3 You are picking a baseball team of 9 randomly chosen playe

3. You are picking a baseball team of 9 randomly chosen players from a total of 2 good players, 8 average players, and 4 bad players.

a. What is the probability that all 4 bad players end up on your team?

b. What is the probability of picking a team consisting of only average and bad players?

c. If your team has at least 1 good player, your team will win by a margin of 1. Otherwise, your team loses by a margin of 2. What is the expected value and variance of your margin of victory (or loss)?

Solution

a)

There are 14C9 ways to choose any 9 players.

If 4 are bad players, then there are 10C5 ways to choose the other 5.

Thus,

P = (10C5)/(14C9) = 0.125874126 [ANSWER]

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b)

There are 12C9 ways to choose only average and bad players. Hence,

P = (12C9)/(14C9) = 0.10989011 [ANSWER]

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c)

Note that

P(at least 1 good) = 1 - P(only average and bad) = 1 - 0.10989011 = 0.89010989

Consider:

Thus,  
  
E(x) = Expected value = mean =    0.67032967 [ANSWER]
Var(x) = E(x^2) - E(x)^2 =    0.880328464 [ANSWER]