2026-04-06

Problem — 2026-04-06

Easy

Easyprobability

A bag contains $3$ red and $2$ blue balls. You draw balls one at a time without replacement. What is the expected number of draws until you get your first red ball?

Express as a fraction or decimal

Show solution

Think of all

5

balls in a random order. We want the expected position of the first red ball. By symmetry, the

3

red balls are equally likely to occupy any of the

\binom{5}{3}

arrangements.

P(\text{1st draw red}) = \frac{3}{5}

P(\text{1st red on draw 2}) = \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}

P(\text{1st red on draw 3}) = \frac{2}{5} \cdot \frac{1}{4} \cdot 1 = \frac{1}{10}

. So

E = 1 \cdot \frac{3}{5} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{10} = \frac{6+6+3}{10} = \frac{3}{2}