Three urn contain 1 white, 2 black balls; 3 white, 1 black balls; 2 white, 3 black balls respectively. One ball is taken from each urn. What is the probability that among the balls drawn, there are 2 white and 1 black balls

We are given three urns, and we want to calculate the probability that exactly two white balls and one black ball are drawn when one ball is randomly selected from each urn.

### Urn compositions:
1. Urn 1: 1 white and 2 black balls
2. Urn 2: 3 white and 1 black balls
3. Urn 3: 2 white and 3 black balls

Let us denote:
- \( W_1 \) as the event of drawing a white ball from Urn 1, \( B_1 \) as the event of drawing a black ball from Urn 1.
- \( W_2 \) as the event of drawing a white ball from Urn 2, \( B_2 \) as the event of drawing a black ball from Urn 2.
- \( W_3 \) as the event of drawing a white ball from Urn 3, \( B_3 \) as the event of drawing a black ball from Urn 3.

We need the probability of drawing exactly two white balls and one black ball. There are three ways this can happen:

1. White from Urn 1, White from Urn 2, and Black from Urn 3: \((W_1, W_2, B_3)\)
2. White from Urn 1, Black from Urn 2, and White from Urn 3: \((W_1, B_2, W_3)\)
3. Black from Urn 1, White from Urn 2, and White from Urn 3: \((B_1, W_2, W_3)\)

Now, let’s calculate the probabilities for each case.

### 1. Probability of \( W_1, W_2, B_3 \):
- Probability of \( W_1 \) from Urn 1: \( P(W_1) = \frac{1}{3} \)
- Probability of \( W_2 \) from Urn 2: \( P(W_2) = \frac{3}{4} \)
- Probability of \( B_3 \) from Urn 3: \( P(B_3) = \frac{3}{5} \)

Thus, the probability of this case is:
\[
P(W_1, W_2, B_3) = \frac{1}{3} \times \frac{3}{4} \times \frac{3}{5} = \frac{9}{60} = \frac{3}{20}
\]

### 2. Probability of \( W_1, B_2, W_3 \):
- Probability of \( W_1 \) from Urn 1: \( P(W_1) = \frac{1}{3} \)
- Probability of \( B_2 \) from Urn 2: \( P(B_2) = \frac{1}{4} \)
- Probability of \( W_3 \) from Urn 3: \( P(W_3) = \frac{2}{5} \)

Thus, the probability of this case is:
\[
P(W_1, B_2, W_3) = \frac{1}{3} \times \frac{1}{4} \times \frac{2}{5} = \frac{2}{60} = \frac{1}{30}
\]

### 3. Probability of \( B_1, W_2, W_3 \):
- Probability of \( B_1 \) from Urn 1: \( P(B_1) = \frac{2}{3} \)
- Probability of \( W_2 \) from Urn 2: \( P(W_2) = \frac{3}{4} \)
- Probability of \( W_3 \) from Urn 3: \( P(W_3) = \frac{2}{5} \)

Thus, the probability of this case is:
\[
P(B_1, W_2, W_3) = \frac{2}{3} \times \frac{3}{4} \times \frac{2}{5} = \frac{12}{60} = \frac{1}{5}
\]

### Total Probability:
The total probability of drawing exactly two white balls and one black ball is the sum of the probabilities of the three cases:
\[
P(\text{2 white, 1 black}) = \frac{3}{20} + \frac{1}{30} + \frac{1}{5}
\]
We first express all terms with a common denominator of 60:
\[
P(\text{2 white, 1 black}) = \frac{9}{60} + \frac{2}{60} + \frac{12}{60} = \frac{23}{60}
\]

Thus, the probability of drawing exactly two white balls and one black ball is \( \frac{23}{60} \).

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Would you like further clarification or additional details? Here are five related questions:

1. What is the probability of drawing exactly 3 white balls?
2. What is the probability of drawing exactly 3 black balls?
3. How would the probability change if Urn 2 had 2 white and 2 black balls?
4. What is the probability of drawing at least one white ball?
5. How would you compute the expected number of white balls drawn?

**Tip:** When solving probability problems, breaking them down into independent events and adding up probabilities for each scenario helps avoid mistakes.

Three urn contain 1 white, 2 black balls; 3 white, 1 black balls; 2 white, 3 black balls respectively. One ball is taken from each urn. What is the probability that among the balls drawn, there are 2 white and 1 black balls?

This probability problem explores drawing balls from three urns with varying compositions of white and black balls. The solution demonstrates how to compute the probability of drawing exactly two white and one black ball using probability rules. Suitable for students in Grades 10-12.

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