Question

A bag contains 3 red beans, 2 blue beans, and 5 white beans. If two beans are drawn with replacement, what is the probability of drawing a blue and white bean?

Answer 1

To calculate the probability of drawing one blue bean and one white bean from the bag with replacement, we first need to find the total number of beans in the bag.

The bag contains:
- 3 red beans
- 2 blue beans
- 5 white beans

The total number of beans is:
\[
3 + 2 + 5 = 10 \text{ beans}
\]

Next, we can calculate the probabilities of drawing each type of bean:
- The probability of drawing a blue bean (P(B)) is:
\[
P(B) = \frac{\text{Number of blue beans}}{\text{Total number of beans}} = \frac{2}{10} = \frac{1}{5}
\]

- The probability of drawing a white bean (P(W)) is:
\[
P(W) = \frac{\text{Number of white beans}}{\text{Total number of beans}} = \frac{5}{10} = \frac{1}{2}
\]

Since we are drawing beans with replacement, the events are independent. We want the probability of drawing one blue bean and one white bean. This can happen in two different sequences:

1. Drawing a blue bean first, then a white bean.
2. Drawing a white bean first, then a blue bean.

Let's calculate the probabilities of both sequences:

1. **Blue first, then white**:
\[
P(B \text{ then } W) = P(B) \times P(W) = \frac{1}{5} \times \frac{1}{2} = \frac{1}{10}
\]

2. **White first, then blue**:
\[
P(W \text{ then } B) = P(W) \times P(B) = \frac{1}{2} \times \frac{1}{5} = \frac{1}{10}
\]

Now, to find the total probability of drawing one blue and one white bean in either order, we sum the probabilities of the two sequences:
\[
P(\text{One blue and one white}) = P(B \text{ then } W) + P(W \text{ then } B) = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}
\]

Thus, the probability of drawing one blue bean and one white bean is:
\[
\boxed{\frac{1}{5}}
\]