The sample space S of a random experiment is given by
𝑆 = {𝑎, 𝑏, 𝑐, 𝑑}
with probabilities 𝑃(𝑎) = 0.2, 𝑃(𝑏) = 0.3, 𝑃(𝑐) = 0.4, and 𝑃(𝑑) = 0.1.
Let A denote the event {𝑎, 𝑏}, and B the event {𝑏, 𝑐, 𝑑} . Determine the
following probabilities:
(a) 𝑃(𝐴); (b) 𝑃(𝐵); (c) 𝑃(𝐴̅); (d) 𝑃(𝐴 ∪ 𝐵); and (e) 𝑃(𝐴 ∩ 𝐵)

We are given a sample space \( S = \{a, b, c, d\} \) with the following probabilities:

- \( P(a) = 0.2 \)
- \( P(b) = 0.3 \)
- \( P(c) = 0.4 \)
- \( P(d) = 0.1 \)

Let:
- Event \( A = \{a, b\} \)
- Event \( B = \{b, c, d\} \)

Now, let’s solve the given probabilities step by step:

### (a) \( P(A) \)
The probability of event \( A \), which is \( \{a, b\} \), is the sum of the probabilities of the elements in \( A \):
\[
P(A) = P(a) + P(b) = 0.2 + 0.3 = 0.5
\]

### (b) \( P(B) \)
The probability of event \( B \), which is \( \{b, c, d\} \), is the sum of the probabilities of the elements in \( B \):
\[
P(B) = P(b) + P(c) + P(d) = 0.3 + 0.4 + 0.1 = 0.8
\]

### (c) \( P(A^c) \)
The complement of event \( A \), denoted \( A^c \), consists of all elements in the sample space \( S \) that are not in \( A \). Thus, \( A^c = \{c, d\} \).

\[
P(A^c) = P(c) + P(d) = 0.4 + 0.1 = 0.5
\]

### (d) \( P(A \cup B) \)
The union of \( A \) and \( B \), denoted \( A \cup B \), consists of all elements that are in either \( A \) or \( B \). So,
\[
A \cup B = \{a, b, c, d\} = S
\]
Since the union covers the entire sample space, the probability is:
\[
P(A \cup B) = P(S) = 1
\]

### (e) \( P(A \cap B) \)
The intersection of \( A \) and \( B \), denoted \( A \cap B \), consists of all elements that are in both \( A \) and \( B \). So,
\[
A \cap B = \{b\}
\]
The probability of this intersection is simply the probability of \( b \):
\[
P(A \cap B) = P(b) = 0.3
\]

### Summary of Results:
- (a) \( P(A) = 0.5 \)
- (b) \( P(B) = 0.8 \)
- (c) \( P(A^c) = 0.5 \)
- (d) \( P(A \cup B) = 1 \)
- (e) \( P(A \cap B) = 0.3 \)

Would you like further clarification on any of the steps?

Here are 5 additional questions that build on this topic:
1. How would you calculate \( P(A^c \cap B) \)?
2. If \( P(B^c) \) were required, how would you approach it?
3. Can you derive the formula for \( P(A \cup B^c) \)?
4. What if event \( A \) and \( B \) were mutually exclusive, how would \( P(A \cup B) \) change?
5. How do we interpret \( P(A^c \cup B^c) \) in terms of this experiment?

**Tip**: The complement rule \( P(A^c) = 1 - P(A) \) is a useful shortcut to quickly find the probability of the complement of an event.

The sample space S of a random experiment is given by S = {a, b, c, d} with probabilities P(a) = 0.2, P(b) = 0.3, P(c) = 0.4, and P(d) = 0.1. Let A denote the event {a, b}, and B the event {b, c, d}. Determine the following probabilities: (a) P(A); (b) P(B); (c) P(A^c); (d) P(A ∪ B); and (e) P(A ∩ B).

Solve a probability problem involving a sample space S = {a, b, c, d} with assigned probabilities for each element. Learn how to calculate P(A), P(B), P(A^c), P(A ∪ B), and P(A ∩ B) step by step using key probability formulas and theorems. This problem is suitable for students in Grades 9-12.

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