Let the probability of a biscuit being broken be denoted by \(p\). Since the mean number of broken biscuits is 2.7, we have:
\(18p = 2.7\)
Solving for \(p\), we get:
\(p = \frac{2.7}{18} = 0.15\)
We need to find the probability that the number of broken biscuits \(X\) is between 2 and 4 inclusive, i.e., \(P(2 \leq X \leq 4)\).
\(X\) follows a binomial distribution \(B(18, 0.15)\). Therefore,
\(P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)\)
Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we calculate:
\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16}\)
\(P(X = 3) = \binom{18}{3} (0.15)^3 (0.85)^{15}\)
\(P(X = 4) = \binom{18}{4} (0.15)^4 (0.85)^{14}\)
Summing these probabilities gives:
\(P(2 \leq X \leq 4) = \binom{18}{2} (0.15)^2 (0.85)^{16} + \binom{18}{3} (0.15)^3 (0.85)^{15} + \binom{18}{4} (0.15)^4 (0.85)^{14}\)
Calculating each term:
\(\binom{18}{2} (0.15)^2 (0.85)^{16} \approx 0.231\)
\(\binom{18}{3} (0.15)^3 (0.85)^{15} \approx 0.287\)
\(\binom{18}{4} (0.15)^4 (0.85)^{14} \approx 0.137\)
Adding these gives:
\(P(2 \leq X \leq 4) = 0.231 + 0.287 + 0.137 = 0.655\)
