Q103 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
For $k=1,2,3$ the box $B_k$ contains ' $k$ ' red balls and $(k+1)$ white balls. Let $P\left(B_1\right)=\dfrac{1}{2}, P\left(B_2\right)=\dfrac{1}{3}$ and $P\left(B_3\right)=\dfrac{1}{6}$. A box is selected at random, and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from $B_2$ is
- Correct answerA$\dfrac{14}{39}$
- B$\dfrac{10}{13}$
- C$\dfrac{35}{78}$
- D$\dfrac{12}{13}$
Solution
Let $R$ be the event that a red ball is drawn. The probabilities of selecting the boxes are $P(B_1) = \dfrac{1}{2}$, $P(B_2) = \dfrac{1}{3}$, and $P(B_3) = \dfrac{1}{6}$.
The probabilities of drawing a red ball from each box are:
$P(R|B_1) = \dfrac{1}{1+2} = \dfrac{1}{3}$
$P(R|B_2) = \dfrac{2}{2+3} = \dfrac{2}{5}$
$P(R|B_3) = \dfrac{3}{3+4} = \dfrac{3}{7}$
Using the law of total probability, the probability of drawing a red ball is:
$P(R) = P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3)$
$P(R) = \left(\dfrac{1}{2} \times \dfrac{1}{3}\right) + \left(\dfrac{1}{3} \times \dfrac{2}{5}\right) + \left(\dfrac{1}{6} \times \dfrac{3}{7}\right)$
$P(R) = \dfrac{1}{6} + \dfrac{2}{15} + \dfrac{1}{14}$
Finding a common denominator, which is $210$:
$P(R) = \dfrac{35}{210} + \dfrac{28}{210} + \dfrac{15}{210} = \dfrac{78}{210} = \dfrac{13}{35}$
Using Bayes' Theorem, the probability that the ball came from $B_2$ given it is red is:
$P(B_2|R) = \dfrac{P(B_2)P(R|B_2)}{P(R)}$
$P(B_2|R) = \dfrac{\dfrac{1}{3} \times \dfrac{2}{5}}{\dfrac{13}{35}} = \dfrac{\dfrac{2}{15}}{\dfrac{13}{35}} = \dfrac{2}{15} \times \dfrac{35}{13} = \dfrac{2 \times 7}{3 \times 13} = \dfrac{14}{39}$
Answer: $\dfrac{14}{39}$
The probabilities of drawing a red ball from each box are:
$P(R|B_1) = \dfrac{1}{1+2} = \dfrac{1}{3}$
$P(R|B_2) = \dfrac{2}{2+3} = \dfrac{2}{5}$
$P(R|B_3) = \dfrac{3}{3+4} = \dfrac{3}{7}$
Using the law of total probability, the probability of drawing a red ball is:
$P(R) = P(B_1)P(R|B_1) + P(B_2)P(R|B_2) + P(B_3)P(R|B_3)$
$P(R) = \left(\dfrac{1}{2} \times \dfrac{1}{3}\right) + \left(\dfrac{1}{3} \times \dfrac{2}{5}\right) + \left(\dfrac{1}{6} \times \dfrac{3}{7}\right)$
$P(R) = \dfrac{1}{6} + \dfrac{2}{15} + \dfrac{1}{14}$
Finding a common denominator, which is $210$:
$P(R) = \dfrac{35}{210} + \dfrac{28}{210} + \dfrac{15}{210} = \dfrac{78}{210} = \dfrac{13}{35}$
Using Bayes' Theorem, the probability that the ball came from $B_2$ given it is red is:
$P(B_2|R) = \dfrac{P(B_2)P(R|B_2)}{P(R)}$
$P(B_2|R) = \dfrac{\dfrac{1}{3} \times \dfrac{2}{5}}{\dfrac{13}{35}} = \dfrac{\dfrac{2}{15}}{\dfrac{13}{35}} = \dfrac{2}{15} \times \dfrac{35}{13} = \dfrac{2 \times 7}{3 \times 13} = \dfrac{14}{39}$
Answer: $\dfrac{14}{39}$
Hence, the correct answer is option: 'A'.