If P(A|B) < P(A), then which one of the following is correct?
P(B|A) < P(B)
This question explores the relationship between conditional probabilities when the occurrence of one event influences the probability of another. We are given a condition involving P(A|B) and P(A) and need to determine the correct implication for P(B|A) and P(B).
Conditional probability is the probability of an event occurring given that another event has already occurred. The formula for the conditional probability of event A given event B is:
$$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$
Similarly, the conditional probability of event B given event A is:
$$ P(B|A) = \frac{P(A \cap B)}{P(A)} $$
Here, $P(A \cap B)$ represents the probability that both events A and B occur.
We are given that the probability of A occurring given B has occurred is less than the probability of A occurring normally. This means that the occurrence of event B makes event A less likely.
The given condition is:
$$ P(A|B) < P(A) $$
Using the formula for $P(A|B)$, we can substitute:
$$ \frac{P(A \cap B)}{P(B)} < P(A) $$
Assuming $P(B) > 0$ (which must be true for $P(A|B)$ to be defined), we can multiply both sides by $P(B)$:
$$ P(A \cap B) < P(A) \cdot P(B) $$
This inequality tells us that the probability of both A and B happening together is less than the product of their individual probabilities. This indicates that events A and B are not independent; specifically, they are negatively dependent in the sense that one event makes the other less likely.
Now, let's look at the expression for $P(B|A)$:
$$ P(B|A) = \frac{P(A \cap B)}{P(A)} $$
We want to compare $P(B|A)$ with $P(B)$. We already established that $P(A \cap B) < P(A) \cdot P(B)$.
Let's substitute the inequality for $P(A \cap B)$ into the expression for $P(B|A)$:
$$ P(B|A) = \frac{P(A \cap B)}{P(A)} $$
Since $P(A \cap B) < P(A) \cdot P(B)$, we can write:
$$ P(B|A) < \frac{P(A) \cdot P(B)}{P(A)} $$
Assuming $P(A) > 0$ (which must be true for $P(B|A)$ to be defined), we can cancel $P(A)$ from the numerator and the denominator:
$$ P(B|A) < P(B) $$
This result shows that if $P(A|B) < P(A)$, then it must also be true that $P(B|A) < P(B)$. This means that if the occurrence of event B makes event A less likely, then the occurrence of event A also makes event B less likely.
Let's compare our finding with the given options:
Therefore, the correct option is the one stating that $P(B|A) < P(B)$.
We started with the condition that $P(A|B) < P(A)$ and through algebraic manipulation using the definition of conditional probability, we proved that this implies $P(B|A) < P(B)$.
| Given Condition | Implied Relationship | Interpretation |
|---|---|---|
| $P(A|B) < P(A)$ | $P(B|A) < P(B)$ | B makes A less likely, and A makes B less likely (negative dependence) |
| $P(A|B) > P(A)$ | $P(B|A) > P(B)$ | B makes A more likely, and A makes B more likely (positive dependence) |
| $P(A|B) = P(A)$ | $P(B|A) = P(B)$ | A and B are independent |
| Term | Notation | Meaning |
|---|---|---|
| Probability of A | $P(A)$ | Likelihood of event A occurring |
| Probability of A and B | $P(A \cap B)$ or $P(A \text{ and } B)$ | Likelihood of both A and B occurring |
| Probability of A given B | $P(A|B)$ | Likelihood of A occurring, given B has occurred |
| Probability of B given A | $P(B|A)$ | Likelihood of B occurring, given A has occurred |
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