SOLUTION: Suppose a football player has a 65% chance of making a goal that he can keep each time he tries to make a goal. What is the probability that he makes a goal for the first three tim

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Question 1176603: Suppose a football player has a 65% chance of making a goal that he can keep each time he tries to make a goal. What is the probability that he makes a goal for the first three times he tries to but not on the fourth try?
Found 3 solutions by ewatrrr, ikleyn, greenestamps:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

Hi Obviously, this presentation is for the interpretation: "he makes a goal for the first three times" as making 3 goals before missing. Obviously making the 1st 3 goals is more exciting. Binomial Distribution: p = .65, n = 4 P(x= 3) = binompdf(4, .65, 3) = .3845 there are four ways x=3 can be done: NG in 1st, 2nd, 3rd, or 4th position P(G,G,G,NG) = .3845/4 = .0961 It is fun to see different ways of approaching a problem: In case of the other tutor: 1st Step: making 3 in a row: should be: 3C3(.65)^3(.35)^0 = .274625 2nd Step: (.35)(.274625) = .0961 same final answer. Wish You the Best in your Studies.

Answer by ikleyn(52806)   (Show Source):
You can put this solution on YOUR website!
.
Suppose a football player has a 65% chance of making a goal that he can keep each time he tries to make a goal.
What is the probability that he makes a goal for the first three times he tries to but not on the fourth try?
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            The solution by @ewatrrr is incorrect.

            I came to bring the correct solution.

This problem asks about the probability to make one goal in first three trials AND do not make it in the fourth trial. So, it is the probability of intersection of two independent events: - first event is to make one goal in first three attepts, and - the second event is do not make a goal in the fourth attempt. For the first event we have a Binomial distribution problem = = = 0.2389 (rounded). For the second event we have the probability = 1 - 0.65 = 0.35. Thus the final formula and the answer to the problem's question are P = = 0.2389*0.35 = 0.0836.

Solved.

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After seeing my post, @ewatrrr made some corrections, but they are partly incomplete and parly incorrect,

so her final answer and the formulas are and remain INCORRECT.

When she comments about different approaches, she makes it WRONG, too.

The right comment is THIS:

- her approach was INCORRECT, - while my approach is a UNIQUALLY RIGHT.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

The responses from tutor @ikleyn are completely irrelevant, despite her assertion that the response from tutor @ewatrrr is incorrect and hers is "uniqually" correct -- whatever that means.

Tutor @ikleyn answered a problem completely different than the given one. The statement of the problem clearly states that the player makes a goal on all three of his first three tries and does not on his fourth try. It does NOT say he scored a goal on one of his first three tries.

The answer from tutor @ewatrrr is the correct answer.

And, as she states in her revised response, there is no need to use binomial probability in the problem, because the sequence of goals made or not made is fixed.

Very simply, this problem is solved as follows:

P(goal on 1st try AND goal on 2nd try AND goal on 3rd try AND no goal on 4th try) = (.65)(.65)(.65)(.35) = 0.9612 to four decimal places.