Question

Consider 6 different-color marbles that are randomly selected. Create a scenario for event A and a scenario for event B , such that P(A and B)=1/30 . State the result of event A , the result of event B , and explain if the two events are independent or dependent.

Answer 1

Let's define two scenarios involving a set of 6 different-color marbles: red, blue, green, yellow, orange, and purple.

### Scenario for Event A
**Event A**: Selecting a marble that is either red or blue.

- The possible outcomes for selecting a marble are {red, blue, green, yellow, orange, purple}.
- The favorable outcomes for event A (selecting red or blue) are {red, blue}.
- Therefore, the probability of event A occurring is:
\[
P(A) = \frac{2}{6} = \frac{1}{3}
\]

### Scenario for Event B
**Event B**: Selecting a marble that is either yellow or orange.

- The favorable outcomes for event B (selecting yellow or orange) are {yellow, orange}.
- Therefore, the probability of event B occurring is:
\[
P(B) = \frac{2}{6} = \frac{1}{3}
\]

### Determine P(A and B)
Now, let's determine \( P(A \text{ and } B) \). In the context described, events A and B involve selecting a single marble, which means:

- To satisfy both events A and B simultaneously under the context of selecting a single marble, we notice that the outcomes of A and B do not overlap (none of the marbles are counted in both events).
- Thus, \( P(A \text{ and } B) = 0 \) given that we cannot have a single marble that is both red/blue (event A) and yellow/orange (event B).

### Truth for P(A and B) = 1/30
To construct the scenario where \( P(A \text{ and } B) = \frac{1}{30} \), we need to adjust our approach:
1. Let’s assume we select two marbles instead of one.

**Revised Event A**: Selecting one marble that is either red or blue.
- Thus, the outcome set remains the same, but now we can select two marbles.

**Revised Event B**: Selecting one marble that is either yellow or orange.

Each marble drawn can be viewed as an independent event since we want to find the combined probability of drawing a red or blue followed by a yellow or orange.

### Calculate Revised P(A and B)
If we select two marbles (first for A and then for B):
- Total outcomes for selecting any 2 from 6 are \( C(6, 2) = 15 \).

1. Selecting red or blue (for A) can be done in \( C(2, 1) = 2 \) ways.
2. Selecting yellow or orange (for B) can also be done in \( C(2, 1) = 2 \) ways.

Now, we multiply the probabilities:
- Number of successful outcomes for (A and B) = 2 (for A) * 2 (for B) = 4.

Hence, the probability \( P(A \text{ and } B) \):
\[
P(A \text{ and } B) = \frac{4}{15}
\]
We need to adjust our required condition \( P(A \text{ and } B) = \frac{1}{30} \) as the events constructed are not achieving this value under standard definitions.

### Conclusion
- With independent events amid drawing two marbles, the adjusted scenario remains on drawing without replacement alterations, which tweaks our outcomes.
- The independence or dependence of these events revolves around how the selection affects subsequent selection. Hence they would typically yield **dependent** probabilities based on their connections within the selection process.

Please observe that adjustments can be made to ensure true alignment and coherent math in adherence to specified conditions!