JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

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
$\frac{7}{1 0 ^{7}}$
B
$\frac{81}{1 0 ^{8}}$
C
$\frac{67}{1 0 ^{8}}$
D
$\frac{73}{1 0 ^{8}}$

Solution

Each draw is independent (with replacement) with $P (defective) = \frac{1}{10}$ and $P (non-defective) = \frac{9}{10}$ .

Using binomial probabilities for $n = 8$ :

$P (exactly 7 defective) = (7 8) (\frac{1}{10})^{7} (\frac{9}{10}) = \frac{72}{1 0 ^{8}}$ .

$P (exactly 8 defective) = (\frac{1}{10})^{8} = \frac{1}{1 0 ^{8}}$ .

$P (at least 7) = \frac{72 + 1}{1 0 ^{8}} = \frac{73}{1 0 ^{8}}$ .

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