
Combinations Problem.
PLEASE HELP
Jim pitches for baseball. He can either:
a. give up no runs.
b. give up 1 or 2 runs.
c. give up 3 or 4 runs.
d. give up at least 5 runs.
1. Jim makes 9 starts. In how many ways can the number of runs be distributed? (order of starts is irrelevant)
I got (9+41 = 12) $\displaystyle _{12}C_{9}$ or $\displaystyle \frac{12!}{9!3!}$
2. Assuming each outcome for the season is equally likely, what is the probability that Jim pitches at least 3 shutouts? (Shutout = Jim gives up no runs)
No idea how to go about doing this one. Help would really be appreciated.