Probability Distribution and Expected Value of X

Question

The city council consists of 6 Democrats, 5 Republicans, and 2 Independents. A committee of 3 councilpersons is randomly selected (to study solid waste removal in St. Moritz, 6 days, 5 nights, all expenses paid).

Let the random variable X represent the number of Republicans on the committee. Write out the probability distribution and find the expected value of X.

Answer 1

The expected value can be determined by computing the probability of 0 R's, 1 R, 2 R's and 3 R's. Let's do 0 and 3 first since those 2 are the most straightforward.

There are 13 total candidates, and 3 total seats. After each of the 3 is selected, 1 fewer candidate remains,since you can't have the same person on more than 1 seat.

E(R = 0): 8/13 * 7/12 * 6/11 = .1958

E(R = 3): 5/13 * 4/12 * 3/11 = .0350

E(R = 1): This is slightly trickier, but we can use the binomial distribution to calculate. There are 3 C 1 ways of choosing 1 republican out of 3, which is equal to 3!/2!1!, which is equal to 3. Note that it does not matter what order the combination of 1 Republican and 2 Non-Republicans occurs.

E(R = 1): 3 * (5/13) * (8/12) * (7/11) = 3 * (.1632) = .4895

E(R= 2): Same principle as above. There are 3 ways of choosing 2 Rs out of 3 total seats, so:

E(R = 2): 3* (5/13) * (4/12) * (8/11) = 3 * (.0816) = .2797

Notice how these 4 different probabilities all sum to 1. So, the expected value of R's on the committee is:

1* .4895 + 2 * .2797 + 3 * .0395 + 0 * .1958 = 1.1674