Let the number of defective batteries in a pack be a binomial random variable, denoted by \(X\), with parameters \(n = 20\) and probability of defect \(p\).
Given the mean number of defective batteries is 1.6, we have:
\(20p = 1.6\)
Solving for \(p\), we get:
\(p = \frac{1.6}{20} = 0.08\)
We need to find \(P(X > 2)\). Using the complement rule:
\(P(X > 2) = 1 - P(X \leq 2)\)
Calculate \(P(X \leq 2)\) using the binomial probability formula:
\(P(X = 0) = \binom{20}{0} (0.08)^0 (0.92)^{20} = (0.92)^{20}\)
\(P(X = 1) = \binom{20}{1} (0.08)^1 (0.92)^{19} = 20 \times 0.08 \times (0.92)^{19}\)
\(P(X = 2) = \binom{20}{2} (0.08)^2 (0.92)^{18} = \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)
Thus,
\(P(X \leq 2) = (0.92)^{20} + 20 \times 0.08 \times (0.92)^{19} + \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)
Substituting the values:
\(P(X \leq 2) = 0.1887 + 0.3281 + 0.2711 = 0.7879\)
Therefore,
\(P(X > 2) = 1 - 0.7879 = 0.212\)
