The problem involves a binomial distribution where the probability of success (a household being satisfied) is 0.66, and the number of trials is 10. We need to find the probability that at least 8 households are satisfied, which means finding the probability of 8, 9, or 10 successes.
The probability of exactly 8 successes is given by:
\(P(X = 8) = \binom{10}{8} (0.66)^8 (0.34)^2\)
The probability of exactly 9 successes is given by:
\(P(X = 9) = \binom{10}{9} (0.66)^9 (0.34)^1\)
The probability of exactly 10 successes is given by:
\(P(X = 10) = \binom{10}{10} (0.66)^{10}\)
Thus, the probability of at least 8 successes is:
\(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\)
Calculating each term:
\(P(X = 8) = \binom{10}{8} (0.66)^8 (0.34)^2 = 45 \times 0.66^8 \times 0.34^2\)
\(P(X = 9) = \binom{10}{9} (0.66)^9 (0.34)^1 = 10 \times 0.66^9 \times 0.34\)
\(P(X = 10) = \binom{10}{10} (0.66)^{10} = 1 \times 0.66^{10}\)
Summing these probabilities gives:
\(P(X \geq 8) = 0.284\)
