RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.1 are part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.1.

## Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.1

Question 1.

If and P(A∩B) = , then find p().

Solution:

We know that

Question 2.

If P(B) = 0.5 and P(A ∩ B) = 0.32, then find p().

Solution:

Question 3.

If 2P(A) = P(B) = and P() = , then find P(A∪B).

Solution:

Question 4.

If P(A) = 0.6, P(B) = 0.3 and P(A ∩B) = 0.2, then find P() and P().

Solution:

Question 5.

If P(A) = 0.8, P(B) = 0.5 and P) = 0.4, then find:

(i) P(A ∩B)

(ii) P()

(iii) P(A ∪B)

Solution:

(iii) P(A ∪B) = P(A) + P(B) -P(A ∩B)

= 0.8 + 0.5 – 0.32

= 1.3 – 1.32

= 0.98

Question 6.

A family has two children. What is the probability that both the children are boys, given that at least one of them is a boy?

Solution:

Let A is event to be at least one boy in a family.

A = {BB, BG, GB}

and B is event to be all two boys.

B= {B, B}

∴ Sample space S = {BB, BG, GB, GG}

∴ Here A ∩ B= {B, B}

∴ n(A)= 3

∴ P(A) =

∵ n(B) = 1

∴ P(B) =

∵ n(A ∩ B) = 4

∴ P(A ∩B) =

Question 7.

Two coins are tossed once. Determine P()

where

(i) A : tail appears on one coin,

B : one coin shows head

(ii) A : no tail appears, B: no head appears

Solution:

(i) The possible outcomes when two coins are tossed once.

S = {HH, HT, TH, TT}

A = tail appears on one coin

= {TH, HT}

and B = One coin shows head

= {HT, TH}

∴ A ∩ B = {HT, TH}

∴ n(A ∩ B) = 2

n(S) = 4

A and B. Find p (A/B)

A : Son on one end B: Father in middle

Question 8.

Mother, father and son line up at random for a family picture. Events related to it are defined as events A and B. Find P() if

A : Son on one end

B: Father in middle

Solution:

Let mother (M), father (F) and son (S) are standing at random.

∴ Number of total ways to be standing all

= ⌊3

= 3 x 2 x 1 = 6

A : Son on one end

∴ A = {(SMF), (SFM), (FMS) (MFS)}

B: Father in middle

B = {(M, F, S), (S, F, M))

∴ A ∩B = {(M, F, S),(S, F, M)}

Question 9.

A fair die is rolled. Consider events A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}, find :

Solution:

(i) Total results on tossing die = 6

A = {1, 3, 5}, B = {2,3}

∴ A ∩ B = {3}

(ii) Given : A = {1, 3, 5}, B = {2, 3, 4, 5}

∴ A ∩ C = {3, 5}

(iii) Given A {1, 3, 5}, B = {2, 3}, C = {2, 3, 4, 5}

∴ A ∩ C = {3, 5}, B ∩C = {2, 3), A ∩ B = {3}

(A ∩B) ∩ C = {3}

Question 10.

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Solution:

Let E = Have different number

F = Sum of both number = 4

F = {(1, 3), (2, 2), (3, 1)}

Question 11.

Ten cards numbered 1 to 10 are placed in a box, mixed up throughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Solution:

Let A : Even number on card drawn and B : Number is greater than 3

We have to determine P(A/B) space

sample S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Then A = {, 4, 6, 8, 10},

B = {4, 5, 6, 7, 8, 9, 10}

and A ∩ B = {4, 6, 8, 10}

Question 12.

In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl.

Solution:

Let A : student chosen randomly studies in class XII and B : The choosen student is girl.

Here P(A/B) is to be find.

Now P(B)= = 0.43

Number of girls studing class XII are 10%.

Question 13.

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once ?

Solution:

Let A : 4 has appeared at least once.

B : Sum of the numbers is observed to be 6.

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

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

We know that P(A) = (∵Total favourable results are 6 x 6 = 36)

Question 14.

Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail?

Solution:

The results of the experiment are expressed by following diagram :

Sample space of experiment is :

S = {(H, H), (H, 7), (T, 1), (1, 2), (1, 3), (T,4), (7,5), (T, 6)}

where (HH) shows that head appears on both throws and (T, 1) shows that Tail on coin and I on die. So the probabilities of 8 fundamental events (H, H), (H, T), (T, 1), (T, 2), (1, 3), (T, 4), (7,5), (T, 6) are

Let A : Number appears more than 4 on die

B: Minimum one tail appears

Then B {(H. T).(T, 1),(T,2),(T,3),(T,4).(T5),(T,6)}

∴ P(B)=P[{(H, T)}] +P[(T. 1)}] + P[(1,2)) + P[(T, 3)] + P[(T, 4)] + P[{T. 5}] + P[{(T, 6)}]

##### RBSE Solutions for Class 12 Maths

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