NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2

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BoardCBSE
TextbookNCERT
ClassClass 12
SubjectMaths
ChapterChapter 13
Chapter NameProbability
ExerciseEx 13.2
Number of Questions Solved18
CategoryNCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2

Ex 13.2 Class 12 Maths Question 1.

If P(A) = frac { 3 }{ 5 } and P(B) = frac { 1 }{ 5 }, find P(A∩B) if A and B are independent events.
Solution:
A and B are independent if P (A ∩ B)
=P(A)times P(B)=frac { 3 }{ 5 } times frac { 1 }{ 5 } =frac { 3 }{ 25 }

Ex 13.2 Class 12 Maths 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:
Number of exhaustive cases = 52
Number of black cards = 26
One black card may be drawn in 26 ways
∴ Probability of getting a black card,
NCERT Solutions for Class 12 Maths Chapter 13 Probability 2

Ex 13.2 Class 12 Maths 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:
S = {12 good oranges, 3 bad oranges),
n(S) = 15
P (a box is approved) = frac { C(12,3) }{ C(15,3) } =frac { 12times 11times 10 }{ 15times 14times 13 } =frac { 44 }{ 91 }

Ex 13.2 Class 12 Maths 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:
When a coin is thrown, head or tail will occur
Probability of getting head P(A) = frac { 1 }{ 2 }
When a die is tossed 1,2,3,4, 5, 6 one of them will appear
∴ Probabillity of getting 3 = P(B) = frac { 1 }{ 6 }
When a die and coin is tosses, total number of cases are
H1,H2,H3,H4,H5,H6
T1,T2,T3,T4,T5,T6
Head and 3 will occur only in 1 way
∴ Probability of getting head and 3=frac { 1 }{ 12 }
NCERT Solutions for Class 12 Maths Chapter 13 Probability 4

Ex 13.2 Class 12 Maths 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:
Even numbers on die are 2,4,6
∴ Probability of getting even number
P(A) = frac { 3 }{ 6 } = frac { 1 }{ 2 }
There are two colours of the die – red and green.
Probability of getting red colour, P (B) = frac { 1 }{ 2 }
Even number in red colour is 2
∴ Probability of getting red colour and even number
vedantu class 12 maths Chapter 13 Probability 5

Ex 13.2 Class 12 Maths Question 6.
Let E and F be the events with P(E) = frac { 3 }{ 5 }, P (F) = frac { 3 }{ 10 } and P (E ∩ F) = frac { 1 }{ 5 }. Are E and F independent?
Solution:
P(E) = frac { 3 }{ 5 } ,P(F) = frac { 3 }{ 10 },
∴ P (E) x P (F) = frac { 3 }{ 5 } times frac { 3 }{ 10 } =frac { 9 }{ 50 }
P(E∩F)≠P(E)xP(F)
∴ The event A and B are not independent.

Ex 13.2 Class 12 Maths Question 7.
Given that the events A and B are such that P(A) = frac { 1 }{ 2 },P(A∪B) = frac { 3 }{ 5 } and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent.
Sol. Let P(A∩B) = x,Now P(A) = frac { 1 }{ 2 } ,P(A∪B) = frac { 3 }{ 5 }, P(B) = P
P(A∪B) = P (A) + P (B) – P (A∩B)
NCERT Solutions for Class 12 Maths Chapter 13 Probability 7

Ex 13.2 Class 12 Maths Question 8.
Let A and B independent events 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:
P (A) = 0.3,
P (B) = 0.4
A and B are independent events
(i) ∴ P (A∩B) = P (A). P (B) = 0.3 x 0.4 = 0.12.
(ii) P(A∪B) = P(A) + P(B) – P(A).P(B)
= 0.3 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58.
vedantu class 12 maths Chapter 13 Probability 8

Ex 13.2 Class 12 Maths Question 9.
If A and B are two events, such that P (A) = frac { 1 }{ 4 }, P(B) = frac { 1 }{ 2 },and P(A∩B) = frac { 1 }{ 8 }.Find P (not A and not B)
Solution:
Event not A and not B = overline { A } cap overline { B }
P(overline { A } cap overline { B } )=1-P(Acup B)
NCERT Solutions for Class 12 Maths Chapter 13 Probability 9

Ex 13.2 Class 12 Maths Question 10
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 Bare independent
Solution:
P(overline { A } cup overline { B } )=1-P(Acap B)
Rightarrow frac { 1 }{ 4 } =1-P(Acap B)
NCERT Solutions for Class 12 Maths Chapter 13 Probability 10

Ex 13.2 Class 12 Maths 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:
(i) A and B are independent events
∴ P(A and B) = P(A∩B) = P(A) x P(B)
= 0.3 x 0.6 [ ∵P (A) 0.3), P (B) = 0.6]
∴ P(A and B) = 0.18
(ii) P(A and B) = and
NCERT Solutions for Class 12 Maths Chapter 13 Probability 11
NCERT Solutions for Class 12 Maths Chapter 13 Probability 11.1

Ex 13.2 Class 12 Maths Question 12
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
S = {1,2,3,4,5,6},n(S) = 6
Let A represents an odd number.
NCERT Solutions for Class 12 Maths Chapter 13 Probability 12

Ex 13.2 Class 12 Maths 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:
S = {10 black balls, 8 red balls}, n (S) = 18
Let drawing of a red ball be a success.
A = {8 red balls}, n (A) = 8
vedantu class 12 maths Chapter 13 Probability 13

Ex 13.2 Class 12 Maths Question 14
Probability of solving specific problem independently by A and B are frac { 1 }{ 2 } and frac { 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:
Probability that A solves the problem = frac { 1 }{ 2 }
=> Probablility that A does not solve the problem
P (A) = 1-frac { 1 }{ 2 } =frac { 1 }{ 2 }
Probability that B solves the problem = frac { 1 }{ 3 }
=> Probability that B does not solved the
NCERT Solutions for Class 12 Maths Chapter 13 Probability 14

Ex 13.2 Class 12 Maths 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:
n(S) = 52
(i) E = {13 spades},P(E) = frac { 13 }{ 52 } = frac { 1 }{ 4 }
F = {4 aces}, P(F) = frac { 4 }{ 52 } = frac { 1 }{ 13 }
NCERT Solutions for Class 12 Maths Chapter 13 Probability 15
NCERT Solutions for Class 12 Maths Chapter 13 Probability 15.1

Ex 13.2 Class 12 Maths 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 probabililty that she reads english newspapers.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
(a) Let H and E represent the event that a student reads Hindi and English newspaper respectively
P (H) = 0.6, P (E) = 0.4, P (H ∩ E) = 0.2
Probability that the student reads at least one paper
vedantu class 12 maths Chapter 13 Probability 16

Choose the correct answer in the following Question 17 and 18:

Ex 13.2 Class 12 Maths Question 17
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
(a) 0
(b) frac { 1 }{ 3 }
(c) frac { 1 }{ 12 }
(d) frac { 1 }{ 36 }
Solution:
(d) n(S) = 36
Let A represents an even prime number one each dice.
NCERT Solutions for Class 12 Maths Chapter 13 Probability 17

Ex 13.2 Class 12 Maths Question 18
Two events A and B are said to 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:
(b) P(A’and B’) = [1 – P(A)]. [1 – P(B) = P(A). P(B’)
Thus option (b) is correct.

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