Q105 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from bag A and put into bag B. However if tail appears then 2 balls are drawn at random from bag A and put into bag B. Now one ball is drawn at random from bag B. Given that the drawn ball from B is white, the probability that head appeared on the coin is
- A$\dfrac{19}{30}$
- Correct answerB$\dfrac{12}{23}$
- C$\dfrac{11}{23}$
- D$\dfrac{23}{30}$
Solution
$P(H) = P(T) = \dfrac{1}{2}$
If Head: 1 ball drawn from Bag A (3W, 2R), put into Bag B (1W).
$P(W_B|H) = \dfrac{3}{5} \times 1 + \dfrac{2}{5} \times \dfrac{1}{2} = \dfrac{3}{5} + \dfrac{1}{5} = \dfrac{4}{5}$
If Tail: 2 balls drawn from Bag A, put into Bag B (1W).
$P(2W) = \dfrac{^3C_2}{^5C_2} = \dfrac{3}{10}$, $P(1W1R) = \dfrac{^3C_1 \times ^2C_1}{^5C_2} = \dfrac{6}{10}$, $P(2R) = \dfrac{^2C_2}{^5C_2} = \dfrac{1}{10}$
$P(W_B|T) = \dfrac{3}{10} \times 1 + \dfrac{6}{10} \times \dfrac{2}{3} + \dfrac{1}{10} \times \dfrac{1}{3} = \dfrac{3}{10} + \dfrac{4}{10} + \dfrac{1}{30} = \dfrac{9+12+1}{30} = \dfrac{22}{30} = \dfrac{11}{15}$
By Bayes' theorem:
$P(H|W_B) = \dfrac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5} + \frac{1}{2} \times \frac{11}{15}} = \dfrac{\frac{4}{5}}{\frac{4}{5} + \frac{11}{15}} = \dfrac{\frac{12}{15}}{\frac{12+11}{15}} = \dfrac{12}{23}$
If Head: 1 ball drawn from Bag A (3W, 2R), put into Bag B (1W).
$P(W_B|H) = \dfrac{3}{5} \times 1 + \dfrac{2}{5} \times \dfrac{1}{2} = \dfrac{3}{5} + \dfrac{1}{5} = \dfrac{4}{5}$
If Tail: 2 balls drawn from Bag A, put into Bag B (1W).
$P(2W) = \dfrac{^3C_2}{^5C_2} = \dfrac{3}{10}$, $P(1W1R) = \dfrac{^3C_1 \times ^2C_1}{^5C_2} = \dfrac{6}{10}$, $P(2R) = \dfrac{^2C_2}{^5C_2} = \dfrac{1}{10}$
$P(W_B|T) = \dfrac{3}{10} \times 1 + \dfrac{6}{10} \times \dfrac{2}{3} + \dfrac{1}{10} \times \dfrac{1}{3} = \dfrac{3}{10} + \dfrac{4}{10} + \dfrac{1}{30} = \dfrac{9+12+1}{30} = \dfrac{22}{30} = \dfrac{11}{15}$
By Bayes' theorem:
$P(H|W_B) = \dfrac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5} + \frac{1}{2} \times \frac{11}{15}} = \dfrac{\frac{4}{5}}{\frac{4}{5} + \frac{11}{15}} = \dfrac{\frac{12}{15}}{\frac{12+11}{15}} = \dfrac{12}{23}$
Hence, the correct answer is option: 'B'.