Consider a collection of 9 bears. There is a family of red bears consisting of one father bear, one mother bear, and one baby bear. There is a similar green bear family, and a similar blue bear family. We draw 5 consecutive times from this collection without replacement (i.e., not returning the bear after each draw). We keep track (in order) of the kind of bears that we get. Let X denote the number of red bears selected. For i=1,2,3,4,5, let Xi=1 if the ith bear selected is red, and Xi=0 otherwise. Refer to the red father bear as red bear #1, and the red mother bear as red bear #2, and the red baby bear as red bear #3. For i=1,2,3, let Yi=1 if the ith red bear is selected (at any time, i.e., on any of the five selections), and Yi=0 otherwise.
12a. Find E(X2) using the probability mass function of X.
12b. Find E(X2) in a different way, namely, using the fact that X=X1+?+X5 (the Xj's are dependent indicators). Expand (X1+?+X5)2 into 25 terms, where 20 of them will behave one way, and the other 5 will behave another way. You are expected to obtain the same answer as 12a.
12c. Find E(X2) in a different way, namely, using the fact that X=Y1+Y2+Y3 (the Yj's are dependent indicators). Expand (Y1+Y2+Y3)2 into 9 terms, where 6 of them will behave one way, and the other 3 will behave another way. You are expected to obtain the same answer as 12a.
12d. Find Var?(X) using your answer to E(X2) and your answer to E(X) from Problem 4 or Problem 8.
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