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?
Answer & explanation
Correct answer: option 2
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: $e^{-5}\frac{5^{50}}{50!}$