There are three urns U1, U2, U3, each with balls of two colours. U1 contains 2 white balls and 3 black balls, U2 contains 3 white balls and 2 black balls and U3 contains 5 white balls and 5 black balls. An urn is chosen at random and a ball is drawn from that urn at random. What is the probability that U2 was chosen given that the ball picked is black in colour?
Probability Problem Setup
This problem involves calculating a conditional probability using Bayes' theorem. We have three urns, each containing a different mix of white and black balls. An urn is chosen randomly, and then a ball is drawn randomly from the chosen urn. We want to find the probability that a specific urn (U2) was chosen, given that the ball drawn is black.
Urn Contents and Probabilities
First, let's list the contents of each urn and the total number of balls:
- Urn U1: 2 white balls, 3 black balls. Total balls = 5.
- Urn U2: 3 white balls, 2 black balls. Total balls = 5.
- Urn U3: 5 white balls, 5 black balls. Total balls = 10.
Since an urn is chosen at random, the probability of choosing any specific urn is equal:
- Probability of choosing U1, \(P(U1) = \frac{1}{3}\)
- Probability of choosing U2, \(P(U2) = \frac{1}{3}\)
- Probability of choosing U3, \(P(U3) = \frac{1}{3}\)
Conditional Probability of Drawing a Black Ball
Now, let's find the probability of drawing a black ball given that a specific urn was chosen. Let B be the event of drawing a black ball.
- From U1: \(P(B | U1) = \frac{\text{Number of black balls in U1}}{\text{Total balls in U1}} = \frac{3}{5}\)
- From U2: \(P(B | U2) = \frac{\text{Number of black balls in U2}}{\text{Total balls in U2}} = \frac{2}{5}\)
- From U3: \(P(B | U3) = \frac{\text{Number of black balls in U3}}{\text{Total balls in U3}} = \frac{5}{10} = \frac{1}{2}\)
Overall Probability of Drawing a Black Ball
To find the overall probability of drawing a black ball, \(P(B)\), we use the Law of Total Probability. A black ball can be drawn if U1 is chosen AND a black ball is drawn from U1, OR if U2 is chosen AND a black ball is drawn from U2, OR if U3 is chosen AND a black ball is drawn from U3.
\[P(B) = P(B | U1) P(U1) + P(B | U2) P(U2) + P(B | U3) P(U3)\] \[P(B) = \left(\frac{3}{5}\right) \left(\frac{1}{3}\right) + \left(\frac{2}{5}\right) \left(\frac{1}{3}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{3}\right)\] \[P(B) = \frac{1}{3} \left(\frac{3}{5} + \frac{2}{5} + \frac{1}{2}\right)\] \[P(B) = \frac{1}{3} \left(\frac{5}{5} + \frac{1}{2}\right)\] \[P(B) = \frac{1}{3} \left(1 + \frac{1}{2}\right)\] \[P(B) = \frac{1}{3} \left(\frac{2}{2} + \frac{1}{2}\right)\] \[P(B) = \frac{1}{3} \left(\frac{3}{2}\right)\] \[P(B) = \frac{3}{6} = \frac{1}{2}\]So, the overall probability of drawing a black ball is \(P(B) = \frac{1}{2}\).
Probability of Choosing U2 Given a Black Ball
We need to find the probability that U2 was chosen given that the ball picked is black, which is \(P(U2 | B)\). We can use Bayes' Theorem:
\[P(U2 | B) = \frac{P(B | U2) P(U2)}{P(B)}\]We already calculated \(P(B | U2) = \frac{2}{5}\), \(P(U2) = \frac{1}{3}\), and \(P(B) = \frac{1}{2}\).
Substitute these values into the formula:
\[P(U2 | B) = \frac{\left(\frac{2}{5}\right) \left(\frac{1}{3}\right)}{\frac{1}{2}}\] \[P(U2 | B) = \frac{\frac{2}{15}}{\frac{1}{2}}\]To divide by a fraction, we multiply by its reciprocal:
\[P(U2 | B) = \frac{2}{15} \times \frac{2}{1}\] \[P(U2 | B) = \frac{4}{15}\]Therefore, the probability that Urn U2 was chosen given that the ball picked is black is \(\frac{4}{15}\).
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