = }}\frac{1}{4}\] \[{\rm{P}}\left( {A' \cup B'} \right) = \frac{1}{4}\] \[P\left( {A \cap B} \right)' = \frac{1}{4}\] \[1 - P\left( {A \cap B} \right) = \frac{1}{4}\] \[P\left( {A \cap B} \right) = 1 - \frac{1}{4}\] \[P\left( {A \cap B} \right) = \frac{3}{4}\] \[P\left( A \right) \cdot P\left( B \right)\] \[ = \frac{1}{2} \times \frac{7}{{12}}\] \[ = \frac{7}{{24}}\] \[ \ne P\left( {A \cap B} \right)\] ∴ A and B are not independent events.Question (11)

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find(i) P(A and B) (ii) P(A and not B) (iii) P(A or B) (iv) P(neither A nor B)

Solution

two independent events A and B such that P(A) = 0.3, P(B) = 0.6.(i) \[{\rm{P ( A and B) = P}}\left( {A \cap B} \right)\] \[{\rm{ = P}}\left( A \right){\rm{P}}\left( B \right)\] \[{\rm{ = 0}}{\rm{.3}} \times {\rm{0}}{\rm{.6}}\] \[{\rm{ = 0}}{\rm{.18}}\] (ii) \[{\rm{P(A and not B)}} = P\left( {A \cap B'} \right)\] \[ = P\left( A \right) - P\left( {A \cap B} \right)\] \[ = 0.3 - 0.18\] \[ = 0.12\] (iii) P(A or B) \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\] \[ = 0.3 + 0.6 - 0.18\] \[ = 0.72\] (iv) \[{\rm{P(neither A nor B) = P}}\left( {A' \cap B'} \right)\] \[ = P\left( {A \cup B} \right)'\] \[ = 1 - P\left( {A \cup B} \right)\] \[ = 1 - 0.72\] \[ = 0.28\]

Question (12)

A die is tossed thrice. Find the probability of getting an odd number at least once.Solution

Let A be the event getting odd number on the first die.P(A) = 1/2, P(A') = 1/2.Let B be event getting odd number on second die. P(B) = 1/2, P(B') = 1/2.

Let C be event getting odd number on third die. P(C) = 1/2, P(C') = 1/2.

Getting odd number on dies is independent events. So A, B and C are independent events.

P( getting an odd number atleastonce.) \[P\left( {A \cup B \cup C} \right) = 1 - P\left( {A \cup B \cup C} \right)'\] \[ = 1 - P\left( {A'} \right)P\left( {B'} \right)P\left( {C'} \right)\] \[ = 1 - \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\] \[ = 1 - \frac{1}{8}\] \[ = \frac{7}{8}\]

Question (13)

. Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red. (ii) first ball is black and second is red. (iii) one of them is black and other is red.Solution

Let A be the event that ball is red.P(A) = 8/18 = 4/9.

The second ball is drawn with replacement. Let B be the event that a ball selected is black.

P(B) = 10/18 = 5/9.

A and B are independent events.

(i) P( Both are red balls) = P(A)× P(A)

\[ = \frac{4}{9} \times \frac{4}{9}\] \[ = \frac{{16}}{{81}}\] (ii) P( first is black and second is red.) \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\] \[ = \frac{4}{9} \times \frac{5}{9}\] \[ = \frac{{20}}{{81}}\] (iii) P(One black and other red) \[P\left[ {\left( {A \cap B} \right) \cup \left( {B \cap A} \right)} \right]\] \[ = P\left( {A \cap B} \right) + P\left( {B \cap A} \right)\] \[ = 2P\left( A \right)P\left( B \right)\] \[ = 2 \times \frac{4}{9} \times \frac{5}{9}\] \[ = \frac{{40}}{{81}}\]

Question (14)

Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.Solution

The probability that A can solve the sum is P(A) = 1/2. P(A') = 1/2The probability that B can solve the sum , =P(B) = 1/3., P(B') = 2/3

A and B are solving it independently.

