There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0.1, then the probability that all the lines are busy?

$\frac{e^{-5}×5^{50}}{50!}$

The correct answer is Option (2) → $\frac{e^{-5}×5^{50}}{50!}$

Given 50 independent telephone lines.

Probability a line is busy = $0.1$

We need $P(\text{all 50 lines busy})$.

Since each line is independent:

$P(\text{all 50 busy}) = (0.1)^{50}$

Using Poisson approximation with $\lambda = np = 50 \times 0.1 = 5$:

$P(X = 50) = e^{-5}\frac{5^{50}}{50!}$

Correct result: $e^{-5}\frac{5^{50}}{50!}$