Probability Part 3

Concept:The problem involves repeated independent trials (with replacement) and asks for the probability of at least 7 successes (defective bulbs).This is a binomial distribution scenario.Formula:For $n$ independent trials, each with success probability $p$, the probability of exactly $k$ successes is:$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$where $q = 1-p$.Solution:Total bulbs = $10$ defective + $90$ non‑defective = $100$.Probability of selecting a defective bulb: $p = \frac{10}{100} = \frac{1}{10}$.Probability of a non‑defective bulb: $q = 1 - p = \frac{9}{10}$.Number of trials: $n = 8$.We need at least $7$ defective bulbs, i.e., $7$ or $8$.$P(X \ge 7) = P(X=7) + P(X=8)$.$P(X=7) = \binom{8}{7} p^7 q^1 = 8 \cdot \left(\frac{1}{10}\right)^7 \cdot \frac{9}{10}$.$P(X=8) = \binom{8}{8} p^8 q^0 = 1 \cdot \left(\frac{1}{10}\right)^8$.Sum: $P = \left(\frac{1}{10}\right)^8 \bigl[ 8 \cdot 9 + 1 \bigr] = \frac{72+1}{10^8} = \frac{73}{10^8}$.Answer:Option A: $\frac{73}{10^8}$.