Question 1176603
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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.<br>
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.<br>
The answer from tutor @ewatrrr is the correct answer.<br>
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.<br>
Very simply, this problem is solved as follows:<br>
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.<br>