Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen. What is E[G]

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Answer 1

This question is incomplete, the complete question is;

Two student council representatives are chosen at random from a group of 7 girls and 3 boys. Let G be the random variable denoting the number of girls chosen.

a) What is E[G] or range of G

b) Give the distribution over the random variable G

Answer:

a) E[G] is [ 0, 1 , 2 ]

b)

the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667

Step-by-step explanation:

Given the data in the question;

Number to be chosen is 2

from 7 girls and 3 boys

a) range of G

number of girls can be; [ 0, 1 , 2 ]

Therefore, E[G] is [ 0, 1 , 2 ]

b) he distribution over the random variable G.

Distribution of the random variable G is Hypergeometric;

so

with P( G=g ) = P( getting g girls from 7 and 2-g boys from 3)

= (([tex]^7C_g[/tex]) × ([tex]^3C_{2-g[/tex])) / [tex]^{10}C_2[/tex]

now since, the range of G is [ 0, 1 , 2 ]

P( G=0 ) = (([tex]^7C_0[/tex]) × ([tex]^3C_{2[/tex])) / [tex]^{10}C_2[/tex]

= [(7!/(0!(7-0)!)) × (3!/(2!(3-2)!))] / (10!/(2!(10-2)!))

= [1 × 3] / 45 = 3/45 = 0.0667

P( G=1 ) = (([tex]^7C_1[/tex]) × ([tex]^3C_{1[/tex])) / [tex]^{10}C_2[/tex]

= [(7!/(1!(7-1)!)) × (3!/(1!(3-1)!))] / (10!/(2!(10-2)!))

= [ 7 × 3 ] / 45 = 21/45 = 0.4667

P( G=2 ) = (([tex]^7C_2[/tex]) × ([tex]^3C_{0[/tex])) / [tex]^{10}C_2[/tex]

= [(7!/(2!(7-2)!)) × (3!/(0!(3-0)!))] / (10!/(2!(10-2)!))

= [ 21 × 1 ] / 45 = 21/45 = 0.4667

Therefore, the distribution over the random variable G.

x     P(x)

0     0.0667

1      0.4667

2     0.4667


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