Question

Suppose you make 90% of your free throws and you attempt 3 free throws. Use the binomial theorem to calculate each probability.
a) You do not make any of them.
b) You only make 1 of them.
c) You only make 2 of them.
d) You make all of them

I have the answers and all of my work, I just need it checked :)

Answer 1

\[_{3}C_{0}(0.90^0)(0.10^3)\] \[_{3}C_{0}=\frac{3!}{0!(3-0)!=1}\]

Answer 2

a) 0.001

Doing the same thing for the next three, I get:
b) 0.027
c) 0.243
d) 0.729

Answer 3

HI!!

Answer 4

none of them
\[.9^3\]

Answer 5

oops that was wrong !!

Answer 6

none of them
\[.1^3\]

Answer 7

one of them
\[3\times .1^2\times .9\]

Answer 8

two of them
\[3\times .1\times .9^2\]

Answer 9

three of them
\[.9^3\]

Answer 10

eh?

Answer 11

oh! okay lol

Answer 12

Thank you!