Question

Caleb makes 70% of his free throws and is shooting free throws after practice. What is the probability that

a) He makes the first free throw he shoots.
b) He does not make a free throw until his 3rd attempt.
c) He does not make a free throw until his 10th attempt.
d) How many free throws should Caleb expect to shoot until he makes his first shot?

Answer 1

@jim_thompson5910

Answer 2

@Frostbite @Hero

Answer 3

thats not the asnwer,i have to answer each part

Answer 4

this sounds like you'll be using the geometric distribution

do you remember how to use this distribution?

Answer 5

hi r u still here @jim_thompson5910

Answer 6

yes do you remember how to use the geometric distribution

Answer 7

hmm... idts

Answer 8

you would use the formula

Pr(X = k) = (1-p)^(k-1)*p

Answer 9

in each and every case for this problem, the value of p stays the same

p = 0.7

Answer 10

ok

Answer 11

how would i do this for a?

Answer 12

for part a), k = 1

for part b) k = 3

for part c) k = 10

part d) ... we'll get to this later

Answer 13

Pr(X = k) = (1-p)^(k-1)*p


Pr(X = k) = (1-0.7)^(k-1)*0.7


Pr(X = 1) = (1-0.7)^(1-1)*0.7


Pr(X = 1) = (0.3)^(0)*0.7


Pr(X = 1) = 1*0.7


Pr(X = 1) = 0.7

Answer 14

So this means that the chances of him making a free throw on the first attempt is 0.7 or 70%

this makes sense because it's given that on any random trial, he makes 70% of his free throws

Answer 15

Ok!

Answer 16

repeat this but instead of k = 1, use k = 3 for part b

Answer 17

So a is:
(a) Pr(X = k) = (1-p)^(k-1)*p
Pr(X = k) = (1-0.7)^(k-1)*0.7
Pr(X = 1) = (0.3)^(0)*0.7
Pr(X = 1) = 1*0.7
Pr(X = 1)
= 0.7 <--

Answer 18

yep

Answer 19

ok, so b

Answer 20

now you're using k = 3

Answer 21

can u write the formula again here?

Answer 22

Pr(X = k) = (1-p)^(k-1)*p


Pr(X = k) = (1-0.7)^(k-1)*0.7


Pr(X = 3) = (1-0.7)^(3-1)*0.7


Pr(X = 3) = ???

Answer 23

umm

Answer 24

its .3^(1,4)

Answer 25

0.1853??

Answer 26

wait no its .063!?

Answer 27

yep, it's 0.063

Answer 28

ok cool! c?

Answer 29

so he has a 6.3% chance of making his first shot on the third attempt

Answer 30

for part c, you use k = 10

Answer 31

Pr(X = k) = (1-p)^(k-1)*p


Pr(X = k) = (1-0.7)^(k-1)*0.7


Pr(X = 10) = (1-0.7)^(10-1)*0.7


Pr(X = 10) = ???

Answer 32

Pr(X=10) = (0.3)^ 0.63 right

Answer 33

no that doesn't look right, one sec

Answer 34

kkk

Answer 35

Pr(X = k) = (1-p)^(k-1)*p


Pr(X = k) = (1-0.7)^(k-1)*0.7


Pr(X = 10) = (1-0.7)^(10-1)*0.7


Pr(X = 10) = (0.3)^(9)*0.7


Pr(X = 10) = 0.000019683*0.7


Pr(X = 10) = 0.0000137781


So the probability that he makes his first shot on the 10th try is 0.0000137781

Answer 36

ok cool

Answer 37

for the last one, you just need to compute 1/p

Answer 38

how do i do that?

Answer 39

p = 0.7

so 1/p = 1/0.7 = ??

Answer 40

1.43

Answer 41

?

Answer 42

now round up to get 2

Answer 43

so he should expect to make 2 shots when the first one will go in (on average)

Answer 44

kk! thanks can u help me with one more prob?

Answer 45

A quarterback for the Seattle Seahawks completes 54% of his passes. Let the random variable X be the number of passes completed in 20 attempts.

a) What is the probability that the quarterback throws three passes before he gets his first completion (on his fourth attempt)?
b) How many passes should the quarterback expect to make before he gets his first completion?
c) What is the probability that he has to throw more than 5 passes to get his first completion?

(a) (b) (c) p= 54 percent = 0.54

Answer 46

so this sounds like another geometric distribution problem

Answer 47

any ideas?