Question

Let \(u_1\) and \(u_2\) be two urns such that \(u_1\) contains 3 white, 2 red balls and \(u_2\) contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from urn \(u_1\) and put into \(u_2\). However, if tail appears, then 2 balls are drawn at random from \(u_1\) and put into \(u_2\). Now, 1 ball is drawn at random from \(u_2\). Then, probability of the drawn ball from \(u_2\) being white is

• A. \(\frac{13}{30}\)
• B. \(\frac{23}{30}\)
• C. \(\frac{19}{30}\)
• D. \(\frac{11}{30}\)

Hint:

Break into cases based on the coin toss outcome and use conditional probability.

Answer 1

The Correct Option is B

To find the probability of drawing a white ball from urn \(u_2\), we need to consider the two scenarios based on the coin toss:

1. Scenario 1: Head appears.
   • One ball is drawn from \(u_1\) (which contains 3 white and 2 red balls) and added to \(u_2\).
   • Probability of drawing a white ball from \(u_1\) is \(\frac{3}{5}\).
   • Probability of drawing a red ball from \(u_1\) is \(\frac{2}{5}\).
   • If a white ball is added to \(u_2\), then \(u_2\) will have 2 white balls.
   • If a red ball is added, \(u_2\) will still have 1 white ball.
2. Scenario 2: Tail appears.
   • Two balls are drawn from \(u_1\) and added to \(u_2\).
   • Total ways to choose 2 balls from 5 is \(\binom{5}{2} = 10\).
   • Ways to choose 2 white balls: \(\binom{3}{2} = 3\).
   • Ways to choose 1 white and 1 red ball: \(\binom{3}{1} \times \binom{2}{1} = 6\).
   • Ways to choose 2 red balls: \(\binom{2}{2} = 1\).

Now, let's analyze the probability of a white ball being drawn from \(u_2\):

1. For Scenario 1 (Head):
   • Probability of head: \(0.5\).
   • Probability that \(u_2\) has 2 white balls: \(0.5 \times \frac{3}{5} = \frac{3}{10}\).
   • Probability that \(u_2\) has 1 white and 1 red ball: \(0.5 \times \frac{2}{5} = \frac{1}{5}\).
2. For Scenario 2 (Tail):
   • Probability of tail: \(0.5\).
   • If two white balls are added, \(u_2\) will have 3 white balls: \(0.5 \times \frac{3}{10} = \frac{3}{20}\).
   • If one white and one red ball are added, \(u_2\) will have 2 white balls: \(0.5 \times \frac{6}{10} = \frac{3}{10}\).
   • If two red balls are added, \(u_2\) will have 1 white ball: \(0.5 \times \frac{1}{10} = \frac{1}{20}\).

Finally, calculate the probability of drawing a white ball from \(u_2\) after all transfers:

• From head (2 white in \(u_2\)): probability of white is \(\frac{3}{10} \times \frac{2}{2} = \frac{3}{10}\), and from (1 white, 1 red): \(\frac{1}{5} \times \frac{1}{2} = \frac{1}{10}\).
• From tail (3 white in \(u_2\)): probability of white is \(\frac{3}{20} \times \frac{3}{3} = \frac{3}{20}\); from (2 white, 1 red): \(\frac{3}{10} \times \frac{2}{3} = \frac{1}{5}\); and from (1 white, 2 red): \(\frac{1}{20} \times \frac{1}{3} = \frac{1}{60}\).

Adding them up, the total probability of drawing a white ball from \(u_2\) is calculated as:

\(\frac{3}{10} + \frac{1}{10} + \frac{3}{20} + \frac{1}{5} + \frac{1}{60} = \frac{23}{30}\).

Thus, the probability of drawing a white ball from \(u_2\) is \(\frac{23}{30}\).