Thread: prob dice question

you throw 1 dice until you get K times the number '6'
X = the number of your throws.
what is P(X) ?

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 $\displaystyle \binom{N-1}{K-1} \left(\frac56\right)^{(N-1)-(K-1)} \left(\frac16\right)^{K-1} \frac16$.

how can i prove that ∑ p(X)=1 ?

Originally Posted by amitbern

how can i prove that ∑ p(X)=1 ?

I'm not quite sure what you want to prove here. This sum has to be 1
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

I know that it's supposed to be 1,
but how can i prove it mathematically (Newton’s binomial )?