(i) the problem solved.

\[P\left( {A \cup B} \right)\] \[ = 1 - P\left( {A \cup B} \right)'\] \[ = 1 - P\left( {A' \cap B'} \right)By\,Demorgan's\,law\] \[ = 1 - P\left( {A'} \right)P\left( {B'} \right)\] \[ = 1 - \frac{1}{2} \times \frac{2}{3}\] \[ = 1 - \frac{1}{3}\] \[ = \frac{2}{3}\] (ii) Exactly one of them solve the sum.

\[P\left[ {\left( {A \cap B'} \right) \cup \left( {A' \cap B} \right)} \right]\] \[ = P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right)\] \[ = P\left( A \right)P\left( {B'} \right) + P\left( {A'} \right)P\left( B \right)\] \[ = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3}\] \[ = \frac{{2 + 1}}{6}\] \[ = \frac{3}{6} = \frac{1}{2}\]

Question (15)

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?(i) E: ‘the card drawn is a spade’

F: ‘the card drawn is an ace’

(ii) E: ‘the card drawn is black’

F: ‘the card drawn is a king’

(iii) E: ‘the card drawn is a king or queen’

F: ‘the card drawn is a queen or jack’

Solution

A card is drawn from a pack of cards., n = 52.(i) E be the event a card drawn is spade.

P(E) = 13/52 = 1/4

F : the card is an ace.

P(F) = 4/52 = 1/13

P(E ∩F ) = 1/52.

\[P\left( E \right)P\left( F \right)\] \[ = \frac{1}{4} \times \frac{1}{{13}}\] \[ = \frac{1}{{52}}\] \[ = P\left( {E \cap F} \right)\] So E and F are independent.

(ii) E be the event a card drawn is black.

P(E) = 26/52 = 1/2

F : the card is king.

P(F) = 4/52 = 1/13

P(E ∩F ) = 2/52 = 1/26.

\[P\left( E \right)P\left( F \right)\] \[ = \frac{1}{2} \times \frac{1}{{13}}\] \[ = \frac{1}{{26}}\] \[ = P\left( {E \cap F} \right)\] So E and F are independent.

(iii) E be the event a card drawn is king or queen.

P(E) = 8/52 = 2/13

F : the card is an queen or jack

P(F) = 8/52 = 2/13

P(E ∩F ) = 4/52 = 1/13.

\[P\left( E \right)P\left( F \right)\] \[ = \frac{2}{13} \times \frac{2}{{13}}\] \[ = \frac{4}{{169}}\] \[ \ne P\left( {E \cap F} \right)\] So E and F are not independent.

Question (16)

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.Solution

Let H be the event that student read Hindi news paper, P(H) = 0.6Let E be the event that student read English news paper, P(E) = 0.4

P( H ∩ E) = 0.2

(a) P( reads neither Hindi nor English newspapers)

\[ = P\left( {H' \cap E'} \right)\] \[ = P\left( {H \cup E} \right)'\] \[ = 1 - P\left( {H \cup E} \right)\] \[ = 1 - [P\left( H \right) + P\left( E \right) - P\left( {H \cap E} \right)]\] \[ = 1 - [0.6 + 0.4 - 0.2]\] \[ = 1 - 0.8\] \[ = 0.2\] (b) P(E/H) \[P\left( {E/H} \right) = \frac{{P\left( {E \cap H} \right)}}{{P\left( H \right)}}\] \[ = \frac{{0.2}}{{0.6}}\] \[ = \frac{1}{3}\] (c) P(H/E) \[P\left( {H/E} \right) = \frac{{P\left( {E \cap H} \right)}}{{P\left( E \right)}}\] \[ = \frac{{0.2}}{{0.4}}\] \[ = \frac{1}{2}\]

Question (17)

The probability of obtaining an even prime number on each die, when a pair of dice is rolled is(A) 0 (B) 1/3 (C) 1/12 (D) 1/36

Solution

When two dies are thrown then number of elements of sample space , n = 36.Let A be the event that even prime number on both dies.

A = {(2,2)}, m = 1

P(A) = m/n = 1/36.

So D is the correct answer.

Question (18)

Two events A and B will be independent, if(A) A and B are mutually exclusive (B) P(A'B') = [1 − P(A)][1 − P(B)] (C) P(A) = P(B) (D) P(A) + P(B) = 1

Solution

P(A'B') = [ 1 - P(A)] [1 - P(B)]P(A' ∩ B') = P(A') P(B')

⇒ A' and B' are independent events.

So A and B are independent events.

So B is the correct opition.