JEE Mathematics 2026 Question with Solution

Given: A lot contains $10$ defective and $90$ non-defective bulbs. $8$ bulbs are selected one by one with replacement.

Find: The probability of getting at least $7$ defective bulbs.

Since selection is with replacement, each trial is independent and the probability of a defective bulb remains constant.

$$P(\text{defective})=\frac{10}{100}=\frac{1}{10}$$

Therefore, the number of defective bulbs selected follows a binomial distribution with

$$n=8,\quad p=\frac{1}{10}$$

“At least $7$ defective bulbs” means

$$P(X\ge 7)=P(X=7)+P(X=8)$$

Now,

$$P(X=7)=\binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right)$$ $$=\frac{72}{10^8}$$

Also,

$$P(X=8)=\binom{8}{8}\left(\frac{1}{10}\right)^8$$ $$=\frac{1}{10^8}$$

Hence,

$$P(X\ge 7)=\frac{72}{10^8}+\frac{1}{10^8}=\frac{73}{10^8}$$

Therefore, the required probability is $\dfrac{73}{10^8}$ and the correct option is B.

Given: Probability of success on each draw is constant because of replacement.

Find: Probability of getting defective bulbs at least $7$ times in $8$ draws.

Use the binomial formula

$$P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$$

Here,

$$n=8,\quad p=\frac{1}{10},\quad 1-p=\frac{9}{10}$$

So we evaluate only the cases $r=7$ and $r=8$.

$$P(X=7)=\binom{8}{7}\left(\frac{1}{10}\right)^7\left(\frac{9}{10}\right)=8\cdot \frac{9}{10^8}=\frac{72}{10^8}$$ $$P(X=8)=\binom{8}{8}\left(\frac{1}{10}\right)^8=\frac{1}{10^8}$$

Adding them,

$$\frac{72}{10^8}+\frac{1}{10^8}=\frac{73}{10^8}$$

Thus, the correct option is B.