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 the second bag. What is the probability that a ball now drawn from the second bag is black?

Answer 1

Answer on Question#38819, Math, Statistics

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 the second bag. What is the probability that a ball now drawn from the second bag is black?

Solution

Let $B_{1}, B_{2}, W_{1}, W_{2}$ represent, respectively, the drawing of a black ball from bag 1, a black ball from bag 2 and a white ball from bag 2. We are interested in the union of the mutually exclusive events $B_{1} \cap B_{2}$ and $W_{1} \cap B_{2}$ . The various possibilities and their probabilities are illustrated in figure.

Now,

$$\begin{array}{l} P \left[ \left(B _ {1} \cap B _ {2}\right) \text {o r} \left(W _ {1} \cap B _ {2}\right) \right] = P \left(B _ {1} \cap B _ {2}\right) + P \left(W _ {1} \cap B _ {2}\right) = P \left(B _ {1}\right) P \left(B _ {2} \mid B _ {1}\right) + P \left(W _ {1}\right) P \left(B _ {2} \mid W _ {1}\right) \\ = \left(\frac {3}{7}\right) \left(\frac {6}{9}\right) + \left(\frac {4}{7}\right) \left(\frac {5}{9}\right) = \left(\frac {3 8}{6 3}\right). \\ \end{array}$$

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