All Sixes

Solution

Let A be the event that the chosen die has 6s on all faces. Let B be the event that the die is rolled three times and all three results are 6. We are given that \[ P(A) = 0.03 \] so \[ P(A^c) = 0.97. \] If the die has 6s on all faces, then every roll must be a 6, so \[ P(B \mid A) = 1. \] If the die is a normal fair die, the probability of rolling three 6s is \[ P(B \mid A^c) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}. \] We want \[ P(A \mid B). \] Using Bayes' theorem, \[ P(A \mid B) = \frac{P(B \mid A)P(A)} {P(B \mid A)P(A) + P(B \mid A^c)P(A^c)}. \] Substitute the values: \[ P(A \mid B) = \frac{1 \cdot 0.03} {1 \cdot 0.03 + \frac{1}{216} \cdot 0.97}. \] Compute the denominator term: \[ \frac{0.97}{216} \approx 0.0044907. \] Thus \[ P(A \mid B) = \frac{0.03}{0.03 + 0.0044907} = \frac{0.03}{0.0344907}. \] Therefore \[ P(A \mid B) \approx 0.8698. \] So after observing three 6s in a row, the probability that the die has 6s on all faces is approximately \( 0.87 \).