Question

If Brian is a 46% free throw shooter, what is the probability of him making at least 3 free throws out of 6?

Answer 1

.46^3
+
.54^3 *.46^3
+
.54^2 * .46^3
+
.54 * .46^3
=
About 19.4%? Is that right?

Answer 2

Can you show me a different way to do it if I'm right?

Answer 3

at least 3 is annoying
3 or 4 or 5 or 6
or you could compute 0 or 1 or 2 and subtact the number from 1
your choice

Answer 4

you are forgetting to multiply by the number of ways to do it

Answer 5

\[\dbinom{n}{k}p^k(1-p)^{n-k}\] is what you need, and it looks like what you have except for the
\[\dbinom{n}{k}\] part

Answer 6

yup satellite is rite,
.46^3+.46^4+.46^5+.46^6

Answer 7

all 6 is
\[.46^6\]
5 out of six is
\[5\times .46^5\times .54\]

Answer 8

why 5 times that?

Answer 9

exactly 4 is
\[\dbinom{6}{4}\times .46^4\times .54^2\]

Answer 10

oh i get it 6 choose 5 = 5

Answer 11

six shots, you want five
there are bernoulli trials

yes six choose five

Answer 12

finally exactly 3 is
\[\dbinom{6}{3}\times.46^3\times .54^3\]
you good from there right? calculate and add

Answer 13

thanks satellite, so it was my equation except i have to multiply by the number of ways to do each term in that?

Answer 14

well it also looks like your exponents were a little messed up

Answer 15

they should add to six for one thing

Answer 16

okay, what i was thinking was that after missing 2 then making 3 it didnt matter what the next one was.

Answer 17

say i want exactly 2 out of 6
i think
\[.46^2\] two successes

\[.54^4\] 4 failures
\[\dbinom{6}{2}\] number of ways to get 2 successes out of 6
so answer would be
\[\dbinom{6}{2}\times .46^2\times .54^4\]

Answer 18

oh but it does matter, because if you are computing exactly 3 successes, another success would make it 4!

Answer 19

but it said at least 3

Answer 20

at least 3 means
3
or 4
or 5
or 6
compute each separately

Answer 21

okay, thanks satellite

Answer 22

yw