Questions: Find the probability that in five tosses of a fair single die that a 3 appears at no time. Round to the nearest tenthousandth if in decimal form and to the nearest hundredth if in percent form.

Solution Steps

We need to find the probability that in five tosses of a fair single die, the number \(3\) appears at no time. This can be modeled using a binomial distribution where the probability of success (getting a \(3\)) in a single toss is \(p = \frac{1}{6}\) and the probability of failure (not getting a \(3\)) is \(q = 1 - p = \frac{5}{6}\).

We are interested in the case where \(k = 0\) (no \(3\)s) in \(n = 5\) tosses. The probability of getting exactly \(0\) successes in \(5\) trials can be calculated using the probability mass function (PMF) of the binomial distribution:

\[ P(X = 0) = \binom{n}{k} p^k q^{n-k} \]

Substituting the values, we have:

\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^{5} = 1 \cdot 1 \cdot \left(\frac{5}{6}\right)^{5} \]

Calculating \( \left(\frac{5}{6}\right)^{5} \):

\[ \left(\frac{5}{6}\right)^{5} \approx 0.401877572 \]

Rounding to four significant digits gives us:

\[ P(X = 0) \approx 0.4019 \]

To express the probability in percentage form, we multiply by \(100\):

\[ P(X = 0) \times 100 \approx 40.1877572\% \]

Rounding to two decimal places gives us:

\[ P(X = 0) \approx 40.19\% \]

Final Answer