JEE Main 2026 — 24 January, Morning Shift — Question 17

Let $p = \frac{10}{100} = 0.1$ be the probability of a defective bulb. Number of trials $n = 8$. We need $P(X \geq 7) = P(X=7) + P(X=8)$. Using binomial distribution: $P(X=k) = \binom{8}{k} (0.1)^k (0.9)^{8-k}$. $P(X=7) = \binom{8}{7} (0.1)^7 (0.9)^1 = 8 \times (0.1)^7 \times 0.9$. $P(X=8) = \binom{8}{8} (0.1)^8 (0.9)^0 = (0.1)^8$. Sum: $8 \times 0.9 \times (0.1)^7 + (0.1)^8 = (7.2 + 0.1) \times (0.1)^7 = 7.3 \times (0.1)^7$. $(0.1)^7 = \frac{1}{10^7}$, so probability = $\frac{7.3}{10^7} = \frac{73}{10^8}$. Thus, the required probability is $\frac{73}{10^8}$.