A bag contains 4 red and 6 blue balls. A ball is drawn at random and put aside without its colour being noted. A second ball is then drawn at random from the remaining balls. The probability that the second ball is red is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{5}\)
The colour of the first ball is not known, so both cases must be counted. Let \(E_1\) and \(E_2\) be the events that the first ball is red and blue, and let A be the event that the second ball is red.
\( \displaystyle P(E_1) = \frac{4}{10},\quad P(A|E_1) = \frac{3}{9} \)
\( \displaystyle P(E_2) = \frac{6}{10},\quad P(A|E_2) = \frac{4}{9} \)
By the theorem of total probability:
\( \displaystyle P(A) = \frac{4}{10}\times\frac{3}{9} + \frac{6}{10}\times\frac{4}{9} \)
\( \displaystyle P(A) = \frac{12}{90}+\frac{24}{90} = \frac{36}{90} = \frac{2}{5} \)
This is the same as the chance that the first ball is red, \(\frac{4}{10}\): a ball put aside unseen does not change the chance for the next draw.
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{3}\) — Incorrect. This is \(\frac{3}{9}\), which assumes the first ball was red.
Option (2) \(\displaystyle\frac{4}{9}\) — Incorrect. This is \(\frac{4}{9}\), which assumes the first ball was blue.
Option (4) \(\displaystyle\frac{2}{15}\) — Incorrect. This is \(\frac{12}{90}\), the chance that both balls are red.