Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from ur...
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Official solution
Let's denote the events as follows :
Let A1,A2, and A3 be the events that urns A, B, and C are chosen, respectively.
Let B be the event that a black ball is drawn.
We need to find the probability that the chosen urn is A given that a black ball is drawn, which is P(A1∣B).
Using Bayes' theorem, we have :
\[ P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)} \]
First, we calculate each term individually :
1. The probability of choosing any urn, since they are chosen at random :
\[ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} \]
2. The probability of drawing a black ball from each urn :
Urn A: \[ P(B|A_1) = \frac{5}{12} \]
Urn B: \[ P(B|A_2) = \frac{7}{12} \]
Urn C: \[ P(B|A_3) = \frac{6}{12} = \frac{1}{2} \]
3. The total probability of drawing a black ball, P(B) :
\[ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) \]
\[ P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} \]
\[ P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} \]
\[ P(B) = \frac{18}{36} = \frac{1}{2} \]
Now, substitute these into Bayes' theorem :
\[ P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)} \]
\[ P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}} \]
\[ P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}} \]
\[ P(A_1|B) = \frac{5}{18} \]
Therefore, the probability that the black ball drawn is from urn A is option C :
\[ \frac{5}{18} \]