Understanding Probability Calculations with Events

This problem involves calculating probabilities related to events, their complements, unions, and conditional probabilities using given information about the probabilities of events A, B, and their intersection.

We are given the following probabilities:

• P(A) = \({0.6}\)
• P(B) = \({0.5}\)
• P(A ∩ B) = \({0.4}\)

We need to evaluate two statements based on these values.

Analyzing Statement 1: P(A̅ ∪ B) = 0.9

Statement 1 asks us to check if the probability of the complement of A union B is 0.9. Let's calculate P(A̅ ∪ B). We can use the formula for the union of two events:

\({P(E \cup F) = P(E) + P(F) - P(E \cap F)}\)

Here, our events are A̅ and B. So, we need P(A̅), P(B), and P(A̅ ∩ B).

First, calculate P(A̅), the probability of the complement of A:

\({P(A̅) = 1 - P(A) = 1 - 0.6 = 0.4}\)

Next, calculate P(A̅ ∩ B), which is the probability that event B occurs but event A does not. This is equivalent to the probability of B minus the probability of the intersection of A and B:

\({P(A̅ \cap B) = P(B) - P(A \cap B)}\)

\({P(A̅ \cap B) = 0.5 - 0.4 = 0.1}\)

Now, substitute these values into the union formula for A̅ and B:

\({P(A̅ \cup B) = P(A̅) + P(B) - P(A̅ \cap B)}\)

\({P(A̅ \cup B) = 0.4 + 0.5 - 0.1}\)

\({P(A̅ \cup B) = 0.9 - 0.1 = 0.8}\)

Statement 1 claims that P(A̅ ∪ B) = 0.9. Our calculation shows P(A̅ ∪ B) = 0.8. Therefore, statement 1 is incorrect.

Analyzing Statement 2: P(B̅ | A̅) = 0.6

Statement 2 asks us to check the conditional probability of the complement of B given the complement of A. The formula for conditional probability is:

\({P(F | E) = \frac{P(E \cap F)}{P(E)}}\)

Here, our events are F = B̅ and E = A̅. So, we need P(B̅ ∩ A̅) and P(A̅).

We already calculated P(A̅):

\({P(A̅) = 0.4}\)

Next, calculate P(B̅ ∩ A̅), which is the probability that neither A nor B occurs. Using De Morgan's laws, this is equivalent to the probability of the complement of the union of A and B:

\({P(B̅ \cap A̅) = P((A \cup B)̅) = 1 - P(A \cup B)}\)

First, calculate P(A ∪ B), the probability of the union of A and B:

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

\({P(A \cup B) = 0.6 + 0.5 - 0.4}\)

\({P(A \cup B) = 1.1 - 0.4 = 0.7}\)

Now, calculate P(B̅ ∩ A̅):

\({P(B̅ \cap A̅) = 1 - P(A \cup B) = 1 - 0.7 = 0.3}\)

Finally, calculate the conditional probability P(B̅ | A̅):

\({P(B̅ | A̅) = \frac{P(B̅ \cap A̅)}{P(A̅)} = \frac{0.3}{0.4}}\)

\({P(B̅ | A̅) = \frac{3}{4} = 0.75}\)

Statement 2 claims that P(B̅ | A̅) = 0.6. Our calculation shows P(B̅ | A̅) = 0.75. Therefore, statement 2 is incorrect.

Conclusion on Statements

Based on our calculations:

• Statement 1: P(A̅ ∪ B) = 0.9 is Incorrect (calculated value is 0.8).
• Statement 2: P(B̅ | A̅) = 0.6 is Incorrect (calculated value is 0.75).

Therefore, neither of the statements is correct.

Revision Table: Key Probabilities Calculated

Additional Information on Probability Concepts

When dealing with probability problems involving multiple events like A and B, it's helpful to understand the basic concepts and formulas:

• Complement (A̅ or A^c): The event that A does not occur. \({P(A̅) = 1 - P(A)}\).
• Intersection (A ∩ B): The event that both A and B occur.
• Union (A ∪ B): The event that either A or B or both occur. \({P(A \cup B) = P(A) + P(B) - P(A \cap B)}\).
• Conditional Probability (P(A | B)): The probability of event A occurring given that event B has already occurred. \({P(A | B) = \frac{P(A \cap B)}{P(B)}}\), provided \({P(B) > 0}\).
• Mutually Exclusive Events: Events that cannot occur at the same time. \({P(A \cap B) = 0}\).
• Independent Events: Events where the occurrence of one does not affect the probability of the other. \({P(A \cap B) = P(A) \times P(B)}\). This was not assumed in this problem; we used the general formulas.
• De Morgan's Laws in Probability: These relate intersections and unions of complements. For example, \({P((A \cup B)̅) = P(A̅ \cap B̅)}\) and \({P((A \cap B)̅) = P(A̅ \cup B̅)}\). We used the first law to calculate P(A̅ ∩ B̅).

Visualizing these concepts with a Venn diagram can also be very helpful in breaking down the sample space into disjoint regions and calculating probabilities of unions and intersections.