Question

Probability theory predicts that there is 77.6% chance of a particular soccer player making 2 penalty shots in a row. If the soccer player taking 2 penalty shots is simulated 2500 times, in about how many simulations would you expect at least 1 missed shot?

Answer 1

@hockeychick23 you there?

Answer 2

yes

Answer 3

If we let p=the probability of making one shot, then q=1-p is the probability of NOT making one shot.

Answer 4

I did the problem but I just wasn't sure if I completed it correctly, i took .224 and multiplied it by 2500 to get 560. But I just wasn't sure if I took the correct steps

Answer 5

and i got .224 by doing 1-.776

Answer 6

.224 is the probability of not making TWO shots, which could be zero or one.

Answer 7

oh ok, so would that mean i would have to multiply my answer by 1/2 since its two shots?

Answer 8

When p is the probability of making ONE shot, the probability of making TWO shots would be p^2=0.776 by the multiplication law.
Can you find p?

Answer 9

yea, p=0.88090862182

Answer 10

Good!

Answer 11

So what is the probability of MISSING one shot?

Answer 12

1-0.88090862182 which would be0.11909137817

Answer 13

Great! You're doing well!

Answer 14

Now, what is the probability of missing TWO shots?

Answer 15

Recall the multiplication rule!

Answer 16

Nooooo!
The multiplication rule says that if q is the probability of missing one shot, the probability of missing TWO shots will be q^2!

Answer 17

oh ok now i got .01418= .11909137817^2

Answer 18

I've got to go.
But if you interpret what (1-q^2) means, and work with n=2500, you would get the answer quite easily.
I'll come back to check on your answer later in the evening.

Answer 19

ok i got 1940 as my final answer

Answer 20

but also considering theres a 77.6% chance they will make 2 in a row it makes much more sense that they would only miss 560 doesn't it?

Answer 21

q is the probability that she will miss one shot. (q=0.11909137817819)
and q^2 is the probability that she will miss both shots.
Therefore 1-q^2 is the probability that she will make at least one shot.
Since there are 2500 trials, each of which has probability 1-q^2 of making at least one shot, then approximately 2500(1-q^2) will be the expected number of trials that she gets at least one shot.
If not clear, please post again.
@hockeychick23

Answer 22

Ye so I got q^2= 0.01418275635638067, then did 1-0.014182756356380670= .98581724364361933 then did.98581724364361933(2500) and got 2464.543109109048325

Answer 23

but it seems way to big to be correct

Answer 24

Indeed!
I have mis-read the question, which asks for "missing at least one shot".
This means that she did not get both shots, or 1-0.776=0.224, as you suggested earlier.
This situation is represented by the shaded part of the figure below:

and the correct answer is 2500(1-0.776)=560
I apologize for having misread the question.