A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted, then another ball is drawn at random. The probability of second ball being red is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4}{9} \)
The first ball is either blue or red, so the theorem of total probability applies:
\( \displaystyle P(\text{first blue})=\frac{5}{9} \)
\( \displaystyle P(\text{second red}\,|\,\text{first blue})=\frac{4}{8} \)
\( \displaystyle P(\text{first red})=\frac{4}{9} \)
\( \displaystyle P(\text{second red}\,|\,\text{first red})=\frac{3}{8} \)
\( \displaystyle P(\text{second red})=\frac{5}{9}\cdot\frac{4}{8}+\frac{4}{9}\cdot\frac{3}{8} \)
\( \displaystyle =\frac{20+12}{72} \)
\( \displaystyle =\frac{32}{72} \)
\( \displaystyle =\frac{4}{9} \)
Shortcut: Since the colour of the first ball is not known, every ball is equally likely to be the second ball drawn, so P(second red) = P(first red) = \( \frac{4}{9} \).