Question

One bag contains 3 white and 2 black balls, another bag contains 5 white and 3 black balls. If a bag is chosen at random and a ball is drawn from it, what is the chance that it is white?

• A. 3/5
• B. 5/8
• C. 49/80
• D. 1/2

Answer 1

Answer: (C)

49/80

Answer 2

Let $B_1$ be the event that the first bag is chosen, and $B_2$ be the event that the second bag is chosen. Let $W$ be the event that a white ball is drawn.

The first bag contains 3 white and 2 black balls, so there are a total of $3 + 2 = 5$ balls. The second bag contains 5 white and 3 black balls, so there are a total of $5 + 3 = 8$ balls.

Since a bag is chosen at random, the probability of choosing either bag is equal: $P(B_1) = \frac{1}{2}$ $P(B_2) = \frac{1}{2}$

The probability of drawing a white ball given that the first bag was chosen is: $P(W|B_1) = \frac{\text{Number of white balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{3}{5}$

The probability of drawing a white ball given that the second bag was chosen is: $P(W|B_2) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{5}{8}$

Using the Law of Total Probability, the overall probability of drawing a white ball is given by: $P(W) = P(W|B_1)P(B_1) + P(W|B_2)P(B_2)$

Substituting the values: $P(W) = \left(\frac{3}{5}\right) \times \left(\frac{1}{2}\right) + \left(\frac{5}{8}\right) \times \left(\frac{1}{2}\right)$ $P(W) = \frac{3}{10} + \frac{5}{16}$

To add these fractions, we find a common denominator, which is 80: $P(W) = \frac{3 \times 8}{10 \times 8} + \frac{5 \times 5}{16 \times 5}$ $P(W) = \frac{24}{80} + \frac{25}{80}$ $P(W) = \frac{24 + 25}{80}$ $P(W) = \frac{49}{80}$