What is the probability that a batsman get run from $15$ balls and no run from $25$ balls before being caught behind in the $40$th balls ? Assuming he has a half-half chance to get run from each ball .

My Attempt:

This is a case of negative binomial distribution

So the distributional form is :

$$P(X=k)=\binom {\alpha+k-1}{k}(1-p)^{\alpha}p^k$$

$\alpha$ is number of failures and $k$ is number of successes.

$$P(X=?)=\binom {40-1}{?}(0.5)^{40}(0.5)^?$$


Not sure about cricket rules, but if the 40th ball has to be no run...

k = number of sucesses = 15

$\alpha$ = number of failures = 25

p = (1 - p) = .5

$P(X=15)=\binom {25 + 15 - 1}{15}(1-p)^{25}p^{15}$

$P(X=15)=\binom {39}{15}(.5)^{40}$

  • $\begingroup$ Law 32.5 If the striker is dismissed Caught, runs from that delivery completed by the batsmen before the completion of the catch shall not be scored ..." and anyway if would be very unusual to complete a run before being caught "behind", i.e. by the wicketkeeper. A bigger problem is the use of "before" in the question. $\endgroup$ – Henry Dec 13 '14 at 20:28

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