Question

An urn contains 4 green balls, five blue balls, and seven red balls. You remove five balls at random without replacement. Let X be the random variable that counts the number of green balls in your sample. a) Find the probability mass function p(x) describing the distribution of X. b) Find the mean and variance of X

Answer 1

An urn contains 4 green balls,5 blue balls, and 7 red balls .

I remove 5 balls at random without replacement.

Let,X be the random variable that counts the number of green balls in the sample.

So,as there are 4 green balls in total,

so in a sample of 5,there can be 0,1,2,3 or 4 green balls.

so,X takes values 0,1,2,3 and 4.

Now,to find the probability of any of these,

all possible cases=(16 c 5)

P(X=0)=(12 c 5)/(16 c 5)=0.181

P(X=1)=(4 c 1)(12 c 4)/(16 c 5)=0.453

P(X=2)=(4 c 2)(12 c 3)/(16 c 5)=0.303

P(X=3)=(4 c 3)(12 c 2)/(16 c 5)=0.060

P(X=4)=(4 c 4)(12 c 1)/(16 c 5)=0.003

(a)

So,the probability mass function of X is

P(X=x)

=0.181 x=0

=0.453 x=1

=0.303 x=2

=0.060 x=3

=0.003 x=4

(b)

Mean of X

=(0*0.181)+(1*0.453)+(2*0.303)+(3*0.060)+(4*0.003)

=0.453+0.606+0.180+0.012

=1.251

Variance of X

=(0.181*1.251²)+(0.453*0.251²)+(0.303*0.749²)+(0.060*1.749²)+(0.003*2.749²)

=0.283+0.028+0.169+0.183+0.022

=0.685