A basketball player has 2 free shots. Probability to take a point both times is 0.6, one time 0.2 and not to take a point 0.2. a) If he has 100 times 2 free shots, find a probability that he’ll take <170 points b) How many times should he do 2 free shots to take at least 100 points with probability 0.9? sorry for bad grammar :)

I don't have a nice way to do this.. but you could do the following using the multinomial dist we get the following let a=#2pointers b=#1pointers c=#0 points \[P(a,b,c)=\frac{100!}{a!b!c!}(.6)^a(.2)^b(.2^c)\] \[=\frac{100!}{a!b!(100-a-b)!}(.6)^a(.2)^b(.2^{(100-a-b)})\] now sum up over all the values of a and b such that 2a+b<170 I get .999958746

looks like 78 or more will give a probability of .90 or more actually ( .912967346753 or more)

I know to solve this now. If you want, I can write the solution here.

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