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) → \(\displaystyle\frac{e^{-5}\, 5^{50}}{50!}\)
Let X be the number of busy lines. X is binomial with \(n = 50\) and \(p = 0.1\). The options are all in Poisson form, so the binomial is approximated by the Poisson distribution with mean
\(\lambda = np = 50 \times 0.1 = 5\)
\( \displaystyle P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \)
“All the lines are busy” means \(r = 50\):
\( \displaystyle P(X = 50) = \frac{e^{-5}\, 5^{50}}{50!} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{e^{-5}\, 5^{50}}{5!}\) — Incorrect. The factorial must be of r = 50, the number of busy lines, not of the mean 5.
Option (3) \(\displaystyle\frac{e^{-50}\, 50^{50}}{50!}\) — Incorrect. This takes the mean as 50, the number of lines; the mean is \(np = 5\).
Option (4) \(\displaystyle\frac{e^{-50}\, 50^{5}}{50!}\) — Incorrect. With \(\lambda = 5\) and r = 50 the factor must be \(e^{-5}\, 5^{50}\); this option has \(e^{-50}\) and \(50^{5}\).