A jeweller has two pouches of beads. The first pouch contains 3 red and 9 green beads, and the second contains 6 red and 4 green beads. One of the two pouches is chosen at random and a bead is drawn from it at random. What is the probability that the bead is red?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{17}{40}\)
The red bead can come through either pouch. Each pouch is chosen with probability \(\frac{1}{2}\); multiply along each route and add the two routes.
First pouch: \(\displaystyle \frac{1}{2}\times\frac{3}{12}=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}\)
Second pouch: \(\displaystyle \frac{1}{2}\times\frac{6}{10}=\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\)
P(red) \(\displaystyle =\frac{1}{8}+\frac{3}{10}=\frac{5+12}{40}=\frac{17}{40}\)
Why the other options are incorrect
Option (1) \(\frac{1}{8}\) — Incorrect. This is only the route through the first pouch.
Option (2) \(\frac{3}{10}\) — Incorrect. This is only the route through the second pouch.
Option (4) \(\frac{9}{22}\) — Incorrect. This puts all 22 beads into one heap (9 red of 22), but the pouches hold different numbers of beads and each pouch is chosen with probability \(\frac{1}{2}\).