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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ NB - number of successes before $n$ failures occur. $\endgroup$ – d.k.o. Jul 15 '15 at 7:43
  • 1
    $\begingroup$ @d.k.o. so is this case I have no way to use NB? $\endgroup$ – gbox Jul 15 '15 at 7:45

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,

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.