Problem 16 - Calculus for Biology and Medicine

Question

Assume that \(P\left(A \cap B^{c}\right)=0.1, P\left(B \cap A^{c}\right)=0.5\), and \(P\left((A \cup B)^{c}\right)=0.2\). Find \(P(A \cap B)\).

Step-by-Step Solution

Answer

The probability \(P(A \cap B)\) is 0.2.

Step 1: Understanding the Properties of Probability

First, we need to understand the components given in the problem. We have: - The probability of the intersection of event A and the complement of event B: \(P(A \cap B^c) = 0.1\). - The probability of the intersection of event B and the complement of event A: \(P(B \cap A^c) = 0.5\). - The probability of the complement of the union of A and B: \(P((A \cup B)^c) = 0.2\). We aim to find \(P(A \cap B)\).

Step 2: Applying the Complement Rule

The probability of the complement of the union of A and B, \(P((A\cup B)^c)\), is also known as none of the events occurring. Thus,\[ P(A \cup B) = 1 - P((A \cup B)^c) = 1 - 0.2 = 0.8. \] This gives us the probability that either A or B or both occur.

Step 3: Using the Formula of Union of Two Events

Recall the formula for the probability of the union of two events A and B:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B). \]Also note from the principle of inclusion-exclusion,\[ P(A \cup B) = P(A \cap B) + P(A \cap B^c) + P(B \cap A^c). \]

Step 4: Substituting Known Probabilities

We substitute our known values into the principle of inclusion-exclusion:\[ 0.8 = P(A \cap B) + 0.1 + 0.5. \]Simplifying the equation:\[ 0.8 = P(A \cap B) + 0.6. \]

Step 5: Solving for P(A ∩ B)

We now solve for \(P(A \cap B)\) by isolating it:\[ P(A \cap B) = 0.8 - 0.6 = 0.2. \]

Step 6: Conclusion

The probability of the intersection of events A and B, \(P(A \cap B)\), is 0.2.

Key Concepts

Complement Rule

In probability, the Complement Rule is a fundamental concept that aids in solving complex problems. It refers to the concept where the probability of an event not occurring is simply the complement of it occurring. If you think about rolling a die, having a 1 in 6 probability to land a 3 means there's a 5 in 6 chance not to land one. But it can also be applied to more complicated scenarios. For instance, when it's declared that the probability of neither A nor B happening is 0.2, this represents the complement of A or B occurring. The formula here is:

• \( P(A \, \cup \, B) = 1 - P((A \, \cup \, B)^c) \).

By calculating this using complements, you can find probabilities indirectly and often more easily, eliminating the need to calculate each outcome from scratch.

Union of Events

Events are said to be in 'union' when one, the other, or both events occur. In probability, calculating the union involves understanding each event's individual probabilities and their overlaps. Think of it as trying to determine the chance that out of two activities, at least one is done.

• The key formula for two events A and B is: \( P(A \, \cup \, B) = P(A) + P(B) - P(A \, \cap \, B) \).

The subtraction of the intersection ensures no double-counting occurs for simultaneous events. For instance, with A as playing a sport and B playing an instrument, both may happen. Accounting for their overlap avoids overstating the probability.
Within our specific problem, knowing the probability of 'neither A nor B' lets us calculate their union by elimination, exemplifying how each formula intertwines.

Inclusion-Exclusion Principle

The principle of inclusion-exclusion helps calculate the probability of overlapping events. Imagine wanting to know the probability of either visiting the park (A) or the zoo (B). You reach a solution by considering individual probabilities but mustn't omit the possibility of visiting both.

• The inclusion-exclusion principle formula becomes: \( P(A \, \cup \, B) = P(A) + P(B) - P(A \, \cap \, B) \).

This balances out overlap to prevent overcounting where both options occur. In our exercise, besides the union probability, specific intersections like \( P(A \, \cap \, B^c) \) and \( P(B \, \cap \, A^c) \) are given. They illustrate cases where only one event happens.
By substituting these into the inclusion-exclusion formula, alongside simplification, we deduce the probability of both A and B occurring together efficiently.