Question

One bag contains 4 white balls and 3 black balls and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in second bag. What is the probability that a ball now drawn from the second bag is black?

Answer 1

Answer on Question #45626 – Math – Statistics and Probability

Question.

One bag contains 4 white balls and 3 black balls and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in second bag. What is the probability that a ball now drawn from the second bag is black?

Solution.

Let event $H_{1}$ – white ball is drawn from the first bag, $H_{2}$ – black ball is drawn from the first bag, $A$ – black ball is drawn from the second bag after passing the ball from the first bag to the second bag. Then $P(H_{1}) = \frac{4}{7}, P(H_{2}) = \frac{3}{7}, P(A / H_{1}) = \frac{5}{9}, P(A / H_{2}) = \frac{6}{9} = \frac{2}{3}$. We use the formula of total probability:

$$P (A) = P (A / H _ {1}) \cdot P (H _ {1}) + P (A / H _ {2}) \cdot P (H _ {2}) = \frac {5}{9} \cdot \frac {4}{7} + \frac {2}{3} \cdot \frac {3}{7} = \frac {38}{63}.$$

Answer. $\frac{38}{63}$.

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