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

Answer:

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

Step-by-step 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,

${11\choose 2}=\frac{11!}{2!(11-2)!} =55$ ways.

The two guests can be selected as follows:

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

The probability of selecting 0 women:

$P(W=0)=\frac{{6\choose 0}{5\choose 2}}{{11\choose 2}}=\frac{1\times10}{55}=0.1818$

The probability of selecting 1 women:

$P(W=1)=\frac{{6\choose 1}{5\choose 1}}{{11\choose 2}}=\frac{6\times5}{55}=0.5455$

The probability of selecting 2 women:

$P(W=2)=\frac{{6\choose 2}{5\choose 0}}{{11\choose 2}}=\frac{15\times1}{55}=0.2727$

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

$E(W)=\sum wP(W=w)\\=[0\times P(W=0)]+[1\times P(W=1)]+[2\times P(W=2)]\\=[0\times0.1818]+[1\times0.5455]+[2\times0.2727]\\=1.0909$

The value of E (W²) is:

$E(W^{2})=\sum w^{2}P(W=w)\\=[0^{2}\times P(W=0)]+[1^{2}\times P(W=1)]+[2^{2}\times P(W=2)]\\=[0\times0.1818]+[1\times0.5455]+[4\times0.2727]\\=1.6363$

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