91Ó°ÊÓ

Mutually exclusive events are a fundamental concept in probability. These are events that cannot occur at the same time. Imagine flipping a coin – you can't get both heads and tails in a single flip. When events are mutually exclusive, the probability of both events occurring together is zero. This is expressed in mathematical terms as \( P(A \cap B) = 0 \).

In our problem, events G and H are mutually exclusive. Thus, \( P(G \cap H) = 0 \). If we were to consider the statement \( P(H|G) = 0.4 \), it implies that the occurrence of event G affects event H. However, since G and H are mutually exclusive, once G happens, the probability of H happening is zero. Therefore, the statement \( P(H|G) = 0.4 \) must be false.

Conditional probability is a measure of the probability of an event occurring, given that another event has already occurred. The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \).

In the context of our problem, \( P(H|G) \) would refer to the probability of H occurring given that G has already occurred. Since G and H are mutually exclusive (they can't happen together), \( P(H \cap G) = 0 \). This simplifies the equation to \( P(H|G) = \frac{0}{P(G)} = 0 \). By definition, if events are mutually exclusive, conditional probabilities like \( P(H|G) \) should be zero, unless the original probability of the conditioning event (here, \( P(G) \)) is zero.

Two events are considered independent if the occurrence of one does not affect the probability of the other. Independence is checked using the formula \( P(A \cap B) = P(A) \times P(B) \). If this holds true, the events are independent.

For mutually exclusive events, they cannot both happen, which automatically makes them dependent. If \( P(A \cap B) = 0 \) (as is the case with mutually exclusive events), the formula \( P(A) \times P(B) \) would not equal zero unless one or both probabilities are zero. Thus, events G and H in our exercise are dependent because knowing that one occurs tells us that the other cannot.