Among 3 bags B1, B_{2} B_{3} there were 4 blue pens and 5 red pens in B., 4 blue pens and 3 red pens in B_{2} 2 blue pens and 4 red pens in B. One pen is drawn from each bag. What is the probability that out of these 3 pens drawn, two are blue and one is red.

To solve this problem, we need to calculate the probability that two of the pens drawn are blue and one is red from the three bags \(B_1\), \(B_2\), and \(B_3\).

### Step 1: Define the Probabilities for Each Bag

Let's denote:

- \(B_1\): 4 blue pens and 5 red pens
- \(B_2\): 4 blue pens and 3 red pens
- \(B_3\): 2 blue pens and 4 red pens

The probabilities of drawing a blue or red pen from each bag are as follows:

1. **Bag \(B_1\):**
   - Probability of drawing a blue pen: \( P(B_1 = \text{blue}) = \frac{4}{9} \)
   - Probability of drawing a red pen: \( P(B_1 = \text{red}) = \frac{5}{9} \)

2. **Bag \(B_2\):**
   - Probability of drawing a blue pen: \( P(B_2 = \text{blue}) = \frac{4}{7} \)
   - Probability of drawing a red pen: \( P(B_2 = \text{red}) = \frac{3}{7} \)

3. **Bag \(B_3\):**
   - Probability of drawing a blue pen: \( P(B_3 = \text{blue}) = \frac{2}{6} = \frac{1}{3} \)
   - Probability of drawing a red pen: \( P(B_3 = \text{red}) = \frac{4}{6} = \frac{2}{3} \)

### Step 2: Calculate the Required Probability

We are asked to find the probability that out of the three pens drawn, two are blue and one is red. There are three scenarios to achieve this:

1. **Scenario 1**: Blue from \(B_1\), Blue from \(B_2\), Red from \(B_3\)
   \[
   P(\text{Blue, Blue, Red}) = P(B_1 = \text{blue}) \cdot P(B_2 = \text{blue}) \cdot P(B_3 = \text{red})
   = \frac{4}{9} \cdot \frac{4}{7} \cdot \frac{2}{3}
   = \frac{32}{189}
   \]

2. **Scenario 2**: Blue from \(B_1\), Red from \(B_2\), Blue from \(B_3\)
   \[
   P(\text{Blue, Red, Blue}) = P(B_1 = \text{blue}) \cdot P(B_2 = \text{red}) \cdot P(B_3 = \text{blue})
   = \frac{4}{9} \cdot \frac{3}{7} \cdot \frac{1}{3}
   = \frac{4}{63}
   \]

3. **Scenario 3**: Red from \(B_1\), Blue from \(B_2\), Blue from \(B_3\)
   \[
   P(\text{Red, Blue, Blue}) = P(B_1 = \text{red}) \cdot P(B_2 = \text{blue}) \cdot P(B_3 = \text{blue})
   = \frac{5}{9} \cdot \frac{4}{7} \cdot \frac{1}{3}
   = \frac{20}{189}
   \]

### Step 3: Adding the Probabilities

Now, we add the probabilities of the three mutually exclusive scenarios:

\[
P(\text{Two blue, one red}) = \frac{32}{189} + \frac{4}{63} + \frac{20}{189}
\]

First, convert \(\frac{4}{63}\) to have a common denominator:

\[
\frac{4}{63} = \frac{4 \times 3}{63 \times 3} = \frac{12}{189}
\]

Now, add all three fractions:

\[
P(\text{Two blue, one red}) = \frac{32}{189} + \frac{12}{189} + \frac{20}{189} = \frac{64}{189}
\]

### Final Answer

The probability that out of the three pens drawn, two are blue and one is red is:

\[
\boxed{\frac{64}{189}}
\]

Would you like more details or have any questions?

### Related Questions

1. What is the probability of drawing three blue pens?
2. What is the probability of drawing three red pens?
3. How would the probability change if the contents of \(B_1\) were altered?
4. What if there were more bags with different numbers of blue and red pens?
5. How does the probability change if pens are drawn without replacement?

### Tip

When calculating probabilities involving multiple events, consider each event's outcome independently and use combinations or permutations if order matters.

Calculate the probability of drawing two blue pens and one red pen from three bags containing different numbers of blue and red pens. Includes step-by-step solution and detailed probabilities for each bag (B1, B2, B3).

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