Hippopotamus

Solution

Answer: 0.244

James selects 10 letters with replacement, so each selection is independent. Let \(X\) be the number of times the letter P is selected. \[ X \sim B(10, 0.25). \] We want the probability that P is selected fewer than 2 times: \[ P(X < 2) = P(X=0) + P(X=1). \] Using the binomial formula \[ P(X=k) = \binom{10}{k}(0.25)^k(0.75)^{10-k}. \] For \(X=0\) \[ P(X=0) = \binom{10}{0}(0.25)^0(0.75)^{10} \] \[ = (0.75)^{10} \approx 0.056314. \] For \(X=1\) \[ P(X=1) = \binom{10}{1}(0.25)^1(0.75)^9 \] \[ = 10(0.25)(0.75)^9. \] \[ (0.75)^9 \approx 0.075085 \] \[ P(X=1) \approx 10 \times 0.25 \times 0.075085 \] \[ P(X=1) \approx 0.187712. \] Adding the probabilities \[ P(X<2) = 0.056314 + 0.187712. \] \[ P(X<2) \approx 0.244. \]