Question.
If a fair die is rolled 10 times, what is the probability that number
5 appears exactly 3 times and the number 6 appears exactly 2 times?
Attempt.
$$
P(E)=\frac{\left(\frac{1}{6}\right)^{3}\left(\frac{1}{6}\right)^{2}\left(\frac{4}{6}\right)^{5}\times10!}{2!3!5!}\cong0.042
$$
Is there a way to solve it using binomial distribution?
Best Answer
Let $A$ be the event that $5$ appears exactly three times. Let $X$ have Binomial$(10,1/6)$ distribution, and let $Y$ have Binomial$(7,1/5)$ distribution. Given $A$, the remaining tosses are independent of each other, and each of them has five equally likely options. Thus $$P(E)=P(A)P(E|A)=P(X=3)P(Y=2)$$ $$ ={10 \choose 3}(\frac16)^3(\frac56)^7 \cdot{7 \choose 2}(\frac15)^2 (\frac45)^5 \,.$$ This agrees with your solution.