In a school, 40% of the students play football, 25% play cricket and 15% play both football and cricket. A student is chosen at random. If the student plays football, then the probability that the student also plays cricket is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{8}\)
Let F be the event that the student plays football and C the event that the student plays cricket. Then
\(P(F) = 0.40,\quad P(C) = 0.25,\quad P(F\cap C) = 0.15\)
The given event is F, so it goes in the denominator of the conditional probability:
\( \displaystyle P(C|F) = \frac{P(F\cap C)}{P(F)} \)
\( \displaystyle = \frac{0.15}{0.40} = \frac{3}{8} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{3}{5}\) — Incorrect. This is \(\frac{0.15}{0.25} = P(F|C)\), the probability of playing football given cricket.
Option (3) \(\displaystyle\frac{3}{20}\) — Incorrect. This is 0.15 itself, the probability of playing both, not divided by P(F).
Option (4) \(\displaystyle\frac{1}{10}\) — Incorrect. This is \(0.40\times 0.25\), which would be the chance of both only if the two games were independent.