Question

A package contains 4 red, 2 green, 8 purple, and 6 blue jelly beans. What is the probability of choosing a purple jelly bean, eating it, and then choosing a blue jelly bean?
StartFraction 1 over 400 EndFraction
StartFraction 1 over 380 EndFraction
StartFraction 3 over 25 EndFraction
StartFraction 12 over 95 EndFraction

Answer 1

To determine the probability of first choosing a purple jelly bean and then a blue jelly bean from the package, we need to consider the total number of jelly beans as well as the number of each color.

### Step 1: Calculate the total number of jelly beans.

The package contains:
- 4 red
- 2 green
- 8 purple
- 6 blue

Total number of jelly beans:

\[
4 + 2 + 8 + 6 = 20
\]

### Step 2: Calculate the probability of choosing a purple jelly bean.

The number of purple jelly beans is 8. The probability of choosing a purple jelly bean is:

\[
P(\text{purple}) = \frac{\text{number of purple jelly beans}}{\text{total number of jelly beans}} = \frac{8}{20} = \frac{2}{5}
\]

### Step 3: After eating the purple jelly bean, calculate the new total and the number of blue jelly beans.

After choosing one purple jelly bean and eating it:
- Total jelly beans left = \(20 - 1 = 19\)
- The number of blue jelly beans remains 6.

### Step 4: Calculate the probability of choosing a blue jelly bean after eating the purple jelly bean.

The probability of choosing a blue jelly bean now is:

\[
P(\text{blue} | \text{purple eaten}) = \frac{\text{number of blue jelly beans}}{\text{total number of jelly beans left}} = \frac{6}{19}
\]

### Step 5: Calculate the combined probability of both events.

The combined probability of first choosing a purple jelly bean and then a blue jelly bean is given by:

\[
P(\text{purple}) \times P(\text{blue} | \text{purple eaten}) = \left(\frac{2}{5}\right) \times \left(\frac{6}{19}\right) = \frac{12}{95}
\]

### Conclusion

The probability of choosing a purple jelly bean, eating it, and then choosing a blue jelly bean is:

\[
\boxed{\frac{12}{95}}
\]