There are 50 telephone lines in an exchange. If the probability that…

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

With many lines and a small probability of each line being busy, the number of busy lines X is taken as a Poisson variable with mean

    \( \lambda=np=50\times 0.1=5 \)

    \( \displaystyle P(X=r)=\frac{e^{-\lambda}\lambda^{r}}{r!} \)

All the lines are busy when r = 50:

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

Why the other options are incorrect

• Option (1) \( \displaystyle\frac{(50)^5\times e^{-5}}{50!} \) — Incorrect. The base and the power are swapped: the numerator is \( \lambda^{r}=5^{50} \), not \( 50^{5} \).

• Option (3) \( \displaystyle\frac{5^{50}\times e^{-50}}{50!} \) — Incorrect. The exponent of e is −λ = −5, not −n = −50.

• Option (4) \( \displaystyle\frac{5^{5}\times e^{-50}}{50!} \) — Incorrect. The power of 5 should be r = 50 and the exponent of e should be −λ = −5.