peyton rolls a standard die 15 times. what is the probability he rolls a two exactly four times?

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. The binomial probability formula is ( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} ), where ( n ) is the number of trials, ( k ) is the number of successes, ( p ) is the probability of success on a single trial. Here, ( n = 15 ), ( k = 4 ), and the probability of rolling a two on a standard die is ( p=\frac{1}{6}), so ( 1 - p=\frac{5}{6} ).

Step2: Calculate the combination

First, calculate the combination ( \binom{n}{k}=\binom{15}{4} ). The formula for combinations is ( \binom{n}{k}=\frac{n!}{k!(n - k)!} ). So ( \binom{15}{4}=\frac{15!}{4!(15 - 4)!}=\frac{15!}{4!11!}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365 ).

Step3: Calculate the probability

Now, substitute the values into the binomial formula: ( P(X = 4)=\binom{15}{4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{11} ). Calculate ( \left(\frac{1}{6}\right)^4=\frac{1}{1296} ) and ( \left(\frac{5}{6}\right)^{11}\approx\frac{48828125}{362797056} ). Then multiply all together: ( 1365\times\frac{1}{1296}\times\frac{48828125}{362797056}\approx1365\times\frac{48828125}{1296\times362797056}\approx1365\times\frac{48828125}{469125799936}\approx\frac{66650492187.5}{469125799936}\approx0.137 ) (approximate value).

Answer:

The probability is approximately ( \boldsymbol{0.137} ) (or can be left in exact fractional form ( \binom{15}{4}\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^{11}=\frac{1365\times48828125}{6^{15}} ))