Question

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 :)

Answer 1

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

Answer 2

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

Answer 3

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