Question

Suppose there are 5 men and 6 women at a party. The task of going to the store for more food and drinks is assigned to 2 party guests, chosen at random. Let W denote the number of women selected. Find E[W] and E[W2].

Answer 1

The value of E (W) = 1.0909 and the value of E (W²) = 1.6363.

Explanation:

The number of men and women at a party are 5 and 6 respectively.

The total number of ways to select 2 party guests is,

`{11\choose 2}=(11!)/(2!(11-2)!) =55` ways.

The two guests can be selected as follows:

S = {(M, M), (W, M) or (W, W)}

The probability of selecting 0 women:

`P(W=0)=\frac{{6\choose 0}{5\choose 2}}{{11\choose 2}}=(1*10)/(55)=0.1818`

The probability of selecting 1 women:

`P(W=1)=\frac{{6\choose 1}{5\choose 1}}{{11\choose 2}}=(6*5)/(55)=0.5455`

The probability of selecting 2 women:

`P(W=2)=\frac{{6\choose 2}{5\choose 0}}{{11\choose 2}}=(15*1)/(55)=0.2727`

Compute the expected value of the number of women selected as follows:

`E(W)=\sum wP(W=w)\\=[0* P(W=0)]+[1* P(W=1)]+[2* P(W=2)]\\=[0*0.1818]+[1*0.5455]+[2*0.2727]\\=1.0909`

The value of E (W²) is:

`E(W^(2))=\sum w^(2)P(W=w)\\=[0^(2)* P(W=0)]+[1^(2)* P(W=1)]+[2^(2)* P(W=2)]\\=[0*0.1818]+[1*0.5455]+[4*0.2727]\\=1.6363`

Thus, the value of E (W) = 1.0909 and the value of E (W²) = 1.6363.