Question

If P(AUB)=0.9 and P(A∩B)=0.4, what is P(A∩B')?

• A. 0.6
• B. 0.4
• C. 0.2
• D. 0.5

Answer 1

Answer: (D)

0.5

Answer 2

To find $P(A \cap B')$, we use the fundamental identity for probabilities:

$P(A) = P(A \cap B) + P(A \cap B')$

This identity states that the probability of event A can be broken down into two disjoint parts: the probability that A and B both occur ($A \cap B$), and the probability that A occurs but B does not ($A \cap B'$).

From this identity, we can express $P(A \cap B')$ as:

$P(A \cap B') = P(A) - P(A \cap B)$

We are given $P(A \cap B) = 0.4$. So, we need to find $P(A)$.

We are also given $P(A \cup B) = 0.9$. We know the formula for the union of two events:

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

Substitute the given values into this formula:

$0.9 = P(A) + P(B) - 0.4$ Rearranging, we get: $P(A) + P(B) = 0.9 + 0.4$ $P(A) + P(B) = 1.3$

At this point, we have one equation with two unknowns ($P(A)$ and $P(B)$), so we cannot uniquely determine $P(A)$ or $P(B)$ from this equation alone. This implies that we might not be able to uniquely determine $P(A \cap B')$.

Let's re-examine the properties using a Venn diagram approach. The event $A \cup B$ can be expressed as the union of three mutually exclusive (disjoint) events:

1. $A \cap B'$ (A occurs, B does not occur - "A only")
2. $B \cap A'$ (B occurs, A does not occur - "B only")
3. $A \cap B$ (A and B both occur - "Intersection")

So, $P(A \cup B) = P(A \cap B') + P(B \cap A') + P(A \cap B)$.

Substitute the given values:

$0.9 = P(A \cap B') + P(B \cap A') + 0.4$ $P(A \cap B') + P(B \cap A') = 0.9 - 0.4$ $P(A \cap B') + P(B \cap A') = 0.5$

This equation tells us the sum of the probabilities of "A only" and "B only". However, it does not allow us to find $P(A \cap B')$ individually, as $P(B \cap A')$ is also unknown.

If the question expects a unique numerical answer, there must be a specific scenario or an implied relationship. Let's consider the options provided. Let's test the possibility that $B$ is a subset of $A$ ($B \subseteq A$). If $B \subseteq A$: Then $A \cup B = A$, so $P(A \cup B) = P(A) = 0.9$. And $A \cap B = B$, so $P(A \cap B) = P(B) = 0.4$.

Now, let's calculate $P(A \cap B')$ using these values: $P(A \cap B') = P(A) - P(A \cap B)$ $P(A \cap B') = 0.9 - 0.4$ $P(A \cap B') = 0.5$

This value ($0.5$) is one of the options. Let's verify if this scenario is consistent with all given information: $P(A) = 0.9$ $P(B) = 0.4$ $P(A \cap B) = 0.4$ (consistent, since $P(B)=0.4$) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.9 + 0.4 - 0.4 = 0.9$ (consistent with given $P(A \cup B)=0.9$).

This scenario ($B \subseteq A$) is a valid interpretation that leads to one of the options. In the context of multiple-choice questions, if a valid scenario yields one of the options, it is often the intended answer. Without additional information, there are multiple possible values for $P(A \cap B')$, but only one matches an option derived from a common special case.