Question

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?

Answer 1

@jim_thompson5910

Answer 2

this is another geometric distribution problem

Answer 3

any ideas here?

Answer 4

hmm not really, sorry

Answer 5

we're using the same formula

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

Answer 6

but now for this problem, the value of p is 0.54

so p = 0.54

for each part a) through c)

Answer 7

in part a), the value of k is k = 4

Answer 8

ok

Answer 9

how do i set it up

Answer 10

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


Pr(X = k) = (1-0.54)^(k-1)*0.54


Pr(X = 4) = (1-0.54)^(4-1)*0.54


Pr(X = 4) = ???

Answer 11

my calculator hasnt been wokring right on these...can u help out plz?

Answer 12

use google as a calculator then

http://www.google.com/search?hl=&q=%281-0.54%29^%284-1%29*0.54&sourceid=navclient-ff&rlz=1B3GGLL_enUS420US420&ie=UTF-8

Answer 13

0.05256144

Answer 14

yep

Answer 15

that's the probability that the quarterback throws three passes before he gets his first completion (on his fourth attempt)

Answer 16

kk! so, thats a? now b

Answer 17

part b

evaluate 1/p = 1/0.54 = ???

and round up

Answer 18

2

Answer 19

2 passes

Answer 20

good

Answer 21

so on average, you'd expect him to make two attempts til he gets a completion

Answer 22

ok! c

Answer 23

?

Answer 24

one sec

Answer 25

k

Answer 26

looking for a calculator that will speed things up (because it's going to a pain to compute the CDF here)

Answer 27

ok

Answer 28

btw, which statements below are true? I had only III but im not sure:

I. The expected value of a geometric random variable is determined by the formula np.
II. If X is a geometric random variable and the probability of success is .85, then the probability distribution of X will be skewed left, since .85 is closer to 1 than 0.
III. An important difference between binomial and geometric random variables is that there are a fixed number of trials in a binomial setting, and the number of trials varies in a geometric setting.

Answer 29

hmm can't find one, we'll just have to do it this way

Answer 30

ok

Answer 31

when k = 1, P(X = k) = 0.54
when k = 2, P(X = k) = 0.2484
when k = 3, P(X = k) = 0.114264
when k = 4, P(X = k) = 0.05256144
when k = 5, P(X = k) = 0.0241782624



these probabilities add to

0.54+0.2484+0.114264+0.05256144+0.0241782624 = 0.9794037024


So basically P(X <= 5) = 0.9794037024


To find P(X > 5), just subtract it from 1 to get

1 - 0.9794037024 = 0.0205962976

So the probability that he has to throw more than 5 passes to get his first completion is 0.0205962976

Answer 32

ok thanks i put P(X less than or equal to 5)
= 0.9794037024
P(X>5)
= 1 - 0.9794037024
= 0.0205963 <---

Answer 33

can u tell me what the answer is to the prob i commented up there? i got III only, but im not sure..

Answer 34

yeah that works, ok one sec

Answer 35

k

Answer 36

good, III looks like the only true statement

Answer 37

kk thanks so much @jim_thompson5910 !!

Answer 38

yw