Probability Distribution Question

Question

On a football team there are 7 quarterbacks and 6 linemen. Coach selects 5 players.

a) Determine the probability that he selects only 1 quarterback

b) Determine the probability that he selects more quarterbacks than linemen.

c) What is the expected number of quarterbacks?

The total possible outcomes are 13 choose 5, so for a) I have [(7 choose 1)(6 choose 4)]/(13/5).

I'm not sure how to do b). And for c), do you multiply the number of trials, by the probability of success? I ended up getting 2.69.

Can someone help me with b)?

Answer 1

For b), you add up the probabilities of 3,4,5 quarterbacks (using the formula from a)), then you have all cases with more quarterbacks.

Answer 2

A)

$$\frac{\binom{7}{1}\cdot\binom{6}{4}}{\binom{13}{5}}$$

B)

$$\frac{\binom{7}{3}\cdot\binom{6}{2}+\binom{7}{4}\cdot\binom{6}{1}+\binom{7}{5}\cdot\binom{6}{0}}{\binom{13}{5}}$$

C)

$$\frac{1\cdot\binom{7}{1}\cdot\binom{6}{4}+2\cdot\binom{7}{2}\cdot\binom{6}{3}+3\cdot\binom{7}{3}\cdot\binom{6}{2}+4\cdot\binom{7}{4}\cdot\binom{6}{1}+5\cdot\binom{7}{5}\cdot\binom{6}{0}}{\binom{13}{5}}$$