These NCERT Solutions for Class 12 Maths Chapter 13 Probability Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Miscellaneous Exercise
Question 1.
A and B are two events such that P (A) ≠ 0. Find P(B|A), if
i. A is a subset of B
ii. A ∩ B = Φ
Solution:
Question 2.
A couple has two children.
Find the probability that both children are males, if it is known that atleast one of the children is male.
Solution:
The sample space, S = {MM, MF, FM, FF},
where M denote male and F denote female.
Let A: both children are males
B : atleast one child is a male
A = {MM},
B = {MM, MF, FM}
\(\mathrm{A} \cap \mathrm{B}=\{\mathrm{MM}\}, \mathrm{P}(\mathrm{A})=\frac{1}{4}\)
\(P(B)=\frac{3}{4}, P(A \cap B)=\frac{1}{4}\)
Question 3.
If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
Solution:
A leap year contains 366 days = 52 weeks + 2 days
The last 2 days can be
i. Monday, Tuesday
ii. Tuesday, Wednesday
iii. Wednesday, Thursday
iv. Thursday, Friday
v. Friday, Saturday
vi. Saturday, Sunday
vii. Sunday, Monday
Of these seven possibilities, (i) & (ii) are favourable to 53 Tuesdays.
∴ P(53 Tuesday) = \(\frac { 2 }{ 7 }\)
Question 4.
An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.
Solution:
Let p be the probability of a success and q the probability of failure.
Then p + q – 1 and p = 2q
Solving p = \(\frac { 2 }{ 3 }\) and q = \(\frac { 1 }{ 3 }\)
Let X be the number of success.
Then X is a binomial distribution with
Question 5.
How many times must a man toss a fair coin so that the probability of having atleast one head is more than 90%?
Solution:
Tossing a coin many times is a Bernoulli trial. Here success is obtaining a Head.
∴ P = \(\frac { 1 }{ 2 }\)
q = 1 – p = 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
Let X be the number of heads obtained
Then X is a binomial distribution B(n, \(\frac { 1 }{ 2 }\))
We know 21 = 2, 2² = 4, 2³ = 8, 24 = 16, 25 = 32 and so on
Hence the minimum value of n is 4 i.e. n > 4
i.e. the man has to toss the coin atleast 4 times.
Question 6.
In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six, Find the expected value of the amount he wins/loses.
Solution:
The game ends in the following ways.
i. The man gets 6 in 1st throw. In this case, he earns ₹ 1.
P(getting 6 ¡n 1st throw) = \(\frac { 1 }{ 6 }\)
ii. The man does not get 6 in 1nd throw and 6 in 2nd throw. In this case he earns ₹ 0
(In 1st throw, he earns ₹ 1 and in 2nd throw he loses ₹ 1)
P(not getting 6 on 1st throw & 6 in 2nd throw) = (\(\frac { 5 }{ 6 }\))(\(\frac { 1 }{ 6 }\)) = \(\frac { 5 }{ 6 }\)
Question 7.
Suppose we have four boxes A, B, C and D containing coloured marbles as given below:
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B?
Solution:
Let E1 : selecting box A
E2 : selecting box B
E3 : selecting box C
E4 : selecting box D
A : selecting a red ball
E1, E2, E3 and E4 are mutually exclusive and exhaustive events.
∴ \(\mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{P}\left(\mathrm{E}_{3}\right) \dot{\mathrm{P}}\left(\mathrm{E}_{4}\right)=\frac{1}{4}\)
\(\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\frac{1}{10}, \quad \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\frac{6}{10}\)
Question 8.
Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Solution:
E1 : a red ball is transferred from bag I to bag II
E2 : a black ball is transferred from bag I to bag II
A : a red ball is taken from bag II after transferring a ball
E1 and E2 are mutually exclusive and exhaustive events
Question 9.
If A 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
Solution:
Question 10.
If P(A|B) > P(A), then which of the following is correct:
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)
Solution:
Question 11.
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
Solution:
