A committee of 4 students is selected at random from a group consisting of 7 boys and 4 girls. Find the probability that there are exactly 2 boys in the committee, given that at least one girl must be there in the committee.

OR

A random variable X has the following probability distribution:

Find the value of C and also calculate mean of the distribution.

Given: In a group there are 7 boys and 4 girls

Let A denote the event that exactly 2 boys will be chosen, and B the event that at least one girl will be chosen.

We require P (A | B).

When a committee of 4 student is selected exactly two boys, it means rest two selection only for girls.

So, the number of way of selection of boys = ⁷C₂

No. of ways of selection of girls = ⁴C₂

So, favourable outcomes = ⁷C₂ × ⁴C₂

Total outcomes = ¹¹C₄

Hence, P(A ∩ B) =

Now, we use the combination formula .i.e.

When at least one girl is selected in committee

There are four cases possible,

case 1:- committee selects 1 girl, 3 boys

number of ways = ⁴C₁ × ⁷C₃

= 4 × 35

= 140

Case 2 :- committee selects 2 girls, 2 boys

Number of ways = ⁴C₂ × ⁷C₂

= 6 × 21

= 126

case 3:- committee selects 3 girls, 1 boy

Number of ways = ⁴C₃ × ⁷C₁

= 4 × 7

= 28

Case 4 :- committee selects, 4 girls, 0 boy

number of ways = ⁴C₄ × ⁷C₀

= 1 [∵ 0! = 1]

Total outcomes = ¹¹C₄

= 330

Hence, P(B)

OR

We know that,

Sum of the probabilities = 1

⇒ C + 2C + 2C + 3C + C² + 2C² + 7C² + C = 1

⇒ 9C + 10C² = 1

⇒ 10C² + 9C – 1 = 0

⇒ 10C² + 10C – C – 1

⇒ 10C (C + 1) – 1(C + 1) = 0

⇒ (10C – 1)(C + 1)

Now, we have to find: E(X)

We know that, μ = E(X)

or

E(X) = ΣXP(X)

= 0 × C + 1 × 2C + 2 × 2C + 3 × 3C + 4 × C² + 5 × 2C² + 6 × (7C² + C)

= 56C² + 21C

= 0.56 + 2.1

= 2.66