Understanding Probability Relationships

The question asks us to compare three different probability terms for two events A and B, given a specific relationship between their probabilities and their range. The terms are the probability of the intersection of A and B, P(A ∩ B), and the two conditional probabilities, P(A|B) and P(B|A).

Analyzing the Given Information

We are given the following conditions:

• 2P(A) = 3P(B)
• 0 < P(A) < P(B) < 1

From the relation 2P(A) = 3P(B), we can express P(A) in terms of P(B):

\begin{equation*} P(A) = \frac{3}{2} P(B) \end{equation*}

This equation implies that \(P(A)\) is 1.5 times \(P(B)\). Since \(P(B) > 0\) (as per the given condition), this means \(P(A) > P(B)\).

We are also given the range for the probabilities: \(0 < P(A) < P(B) < 1\). This tells us that both \(P(A)\) and \(P(B)\) are greater than 0 and less than 1. It also states that \(P(A) < P(B)\). Note that the condition \(2P(A) = 3P(B)\) implies \(P(A) > P(B)\), which contradicts the condition \(P(A) < P(B)\). However, we will proceed by using the relationship \(2P(A) = 3P(B)\) to derive the key relationship between \(P(A)\) and \(P(B)\), which is \(P(A) > P(B)\), and also use the fact that \(P(A) < 1\) and \(P(B) < 1\) from the given range.

Comparing Conditional Probabilities

The formulas for the conditional probabilities are:

• P(A|B) = \frac{P(A \cap B)}{P(B)}
• P(B|A) = \frac{P(A \cap B)}{P(A)}

We want to compare \(P(A|B)\) and \(P(B|A)\). Both expressions have the same numerator, \(P(A \cap B)\). Their denominators are \(P(B)\) and \(P(A)\) respectively.

We derived from \(2P(A) = 3P(B)\) that \(P(A) > P(B)\). When comparing two fractions with the same positive numerator, the fraction with the larger denominator is smaller. Assuming \(P(A \cap B) > 0\) (which must be true for the conditional probabilities to be positive and for strict inequalities to hold in the options), since \(P(A) > P(B)\), it follows that:

\begin{equation*} \frac{P(A \cap B)}{P(A)} < \frac{P(A \cap B)}{P(B)} \end{equation*}

Therefore, \(P(B|A) < P(A|B)\).

Comparing with the Intersection Probability

Now let's compare \(P(A \cap B)\) with \(P(B|A)\) and \(P(A|B)\).

Consider \(P(B|A) = \frac{P(A \cap B)}{P(A)}\). From the given condition \(0 < P(A) < P(B) < 1\), we know that \(P(A) < 1\). If \(P(A \cap B) > 0\), dividing \(P(A \cap B)\) by a number less than 1 (\(P(A)\)) results in a value greater than \(P(A \cap B)\). Thus, \(P(A \cap B) < P(B|A)\).

Consider \(P(A|B) = \frac{P(A \cap B)}{P(B)}\). From the given condition \(0 < P(A) < P(B) < 1\), we know that \(P(B) < 1\). If \(P(A \cap B) > 0\), dividing \(P(A \cap B)\) by a number less than 1 (\(P(B)\)) results in a value greater than \(P(A \cap B)\). Thus, \(P(A \cap B) < P(A|B)\).

Combining the Inequalities

We have established the following relationships:

• \(P(B|A) < P(A|B)\)
• \(P(A \cap B) < P(B|A)\)

Combining these two inequalities, we get the complete ordering:

\begin{equation*} P(A \cap B) < P(B|A) < P(A|B) \end{equation*}

This order assumes \(P(A \cap B) > 0\), which is necessary for the strict inequalities in the options to hold true for non-zero probabilities.

Let's summarize the comparison:

Therefore, the correct ordering is \(P(A \cap B) < P(B|A) < P(A|B)\).