A student goes to school by bus, by cycle or on foot with probabilities \(\displaystyle\frac{1}{2}\), \(\displaystyle\frac{3}{10}\) and \(\displaystyle\frac{1}{5}\) respectively. The probabilities that she reaches late are \(\displaystyle\frac{1}{10}\) by bus, \(\displaystyle\frac{1}{6}\) by cycle and \(\displaystyle\frac{1}{4}\) on foot. On a certain day she reaches late. The probability that she came by bus that day is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{3}\)
Let \(E_1, E_2, E_3\) be the events that she comes by bus, by cycle and on foot, and let L be the event that she is late.
\( \displaystyle P(E_1)P(L|E_1) = \frac{1}{2}\times\frac{1}{10} = \frac{1}{20} \)
\( \displaystyle P(E_2)P(L|E_2) = \frac{3}{10}\times\frac{1}{6} = \frac{1}{20} \)
\( \displaystyle P(E_3)P(L|E_3) = \frac{1}{5}\times\frac{1}{4} = \frac{1}{20} \)
By the theorem of total probability:
\( \displaystyle P(L) = \frac{1}{20}+\frac{1}{20}+\frac{1}{20} = \frac{3}{20} \)
By Bayes’ theorem:
\( \displaystyle P(E_1|L) = \frac{\frac{1}{20}}{\frac{3}{20}} = \frac{1}{3} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the chance of coming by bus on an ordinary day, before we know she is late.
Option (3) \(\displaystyle\frac{3}{20}\) — Incorrect. This is P(L), the chance that she is late on a given day.
Option (4) \(\displaystyle\frac{1}{10}\) — Incorrect. This is \(P(L|E_1)\), the chance of being late when she comes by bus; the condition is the wrong way round.