In a scenario involving 6 different-color marbles—red, blue, green, yellow, orange, and purple—let's define two events. Event A could be the selection of a blue marble, while event B could be the selection of a green marble. In this situation, when we randomly select a marble from the collection, we can calculate the individual probabilities of these events. The probability of selecting the blue marble, P(A), is 1 out of 6, or \( \frac{1}{6} \), since there are 6 distinct marbles. Similarly, the probability of selecting the green marble, P(B), is also \( \frac{1}{6} \).
Now, to find \( P(A \text{ and } B) \), we need to consider the combined scenario of selecting both a blue and a green marble. However, since we are only drawing one marble at a time, these two events cannot occur simultaneously in a single draw, implying that \( P(A \text{ and } B) \) is actually 0. Hence, we assert that the occurrence of event A has no effect on the occurrence of event B, leading us to conclude that the events are independent when considering alternative selections.
For instance, if we instead designed a scenario with a sequence of draws where we first draw a marble and keep it aside, then replace it back into the pool before making a second choice, we might frame the events differently. Nevertheless, as it stands with the definitions above and under simple random selection, event A's outcome does not influence event B's outcome. Thus, in summary, selecting a blue marble does not change the probability of later selecting a green marble, affirming that events A and B are independent.
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