Question

A team of 3 is to be chosen from 4girls and 6 boys in the team find the probability distribution distribution of x hence determine the mean and the variance of x

Answer 1

Let x will be the random variable “the number of girls in the team”.

Let y will be the random variable “the number of boy in the team”.

x=0,1,2,3

Total persons = 4+6 =10

Sample space

$S = \frac{10!}{3!(10-3)!}=120 \;ways$

x=0 (only boys will be in a team)

$n(x=0, y=3) = \frac{4!}{0!(4-0)!} \times \frac{6!}{3!(6-3)!} = 1 \times 20 =20 \;ways \\ P(X=0) = \frac{20}{120} = 0.1666$

x=1 (one girl and two boys will be in a team)

$n(x=1,y=2) = \frac{4!}{1!(4-1)!} \times \frac{6!}{2!(6-2)!} = 4 \times 15 =60 \;ways \\ P(x=1) = \frac{60}{120} = 0.5$

x=2 (two girls and one boy will be in a team)

$n(x=2,y=1) = \frac{4!}{2!(4-2)!} \times \frac{6!}{1!(6-1)!} = 6 \times 6 =36 \;ways \\ P(x=2) = \frac{36}{120} = 0.3$

x=3 (only three girls will be in a team)

$n(x=3,y=0) = \frac{4!}{3!(4-3)!} \times \frac{6!}{0!(6-0)!} = 4 \times 1 =4 \;ways \\ P(x=3) = \frac{4}{120} = 0.0333$

Probability distribution:

The mean

$\mu = \sum x \times P(x) \\ \mu = 0 \times 0.1666 + 1 \times 0.5 + 2 \times 0.3 + 3 \times 0.0333 = 1.1999 \\ \sum x^2 \times P(x) = 0^2 \times 0.1666+ 1^2 \times 0.5 + 2^2 \times 0.3 + 3^2 \times 0.0333 \\ = 0 + 0.5+ 1.2 + 0.2997 =1.9997$

The variance

$\sigma^2 = \sum x^2 \times P(x) -\mu^2 \\ = 1.9997 -(1.1999)^2 \\ = 1.9997 -1.4397 \\ =0.56$