Question 1176603
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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.



<pre>
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

    {{{P[1]}}} = {{{C[3]^1*0.65^1*(1-0.65)^2}}} = {{{3*0.65*0.35^2}}} = 0.2389   (rounded).


For the second event we have the probability

    {{{P[2]}}} = 1 - 0.65 = 0.35.


Thus the final formula and the answer to the problem's question are

    P = {{{P[1]*P[2]}}} = 0.2389*0.35 = 0.0836.
</pre>

Solved.



/\/\/\/\/\/\/\/


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:


<pre>
    - her approach was INCORRECT,

    - while my approach is a UNIQUALLY RIGHT.
</pre>