Question 51

P(E ∪ G) cannot equal 0.9.

The union of events E and G, denoted as P(E ∪ G), represents the probability of either event occurring. Given that P(E) = 0.6 and P(G) = 0.2, the maximum possible value for P(E ∪ G) is 0.8, making 0.9 an impossible value.

A) 0.6

This value is possible for P(E ∪ G) because it is equal to the probability of event E alone, and thus event G could either occur or not occur without exceeding the limits set by the individual probabilities.

B) 0.7

P(E ∪ G) can equal 0.7, as it falls within the range of possible probabilities derived from the individual probabilities of events E and G. It represents a scenario where both events contribute to the total probability without exceeding the sum of their individual probabilities.

C) 0.8

This value is the maximum possible for P(E ∪ G) given the probabilities of P(E) and P(G). It reflects the situation where both events E and G occur together without exceeding the total probability limit.

D) 0.9

This value is impossible for P(E ∪ G) because it exceeds the combined maximum of P(E) and P(G), which is 0.8. Thus, it cannot be a valid probability for the union of these two events.

Conclusion

The impossible value for P(E ∪ G) is 0.9, as it surpasses the maximum probability derived from the individual events' probabilities. Options A, B, and C are all viable probabilities that lie within the acceptable range, while D clearly exceeds the permissible limits, confirming it as the only impossible option.