Question

Suppose you make 90% of your free throws and you attempt 3 free throws. Use 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

Answer 1

C

Answer 2

How? @sangya21 Please explain.

Answer 3

It can't be C. It is four parts of one problem that I have to solve but I don't know how

Answer 4

I am sorry.I thought it was an option question
plus
90% of 3 = 2.7
as you are not getting 100% so you cant have all three

Answer 5

That's okay @sangya21 can you help me with the problem?

Answer 6

Yeah just need a little time

Answer 7

Binomial formula
\[P(k\space success\space for \space n \space trials) = \left(\begin{matrix}n \\ k\end{matrix}\right) *p^k*q^{n-k}\]

Answer 8

Can you walk me through it?

Answer 9

n = no. of trials
k = success
p = success probability = 0.9
q = failure = 0.1

1 n= 3; k = 0 ; p = 0.9 and q = 0.1
so P = ?

Answer 10

You just said p=0.9

Answer 11

its given that you find success 90% of time. so thats success probability

Answer 12

I need to find the probability of making 1, 2, and 3 of the free throws.

Answer 13

1.
\[ P\space = \left(\begin{matrix}3 \\ 0\end{matrix}\right) (0.9)^0 (0.1)^{3-0}\]
\[P\space= 3!/0!(3-0)! * 1* (0.1)^3\]
\[P\space= 3!/3! * 1* (0.001)\]
\[P\space= (0.001)\]

Answer 14

Are you sure that's what it is?

Answer 15

yes

Answer 16

for a part. when you have 0 success

Answer 17

What about b?

Answer 18

in b put k = 1

Answer 19

You got it?

Answer 20

so is it \[P=\left(\begin{matrix}3 \\ 1\end{matrix}\right)(0.9)^{1}(0.1)^{3-1}\]

Answer 21

Correct.

Answer 22

\[P=3!/1!(3-1)!*1*(0.1)^{3}\]

Answer 23

What is next @sangya21

Answer 24

0.9^1 = 0.9
so you have \[P \space = 3!/2! *0.9 *0.01\]

Answer 25

P= 0.027?

Answer 26

correct

Answer 27

And c is \[P=\left(\begin{matrix}3 \\ 2\end{matrix}\right)(0.9)^{2}(0.1)^{3-2}\]

Answer 28

\[P=3!/2!(3-2)!*1*(0.1)^{3}\]

Answer 29

\[P=3!/2!*0.9*0.001\]

Answer 30

@sangya21

Answer 31

\[P = 3!/2!*1! *(0.9)^2 *0.1\]
\[P = 3 *(0.9)^2 *0.1\]

Answer 32

\[P = 3*0.81 *0.1\]
\[P = 0.243\]

Answer 33

d part
\[P = (3!/3!*0!)*(0.9)^3 *(0.1)^0\]
\[P = 1* (0.9)^3 *1\]
because 3!/3! = 1 and (0.1)^0 = 1
\[P = 0.729\]