Question

a die is rolled eight times. what is the probability of rolling a perfect square a maximum of three times?

Explanation:

Step1: Identify perfect squares on a die

A die has numbers 1, 2, 3, 4, 5, 6. Perfect squares here are 1 (\(1^2\)) and 4 (\(2^2\)). So number of successful outcomes (rolling a perfect square) \(p=\frac{2}{6}=\frac{1}{3}\), failure \(q = 1 - \frac{1}{3}=\frac{2}{3}\), number of trials \(n = 8\).

Step2: Use binomial probability formula

We need \(P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)\), where binomial probability formula is \(P(X=k)=\binom{n}{k}p^{k}q^{n - k}\), \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\).

For \(k = 0\):

\(\binom{8}{0}(\frac{1}{3})^{0}(\frac{2}{3})^{8}=\frac{8!}{0!8!}\times1\times(\frac{256}{6561}) = 1\times1\times\frac{256}{6561}=\frac{256}{6561}\)

For \(k = 1\):

\(\binom{8}{1}(\frac{1}{3})^{1}(\frac{2}{3})^{7}=\frac{8!}{1!7!}\times\frac{1}{3}\times(\frac{128}{2187}) = 8\times\frac{1}{3}\times\frac{128}{2187}=\frac{1024}{6561}\)

For \(k = 2\):

\(\binom{8}{2}(\frac{1}{3})^{2}(\frac{2}{3})^{6}=\frac{8!}{2!6!}\times\frac{1}{9}\times(\frac{64}{729}) = 28\times\frac{1}{9}\times\frac{64}{729}=\frac{1792}{6561}\)

For \(k = 3\):

\(\binom{8}{3}(\frac{1}{3})^{3}(\frac{2}{3})^{5}=\frac{8!}{3!5!}\times\frac{1}{27}\times(\frac{32}{243}) = 56\times\frac{1}{27}\times\frac{32}{243}=\frac{1792}{6561}\)

Step3: Sum the probabilities

\(P(X\leq3)=\frac{256 + 1024+1792 + 1792}{6561}=\frac{256+1024 = 1280; 1280+1792 = 3072; 3072+1792 = 4864}{6561}=\frac{4864}{6561}\approx0.741\) (or keep as fraction)

Answer:

\(\frac{4864}{6561}\) (or approximately \(0.741\))