Let's help Luca with the following probability questions based on the given information that events $A$ and $B$ are mutually exclusive.

Given:

• $P(A) = 0.78$
• $P(B) = 0.17$
• Since $A$ and $B$ are mutually exclusive, they cannot both occur simultaneously, meaning $P(A \cap B) = 0$.

1. $P(A \cap B)$ (Intersection of A and B):

Since $A$ and $B$ are mutually exclusive, the probability that both $A$ and $B$ occur at the same time is zero.

$P(A \cap B) = 0$

2. $P(A \cup B)$ (Union of A and B):

The probability of the union of two mutually exclusive events is the sum of their individual probabilities. Since $P(A \cap B) = 0$, the formula simplifies to:

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

Substituting the values:

$P(A \cup B) = 0.78 + 0.17 = 0.95$

3. $P(A | B)$ (Conditional Probability of A given B):

The conditional probability formula is:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Since $P(A \cap B) = 0$ (because the events are mutually exclusive), we have:

$P(A | B) = \frac{0}{0.17} = 0$

Summary:

1. $P(A \cap B) = 0$
2. $P(A \cup B) = 0.95$
3. $P(A | B) = 0$

Would you like further clarification on any of these steps?

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Here are 5 related questions to consider:

1. What is $P(B | A)$, given that $A$ and $B$ are mutually exclusive?
2. How would $P(A \cup B)$ change if $A$ and $B$ were not mutually exclusive?
3. What would $P(A \cap B)$ be if $A$ and $B$ were independent?
4. How does the fact that $A$ and $B$ are mutually exclusive affect the calculation of $P(A \cup B)$?
5. Can two independent events ever be mutually exclusive?

Tip: For mutually exclusive events, the intersection probability is always zero, simplifying many calculations!