Probability with multiple dice If you roll $24$ six-sided dice (assuming they're all "fair"), what is the probability that exactly $10$ dice will land on $3$?
 A: Choose $10$ dice from the $24$ that will land on $3$. There are $\binom{24}{10}$ ways to do this. The probability of landing $3$ is $1/6$. For the remaining $14$ dice, the probability that they do not come up $3$ is $5/6$.
Can you put this all together and conclude?
ADDENDUM: What you have here (assuming the rolls are independent) is a binomial distribution https://en.wikipedia.org/wiki/Binomial_distribution.
In general, such a distribution tells you the probability of getting $k$ "successes" in $n$ independent "trials". Here, your trials are rolls of dice and there are $n = 24$ of them. Success, for you, is rolling a $3$, and you want the probability of $k = 10$ successes. Th probability of success (rolling a 3) is $p = 1/6$ and the probability of failure (not rolling a 3) is $(1-p) = 5/6$.
The probability of a binomial is given by 
$$ P(\text{# of successes is k}) = \binom{n}{k}p^{k}(1-p)^{n-k} $$
There are $\binom{n}{k}$ different ways that exactly $k$ successes occur (the first $10$ rolls land $3$, the last $10$ rolls land $3$, etc.). For each of the $k$ successes, its probability of occurring is $p$; for each of the $n-k$ failures, its probability of occurring is $1-p$.
