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. What is the probability of second ball being blue?

$\frac{5}{9}$

The correct answer is Option (3) → $\frac{5}{9}$ ##

A box $= \{5 \text{ blue}, 4 \text{ red}\}$

Let $E_1$ be the event that first ball drawn is blue, $E_2$ be the event that first ball drawn is red and $E$ be the event that second ball drawn is blue.

$∴P(E) = P(E_1) \cdot P(E | E_1) + P(E_2) \cdot P(E | E_2)$

$= \frac{5}{9} \cdot \frac{4}{8} + \frac{4}{9} \cdot \frac{5}{8} = \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9}$