Question

7.
Suppose you choose a marble from a bag containing 3 red marbles, 3 white marbles, and 5 blue marbles. You return the first marble to the bag and then choose again. Find P (red and blue).

Answer 1

To find the probability of choosing one red marble and one blue marble when you select from the bag, we can break the problem down into two scenarios:

1. Draw a red marble first and then a blue marble.
2. Draw a blue marble first and then a red marble.

Let's calculate the probability for each scenario and then sum them up.

### Total marbles in the bag:
- Red marbles: 3
- White marbles: 3
- Blue marbles: 5
- **Total marbles** = 3 + 3 + 5 = 11

### Scenario 1: Red first, then blue
1. Probability of drawing a red marble first:
\[
P(\text{Red first}) = \frac{3}{11}
\]
2. After returning the red marble, draw a blue marble:
\[
P(\text{Blue second}) = \frac{5}{11}
\]
3. Combined probability for this scenario:
\[
P(\text{Red first, Blue second}) = P(\text{Red first}) \times P(\text{Blue second}) = \frac{3}{11} \times \frac{5}{11} = \frac{15}{121}
\]

### Scenario 2: Blue first, then red
1. Probability of drawing a blue marble first:
\[
P(\text{Blue first}) = \frac{5}{11}
\]
2. After returning the blue marble, draw a red marble:
\[
P(\text{Red second}) = \frac{3}{11}
\]
3. Combined probability for this scenario:
\[
P(\text{Blue first, Red second}) = P(\text{Blue first}) \times P(\text{Red second}) = \frac{5}{11} \times \frac{3}{11} = \frac{15}{121}
\]

### Total probability
Now we add the probabilities of both scenarios:
\[
P(\text{Red and Blue}) = P(\text{Red first, Blue second}) + P(\text{Blue first, Red second}) = \frac{15}{121} + \frac{15}{121} = \frac{30}{121}
\]

Therefore, the probability of selecting one red marble and one blue marble in two draws (with replacement) is:
\[
\boxed{\frac{30}{121}}
\]

Answer 2

what does p=