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 ].

E(W)= 1.0909 ; E(W²)= 1.6362

Step-by-step explanation:

total number of guests = 5men + 6women =11

number if ways 2 guests are selected from 11 guests = ¹¹C₂ = 55

Case 1: When no women is selected

P(W=0) = (⁶C₀ . ⁵C₂)/55  = 0.1818

W*PW  = 0

W²*PW = 0

Case 2: When 1 women is selected

P(W=1) = (⁶C₁ . ⁵C₁)/55    = 0.5455

W*PW  = 0.5455

W²*PW = 0.5454

Case 3: When 2 women are selected

P(W=2) = (⁶C₂ . ⁵C₀)/55  = 0.2727

W*PW  = 0.5454

W²*PW = 1.0908

∑W*PW   = 1.0909 = E(W)

∑W²*PW  = 1.6362 = E(W²)

E(W)= ∑WP(W)=1.0909

E(W^{2})= ∑W^{2}P(W)=1.6363

Step-by-step explanation:

Here W can take value 0, 1, 2. There are total 5+6= 11 guests in the party. So total number of ways of selecting 2 guests out of 11 is

\binom{11}{2}=55

For W = 0 means both selected guests are men. So

P(W=0)=\frac{\binom{5}{2}}{55}=\frac{10}{55}=0.1818

For W = 1 means one man and one woman selected. So

P(W=1)=\frac{\binom{5}{1}\binom{6}{1}}{55}=\frac{30}{55}=0.5455

For W = 2 means two women selected. So

P(W=2)=\frac{\binom{6}{2}}{55}=\frac{15}{55}=0.2727

Following table shows the calcualtion for E(W) and E(W^2):

WP(W)W*P(W)W^2*P(W)

00.181800

10.54550.54550.5455

20.27270.54541.0908

Total11.09091.6363

So

E(W)= ∑WP(W)=1.0909

E(W^{2})= ∑W^{2}P(W)=1.6363