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:
- 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 \( C(6, 2) = 15 \).
- Selecting red or blue (for A) can be done in \( C(2, 1) = 2 \) ways.
- 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!