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
