If a fair die is rolled 10 times, what is the probability that number 5 appears 3 times and the number 6 appears 2 times?

Question

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?

Answer 1

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.

Answer 2

Alternative approach:

Instead of regarding the problem as a probability of events problem, regard it as a Combinatorics problem, which is my favorite way of attacking such problems. The probability is expressed as

$$\frac{N}{D}, \tag1 $$

where $~D = 6^{(10)},~$ which equals the total number of different results that can occur, when a die is rolled $(10)$ times.

Then

$$N = \binom{10}{3} \times \binom{7}{2} \times 4^5. \tag2$$

In (2) above, the first factor represents the number of distinct ways that the three $(5)$'s can be placed. Then, the second factor represents the number of distinct ways that the two $(6)$'s can be placed, in the $7$ remaining positions, after the three $(5)$'s are placed.

Then, in (2) above, the third factor represents the number of ways that the five other (unrestricted) postions can occur, where there are four different numbers possible for each of these unrestricted positions.

So

$$N = 120 \times 21 \times 1024 = 2580480$$

and

$$D = 6^{(10)} = 60466176.$$

Therefore, the probability is

$$\frac{N}{D} = \frac{2580480}{60466176} \approx (0.042676).$$