Analyzing Conditional Probability with Given Probabilities

The problem asks us to find the correct relationship for the conditional probability \(P(A|B)\) given the probabilities of two events A and B, \(P(A) = L\) and \(P(B) = M\). Conditional probability is a fundamental concept in probability theory, describing the probability of an event occurring given that another event has already occurred.

Understanding Conditional Probability

The formula for the conditional probability of event A given event B is defined as:

\(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\)

where \(P(A \cap B)\) is the probability of both events A and B occurring (the probability of their intersection), and \(P(B)\) is the probability of event B occurring. For this formula to be defined, we must have \(P(B) > 0\), which means M > 0.

Relating Intersection and Union Probabilities

We know the formula for the probability of the union of two events A and B:

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)

We can rearrange this formula to express the probability of the intersection \(P(A \cap B)\) in terms of the probabilities of A, B, and their union:

\(P(A \cap B) = P(A) + P(B) - P(A \union B)\)

Substituting the given probabilities, we get:

\(P(A \cap B) = L + M - P(A \cup B)\)

Establishing Bounds for the Intersection Probability

The probability of any event must lie between 0 and 1, inclusive. Thus, \(0 \le P(A \cup B) \le 1\). This property helps us find bounds for \(P(A \cap B)\).

From \(P(A \cup B) = L + M - P(A \cap B)\) and the upper bound \(P(A \cup B) \le 1\), we have:

\(L + M - P(A \cap B) \le 1\)

Rearranging this inequality to isolate \(P(A \cap B)\):

\(L + M - 1 \le P(A \cap B)\)

Also, since \(P(A \cap B)\) is a probability, it must be non-negative:

\(P(A \cap B) \ge 0\)

Combining these, we know that \(P(A \cap B)\) must be at least the maximum of 0 and \(L+M-1\). However, for the purpose of finding a lower bound for \(P(A|B)\) in terms of the expression \(\dfrac{L+M-1}{M}\), the inequality \(P(A \cap B) \ge L + M - 1\) is directly useful, provided \(L+M-1\) is a possible lower bound greater than or equal to 0.

Deriving the Inequality for P(A|B)

Now, let's substitute the inequality \(P(A \cap B) \ge L + M - 1\) into the definition of conditional probability \(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\).

Assuming \(P(B) = M > 0\), we can divide the inequality \(P(A \cap B) \ge L + M - 1\) by M without changing the direction of the inequality sign:

\(\dfrac{P(A \cap B)}{M} \ge \dfrac{L + M - 1}{M}\)

Since \(P(A|B) = \dfrac{P(A \cap B)}{M}\), we have:

\(P(A|B) \ge \dfrac{L + M - 1}{M}\)

This inequality provides a lower bound for the conditional probability \(P(A|B)\) based on the probabilities L and M, and the fact that the probability of the union is at most 1.

Evaluating the Options

Let's compare our derived inequality with the given options:

1. \(P(A|B)<\dfrac{L+M-1}{M}\)
2. \(P(A|B)>\dfrac{L+M-1}{M}\)
3. \(P(A|B)\ge\dfrac{L+M-1}{M}\)
4. \(P(A|B)=\dfrac{L+M-1}{M}\)

Our derived inequality \(P(A|B) \ge \dfrac{L + M - 1}{M}\) exactly matches option 3.

Conclusion

Based on the relationship between the probability of intersection and union of events and the properties of probability measures, we derived that \(P(A|B)\) must be greater than or equal to the expression \(\dfrac{L+M-1}{M}\), assuming \(P(B) = M > 0\). This confirms that option 3 is the correct statement.

Probability Concepts Revision Table

Additional Information on Event Probabilities

It's important to remember the fundamental rules that govern probabilities of events. For any events A and B in a sample space S:

• The probability of any event is between 0 and 1: \(0 \le P(A) \le 1\).
• The probability of the impossible event is 0, \(P(\emptyset) = 0\).
• The probability of the sure event (sample space) is 1, \(P(S) = 1\).
• For mutually exclusive events (A and B cannot happen at the same time, \(A \cap B = \emptyset\)), the probability of their union is the sum of their probabilities: \(P(A \cup B) = P(A) + P(B)\).

The inequality \(P(A \cap B) \ge L + M - 1\) arises from the fact that \(P(A \cup B)\) cannot exceed 1. The intersection \(A \cap B\) must be "large enough" for the union \(A \cup B\) (which is \(A \cup B\)) to be "small enough" (less than or equal to 1), given the individual probabilities \(P(A)\) and \(P(B)\).