A box contains 3 red and 2 black balls and another box contains 2 red and 4 black balls. One box chosen at random. From the selected box one ball is drawn. The probability of getting a red ball is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{7}{15} \)
Let \( E_1 \), \( E_2 \) be the events that the first or second box is chosen, and R that the ball is red. By the theorem of total probability:
\( \displaystyle P(R)=P(E_1)P(R|E_1)+P(E_2)P(R|E_2) \)
\( \displaystyle =\frac{1}{2}\cdot\frac{3}{5}+\frac{1}{2}\cdot\frac{2}{6} \)
\( \displaystyle =\frac{3}{10}+\frac{1}{6} \)
\( \displaystyle =\frac{9+5}{30}=\frac{7}{15} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{3}{5} \) — Incorrect. \( \frac{3}{5} \) is the probability of red from the first box only.
Option (2) \( \displaystyle\frac{1}{3} \) — Incorrect. \( \frac{1}{3} \) is the probability of red from the second box only.
Option (4) \( \displaystyle\frac{14}{15} \) — Incorrect. \( \frac{14}{15}=\frac{3}{5}+\frac{1}{3} \), the two conditional probabilities added without the factor \( \frac{1}{2} \) for choosing the box.