document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #803842 by greenestamps(13203) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "The answer from tutor @ewatrrr is the correct answer.

\n" ); document.write( "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.

\n" ); document.write( "Very simply, this problem is solved as follows:

\n" ); document.write( "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.

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