Let the probability that a water pistol does not work properly be denoted by \(p = 0.08\). The probability that a water pistol works properly is \(1 - p = 0.92\).
We are dealing with a binomial distribution where \(n = 19\) and \(p = 0.08\). We need to find the probability that at most 2 do not work properly, i.e., \(P(X \leq 2)\).
The probability \(P(X \leq 2)\) is given by:
\(P(X = 0) + P(X = 1) + P(X = 2)\)
Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:
\(P(X = 0) = \binom{19}{0} (0.08)^0 (0.92)^{19} = (0.92)^{19}\)
\(P(X = 1) = \binom{19}{1} (0.08)^1 (0.92)^{18} = 19 \times 0.08 \times (0.92)^{18}\)
\(P(X = 2) = \binom{19}{2} (0.08)^2 (0.92)^{17} = \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)
Summing these probabilities gives:
\(P(X \leq 2) = (0.92)^{19} + 19 \times 0.08 \times (0.92)^{18} + \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)
Calculating this expression yields \(P(X \leq 2) = 0.809\).
