you throw 1 dice until you get K times the number '6'

X = the number of your throws.

what is P(X) ?

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- May 6th 2006, 12:45 PMamitbernprob dice question
you throw 1 dice until you get K times the number '6'

X = the number of your throws.

what is P(X) ? - May 6th 2006, 01:37 PMrgep
The probability that N throws contains (K-1) '6's somewhere in the first (N-1) throws, with a '6' on the N-th throw is $\displaystyle \binom{N-1}{K-1} \left(\frac56\right)^{(N-1)-(K-1)} \left(\frac16\right)^{K-1} \frac16$.

- May 6th 2006, 08:54 PMamitbern
how can i prove that ∑ p(X)=1 ?

- May 6th 2006, 10:41 PMCaptainBlackQuote:

Originally Posted by**amitbern**

since with probability 1 you will eventually reach K sixes. So the sum

of the probabilities of the number of throws needed for this is 1.

RonL - May 6th 2006, 10:54 PMamitbern
I know that it's supposed to be 1,

but how can i prove it mathematically (Newton’s binomial )?