A bag contains 2 white and 3 black balls. A coin is tossed. If it shows a head, one white ball is added to the bag; if it shows a tail, one black ball is added to the bag. A ball is then drawn at random from the bag. The probability that the ball drawn is white is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{5}{12}\)
Let \(E_1\) be the event that the coin shows a head and \(E_2\) the event that it shows a tail, and let W be the event that the ball drawn is white.
\( \displaystyle P(E_1) = P(E_2) = \frac{1}{2} \)
After a head the bag has 3 white and 3 black balls; after a tail it has 2 white and 4 black balls:
\( \displaystyle P(W|E_1) = \frac{3}{6},\quad P(W|E_2) = \frac{2}{6} \)
By the theorem of total probability:
\( \displaystyle P(W) = \frac{1}{2}\times\frac{3}{6}+\frac{1}{2}\times\frac{2}{6} \)
\( \displaystyle P(W) = \frac{3}{12}+\frac{2}{12} = \frac{5}{12} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{5}\) — Incorrect. This is the chance of white from the bag before any ball is added.
Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the chance of white only when the coin shows a head.
Option (4) \(\displaystyle\frac{7}{12}\) — Incorrect. This is \(1-\frac{5}{12}\), the chance that the ball drawn is black.