Say you have an unfair 3-sided die.
Side 1 will be rolled with probability $p_1$.
Side 2 will be rolled with probability $p_2$.
Side 3 will be rolled with probability $p_3$.
Find a formula for the probability that the 8th 3 will be rolled on the 10th roll.
My theory on this is to use a binomial distribution, but I am pretty lost. Could anyone with more experience in probability help me out with this? I was just trying to work through a book I had and I want to figure this out.
Edit: Using Robert's advice this is the direction I'm going:
$9\choose 7$$(p_3)^7$$(1-p_3)^2*p_3$
Any advice?
Hint. There are $\binom{9}{7}$ ways to distribute the seven $3$s rolled before the $10$th roll. The probability to get a $3$ is $p_3$. The probability to get a different number is $1-p_3$.
Please edit your question and write your attempt.