Question

A certain hockey player scores on 18% of his shots.

What is the probability he will score for the first time on his fifth shot?
What is the probability that it will take him fewer than 3 shots to score?
How many shots should he expect to take before scoring?

Answer 1

The probability that Sam will score above a 90 on a mathematics test is 4/5.
What is the probability that she will score above a 90 on exactly 3 of the 4 tests this quarter?
P(Sam will score >90 on exactly 3 of the 4 tests) =
P(Sam fails in 1st and succeeds in others : 2, 3, 4) +
P(Sam fails in 2nd and succeeds in others : 1, 3: 4) +
P(Sam fails in 3rd and succeeds in others : 1, 2, 4) +
P(Sam fails in 4th and succeeds in others : 1, 2, 3) +
each term has the same value :
P(Sam will score >90 on exactly 3 of the 4 tests) = 4 x (4/5)(1-4/5)^3
= 4² /5^4 = 16/625 = 0.64

2)
A certain hockey player scores on 18% of his shots
let's pose : a = 0.18
P(he will score for the first time on his fifth shot) =
P(he fails on shots 1 to 4) = (1- 0.18)^4 # 0.452

P(it will take him fewer than 3 shots to score) =
P( 1st shot ok) + P(1st shot fails and 2nd ok) =
a + (1-a)a = 0.18 + 0.82 x 0.18 = 03276 = 32.76%

How many shots should he expect to take before scoring?
this question is a bit more difficult :
P(player fails n times before first success) = a(1-a)^n
this introduces the random variable X which is the amount of successive failures before 1st success
P(X=n) = a(1 - a)^n
and so the definition of the mathematical expectation is
E(X) = sigma from n=1 to infinite of ( n a (1 - a)^n )
the calculation (it's a "classical" known calculation) gives
E(X) = 1/a # 5.55

hope it'll help!!

Answer 2

I still don't understand how to do the last part...

Answer 3

which one are we doing?

Answer 4

or all?

Answer 5

The hockey player question, I'm on the last part and unsure how to even start the answer

Answer 6

the expected number of shots?

Answer 7

Yes

Answer 8

if the probability he makes the shot is \(p\) then the expected waiting time is \(\frac{1}{p}\)

Answer 9

so your answer is \[\frac{1}{.18}\] whatever that is

Answer 10

why is there a waiting time though?

Answer 11

and why is it 1 divided by P?

Answer 12

it is just called "waiting time" is all
how many shots do you expect to have to make before you make one

Answer 13

for example if you roll one die, the probability you get a "2" is \(\frac{1}{6}\) so you expect to have to roll \(6\) times to get a \(2\)

Answer 14

if you are asking why, to derive the formula requires a summation trick from calculus
probably not a able to explain it unless you have had one semester of calc at least

Answer 15

ok I'm just a little confused because 1 divided by 0.18 is 5.55 but you can't have a decimal point in the number of shots you're taking. Do I just round or...?

Answer 16

an expected value is a kind of average

Answer 17

if you have a bunch of numbers, the average of those numbers does not have to be a whole number or even one of the numbers given

Answer 18

Oh. Okay thanks

Answer 19

yw