An unbiased 8-sided die (a d8) is rolled until an 8 appears uppermost. What is the probability that 15 rolls are required until an 8 first appears?

Question

So far I have, $n=15, p = 1/5, k = 1.$ Don't know if that is correct. And then if it is, I have: $$\frac{15!}{1!(15-1)!}\times (1/15)^1 \times (14/15)^{15-1} = 0.38\%$$

Is that correct?

Answer 1

Your calculation is not correct. First you need the probability that you don't roll an $8$ on any of your first $14$ rolls. That probability is $\left( \frac 78 \right)^{14}$. Then you need the probability that your $15$th roll is an $8$. That probability is $\frac 18$. So your answer is $\frac{7^{14}}{8^{15}}$, which is approximately $1.93$%.

Answer 2

First of all, please use mathJax to format math on this site, as discussed at https://math.stackexchange.com/help/notation.

I had trouble deciphering "p = 1/5 k = 1", and you did show work, so I'll just give you the answer.

Let $A$ denote the probability that the 1st 14 rolls are all 7 or less. Let $B$ denote the probability of getting an 8 on the 15th roll.

Then the probability of both of these (independent) events occurring is $A \times B.$

$p(A) = \left(\frac{7}{8}\right)^{(14)}.$
$p(B) = \frac{1}{8}.$