If a fair die is rolled 5 times what is the probability that the number shown will be less than 3 exactly 2 times. I have tried to work problem using formula from textbook and still get it wrong
[Math] Probability of exactly 2 low rolls in 5 throws of a die
probability
Best Answer
Since there are two possibilities, we use the binomial probability formula
$$\binom{n}{k}p^kq^{n - k}$$
where $p$ is the probability of the event, $q = 1 - p$ is the probability of the complement of the event, $n$ is the number of trials, and $k$ is the number of times the event occurs during those $n$ trials.
Here, the probability that the outcome of a roll is less than $3$ is $1/3$ since two of the six outcomes are less than $3$. That means the probability that a roll will produce an outcome of at least $3$ is $2/3$. Since we want the probability that exactly $2$ of the five rolls are less than $3$, $n = 5$, $k = 2$, and $n - k = 3$. Hence, the probability of exactly two rolls with an outcome less than $3$ in five rolls of the die is
$$P = \binom{5}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^3 = \frac{5!}{2!3!}\left(\frac{1}{9}\right)\left(\frac{8}{27}\right) = \frac{5 \cdot 4}{2 \cdot 1}\left(\frac{8}{243}\right) = \frac{80}{243}$$