Q87 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
Three bags contain a number of red and white balls are as follows.
Bag I: 3 red balls
Bag II: 2 red balls and 1 white ball
Bag III: 3 White balls
The probability that bag $i$ will be chosen and a ball is selected from it is $\dfrac{i}{6}, i=1,2,3$. if a white ball is selected, what is the probablity that it came from Bag III
- Correct answerA$\dfrac{9}{11}$
- B0
- C$\dfrac{1}{11}$
- D$\dfrac{2}{11}$
Solution
Let $B_1, B_2, B_3$ be the events of choosing Bag I, Bag II, and Bag III respectively. Let $W$ be the event of selecting a white ball.
The probabilities of choosing the bags are given by $P(B_1) = \dfrac{1}{6}$, $P(B_2) = \dfrac{2}{6} = \dfrac{1}{3}$, and $P(B_3) = \dfrac{3}{6} = \dfrac{1}{2}$.
The conditional probabilities of selecting a white ball from each bag are:
$P(W|B_1) = \dfrac{0}{3} = 0$
$P(W|B_2) = \dfrac{1}{3}$
$P(W|B_3) = \dfrac{3}{3} = 1$
Using the law of total probability, the probability of selecting a white ball is:
$P(W) = P(B_1)P(W|B_1) + P(B_2)P(W|B_2) + P(B_3)P(W|B_3)$
$P(W) = \left(\dfrac{1}{6} \times 0\right) + \left(\dfrac{2}{6} \times \dfrac{1}{3}\right) + \left(\dfrac{3}{6} \times 1\right)$
$P(W) = 0 + \dfrac{2}{18} + \dfrac{3}{6} = \dfrac{1}{9} + \dfrac{1}{2} = \dfrac{2 + 9}{18} = \dfrac{11}{18}$
Using Bayes' theorem, the probability that the white ball came from Bag III is:
$P(B_3|W) = \dfrac{P(B_3)P(W|B_3)}{P(W)}$
$P(B_3|W) = \dfrac{\dfrac{3}{6} \times 1}{\dfrac{11}{18}} = \dfrac{1}{2} \times \dfrac{18}{11} = \dfrac{9}{11}$
Answer: $\dfrac{9}{11}$
The probabilities of choosing the bags are given by $P(B_1) = \dfrac{1}{6}$, $P(B_2) = \dfrac{2}{6} = \dfrac{1}{3}$, and $P(B_3) = \dfrac{3}{6} = \dfrac{1}{2}$.
The conditional probabilities of selecting a white ball from each bag are:
$P(W|B_1) = \dfrac{0}{3} = 0$
$P(W|B_2) = \dfrac{1}{3}$
$P(W|B_3) = \dfrac{3}{3} = 1$
Using the law of total probability, the probability of selecting a white ball is:
$P(W) = P(B_1)P(W|B_1) + P(B_2)P(W|B_2) + P(B_3)P(W|B_3)$
$P(W) = \left(\dfrac{1}{6} \times 0\right) + \left(\dfrac{2}{6} \times \dfrac{1}{3}\right) + \left(\dfrac{3}{6} \times 1\right)$
$P(W) = 0 + \dfrac{2}{18} + \dfrac{3}{6} = \dfrac{1}{9} + \dfrac{1}{2} = \dfrac{2 + 9}{18} = \dfrac{11}{18}$
Using Bayes' theorem, the probability that the white ball came from Bag III is:
$P(B_3|W) = \dfrac{P(B_3)P(W|B_3)}{P(W)}$
$P(B_3|W) = \dfrac{\dfrac{3}{6} \times 1}{\dfrac{11}{18}} = \dfrac{1}{2} \times \dfrac{18}{11} = \dfrac{9}{11}$
Answer: $\dfrac{9}{11}$
Hence, the correct answer is option: 'A'.