https://file.aiquickdraw.com/mathbot/file/1726261879023cdstca2k.jpg

Let's go over the problems in the image and solve them one by one.

### (a) Given that A and B are mutually exclusive events and:
- \( P(A \cup B) = 0.50 \)
- \( P(A) = 0.05 \)

Find \( P(B) \) and \( P(A \cap B) \).

#### Solution:

1. **Mutually exclusive events**: For mutually exclusive events, \( P(A \cap B) = 0 \). This is because mutually exclusive events cannot happen at the same time.
   
2. **Formula for union of two events**:
   \[
   P(A \cup B) = P(A) + P(B) - P(A \cap B)
   \]
   Since \( P(A \cap B) = 0 \), the formula simplifies to:
   \[
   P(A \cup B) = P(A) + P(B)
   \]
   Substituting the known values:
   \[
   0.50 = 0.05 + P(B)
   \]
   Solving for \( P(B) \):
   \[
   P(B) = 0.50 - 0.05 = 0.45
   \]
   
   So, \( P(B) = 0.45 \) and \( P(A \cap B) = 0 \).

---

### (b) Suppose \( Q \) and \( S \) are independent events such that:
- The probability that at least one of them occurs is \( \frac{1}{3} \).
- The probability that \( Q \) occurs but \( S \) does not occur is \( \frac{1}{9} \).

Find \( P(S) \).

#### Solution:

Let’s define the given probabilities:
- The probability that at least one occurs is \( P(Q \cup S) = \frac{1}{3} \).
- The probability that \( Q \) occurs and \( S \) does not is \( P(Q \cap S^c) = \frac{1}{9} \).

We know that for independent events:
\[
P(Q \cap S) = P(Q) \cdot P(S)
\]

Also, \( P(Q \cap S^c) = P(Q) - P(Q \cap S) = \frac{1}{9} \).

Using the formula for the union of two events:
\[
P(Q \cup S) = P(Q) + P(S) - P(Q \cap S)
\]
Substitute \( P(Q \cup S) = \frac{1}{3} \):
\[
\frac{1}{3} = P(Q) + P(S) - P(Q) \cdot P(S)
\]
Also, we have the equation \( P(Q) - P(Q) \cdot P(S) = \frac{1}{9} \).

Now, we can solve this system of equations to find \( P(S) \).

---

### (c) Given \( P(A) = 0.55 \) and \( P(A \cup B) = 0.85 \), find \( P(B) \) for the following cases:

#### (i) \( A \) and \( B \) are mutually exclusive:
Since \( A \) and \( B \) are mutually exclusive, \( P(A \cap B) = 0 \).

Using the formula for the union:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substitute the given values:
\[
0.85 = 0.55 + P(B) - 0
\]
Solving for \( P(B) \):
\[
P(B) = 0.85 - 0.55 = 0.30
\]

So, \( P(B) = 0.30 \).

#### (ii) \( A \) and \( B \) are independent:
For independent events, \( P(A \cap B) = P(A) \cdot P(B) \).

Using the formula for the union:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Substitute the given values:
\[
0.85 = 0.55 + P(B) - (0.55 \cdot P(B))
\]
Now solve for \( P(B) \).

#### (iii) \( P(A \mid B) = 0.45 \):
We are given \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} = 0.45 \).

We can also use the formula for the union:
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
Now use these two equations to solve for \( P(B) \).

---

Would you like me to solve part (b) and (c) further with the actual steps, or explain any part in more detail?

### Relative Questions:
1. How can you calculate probabilities when two events are independent?
2. What is the relationship between mutually exclusive and independent events?
3. How do you calculate conditional probability \( P(A \mid B) \)?
4. Can mutually exclusive events be independent?
5. What changes in the formula for union when events are independent?

### Tip:
Mutually exclusive events can never be independent, as the occurrence of one event excludes the occurrence of the other.

Given that A and B are mutually exclusive events, P(A ∪ B) = 0.50, P(A) = 0.05, find P(B) and P(A ∩ B). Suppose Q and S are independent events such that the probability that at least one of them is 1/3 and the probability that Q occurs but S does not occur is 1/9, what is P(S)? Given P(A) = 0.55 and P(A ∪ B) = 0.85, find P(B) when (i) A, B are mutually exclusive (ii) A, B are independent (iii) P(A|B) = 0.45.

Learn how to solve probability problems involving mutually exclusive and independent events. This includes calculating P(A ∪ B), P(B), P(A|B), and understanding how to use key formulas like the addition rule and conditional probability for these types of events. Suitable for undergraduate students.

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