Q27 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is
- A$\frac{67}{10^{8}}$
- B$\frac{7}{10^{7}}$
- C$\frac{81}{10^{8}}$
- Correct answerD$\frac{73}{10^{8}}$
Solution
Binomial distribution with $n=8$, $p=\frac{1}{10}$, $q=\frac{9}{10}$.
$P(X \ge 7) = P(X=7) + P(X=8)$
$P(X=7) = \binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right) = \frac{72}{10^8}$
$P(X=8) = \left(\frac{1}{10}\right)^8 = \frac{1}{10^8}$
$P(X \ge 7) = \frac{73}{10^8}$
$P(X \ge 7) = P(X=7) + P(X=8)$
$P(X=7) = \binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right) = \frac{72}{10^8}$
$P(X=8) = \left(\frac{1}{10}\right)^8 = \frac{1}{10^8}$
$P(X \ge 7) = \frac{73}{10^8}$
Hence, the correct answer is option: 'D'.