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 urn $\mathrm{A}$ is :

<p>Let's denote the events as follows :</p> <p>Let $A_1, A_2,$ and $A_3$ be the events that urns A, B, and C are chosen, respectively.</p> <p>Let $B$ be the event that a black ball is drawn.</p> <p>We need to find the probability that the chosen urn is A given that a black ball is drawn, which is $P(A_1|B)$.</p> <p>Using Bayes' theorem, we have :</p> <p>$P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)}$</p> <p>First, we calculate each term individually :</p> <p>1. The probability of choosing any urn, since they are chosen at random :</p> <p>$P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}$</p> <p>2. The probability of drawing a black ball from each urn :</p> <p>Urn A: $P(B|A_1) = \frac{5}{12}$</p> <p>Urn B: $P(B|A_2) = \frac{7}{12}$</p> <p>Urn C: $P(B|A_3) = \frac{6}{12} = \frac{1}{2}$</p> <p>3. The total probability of drawing a black ball, $P(B)$ :</p> <p>$P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3)$</p> <p>$$ P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} $$</p> <p>$P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36}$</p> <p>$P(B) = \frac{18}{36} = \frac{1}{2}$</p> <p>Now, substitute these into Bayes' theorem :</p> <p>$P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)}$</p> <p>$P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}}$</p> <p>$P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}}$</p> <p>$P(A_1|B) = \frac{5}{18}$</p> <p>Therefore, the probability that the black ball drawn is from urn A is option C :</p> <p>$\frac{5}{18}$</p>