probability distribution of a random variable

Question

We roll a dice until we get 6. Knowing that we have rolled 10 times, evaluate the probability that in the next 20 rolls there will be no 6. So in this question are we supposed to use binomial distribution or geometric distribution?

Answer 1

Suppose the dice rolls are independent, so previous rolls have no impact on future rolls. The probability of getting a 6 once is $p_6$, so the probability of not getting a 6 is $1-p_6$. The probability of getting no 6s in 20 rolls is $$ pr[d_1 \neq 6, d_2 \neq 6, ..., d_{20} \neq 6] =pr[d_1 \neq 6] pr[d_2 \neq 6] ...pr[d_{20} \neq 6] = (1-p_6)^{20} $$ if $d_k$ is the outcome of the $k$-th roll. For the same reason, the first 10 rolls are irrelevant for the outcome of the next 20.

Answer 2

Yes, you can use the binomial distribution in this question. However, two of the key principles in a binomial distribution is that each observation is independent AND the probability of either success or failure is the same for each outcome. Hence, if you rolled 10 times and got a 6 - it has no effect on the next roll.

To your question if using binomial:

${n\choose k}p^k(1-p)^{n-k}$

where in 20 rounds you wish to have any number but 6:

${20 \choose 20}* \left(\frac{5}{6}\right)^{20}*\left(\frac{1}{6}\right)^0=0.026$