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A Basketball player sink a ball in probability of 60%, independently. What is the probability he will success in 5 out 6 shots?

Every event is independent and has $p=0.6$, and $q=0.4$, and we want 5 out of 6 (no matter the order).
is it $Bin\sim (6,0.6)$ therefore it is ${6\choose 5}*0.6^5*0.4=0.186$

or is it $BN\sim (5,0.6)$? in general when should I use Negative binomial distribution rather than binomial distribution?

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  • $\begingroup$ You can compute it without referring to any "distribution"... (Binomial - number of successes in $n$ trials) $\endgroup$ – d.k.o. Jul 15 '15 at 7:24
  • $\begingroup$ yes it is $\frac{{6\choose 5}{2\choose 1}}{2^6}$, I am trying to understand when to use negative binomial distribution vs binomial distribution $\endgroup$ – gbox Jul 15 '15 at 7:40
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    $\begingroup$ NB - number of successes before $n$ failures occur. $\endgroup$ – d.k.o. Jul 15 '15 at 7:43
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    $\begingroup$ @d.k.o. so is this case I have no way to use NB? $\endgroup$ – gbox Jul 15 '15 at 7:45
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The negative binomial is not useful here. That's the distribution for the number of successes before the $n$-th failure.

Here you should use the binomial distribution,

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