Step 1: Understanding the Concept:
This question is about calculating probability. Probability is the ratio of the number of favorable outcomes to the total number of possible outcomes. The event we are interested in is selecting a jelly bean that is "neither red nor purple".
Step 2: Key Formula or Approach:
The probability of an event E is given by:
\[ P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \]
We can solve this in two ways:
1. Directly count the number of jelly beans that are not red or purple.
2. Use the complement rule: find the probability of the opposite event (picking a red or purple jelly bean) and subtract it from 1.
Step 3: Detailed Explanation (Method 1: Direct Counting):
- Total number of jelly beans = 60. This is the total number of possible outcomes.
- We want a jelly bean that is "neither red nor purple". The number of favorable outcomes is the count of all other colors.
- Number of favorable jelly beans = (white) + (green) + (yellow) = 22 + 18 + 11 = 51.
- Alternatively, number of unfavorable jelly beans (red or purple) = 5 (red) + 4 (purple) = 9.
- Number of favorable jelly beans = Total - Unfavorable = 60 - 9 = 51.
- Now, calculate the probability:
\[ P(\text{neither red nor purple}) = \frac{51}{60} \]
- The answer choices are in decimal form, so we need to convert this fraction:
\[ \frac{51}{60} = \frac{17 \times 3}{20 \times 3} = \frac{17}{20} = \frac{17 \times 5}{20 \times 5} = \frac{85}{100} = 0.85 \]
Step 3: Detailed Explanation (Method 2: Complement Rule):
- First, find the probability of the opposite event: picking a red or purple jelly bean.
- Number of red or purple jelly beans = 5 + 4 = 9.
\[ P(\text{red or purple}) = \frac{9}{60} = \frac{3}{20} = 0.15 \]
- The probability of an event not happening is 1 minus the probability that it does happen.
\[ P(\text{neither red nor purple}) = 1 - P(\text{red or purple}) \]
\[ = 1 - 0.15 = 0.85 \]
Step 4: Final Answer:
Both methods yield a probability of 0.85. Therefore, the correct answer is (D).