Question

In a bag, there are 4 blue marbles, 5 red marbles, and 3 green marbles. What is the probability of selecting a blue marble and not replacing it, then a red marble?

Answer 1

To find the probability of selecting a blue marble and then a red marble without replacement, we can follow these steps:

1. **Total number of marbles initially**:
\[
\text{Total} = 4 \text{ (blue)} + 5 \text{ (red)} + 3 \text{ (green)} = 12 \text{ marbles}
\]

2. **Probability of selecting a blue marble first**:
The probability of selecting a blue marble on the first draw is the number of blue marbles divided by the total number of marbles:
\[
P(\text{Blue first}) = \frac{4 \text{ (blue)}}{12 \text{ (total)}} = \frac{4}{12} = \frac{1}{3}
\]

3. **Total number of marbles after removing one blue marble**:
After the blue marble is removed, there are:
\[
12 - 1 = 11 \text{ marbles left}
\]

4. **Probability of selecting a red marble second**:
Now, the number of red marbles is still 5. Thus, the probability of selecting a red marble after having removed one blue marble is:
\[
P(\text{Red second} \mid \text{Blue first}) = \frac{5 \text{ (red)}}{11 \text{ (remaining total)}} = \frac{5}{11}
\]

5. **Combined probability**:
The total probability of both events happening (selecting a blue marble first and then a red marble) is found by multiplying the probability of both events:
\[
P(\text{Blue and then Red}) = P(\text{Blue first}) \times P(\text{Red second} \mid \text{Blue first}) = \left(\frac{1}{3}\right) \times \left(\frac{5}{11}\right) = \frac{5}{33}
\]

Thus, the probability of selecting a blue marble and then a red marble without replacement is \(\frac{5}{33}\).