Question

Given that the sample space consists of the following outcomes {1,2,3,4,5,6,7,8,9,10} and the event A={1,3,5,7} and B={2,3, 4,6,8} . Find the following:
a) (A or B)= 8/10
(A and B)= 1/10
(A and Bc)= 3/10
(A or B)c= 2/10

Answer 1

To solve the problem, let's first clarify the definitions and some basic concepts involved:

1. **Sample Space (S)**: The set of all possible outcomes. Here, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

2. **Event A**: The outcomes in event A are {1, 3, 5, 7}.

3. **Event B**: The outcomes in event B are {2, 3, 4, 6, 8}.

4. **Complement of an event (Bc)**: The outcomes not in event B.

Now let's find each of the events indicated:

### a) \( P(A \text{ or } B) \)
This refers to the union of A and B, denoted \( A \cup B \).

- The outcomes in A are {1, 3, 5, 7}.
- The outcomes in B are {2, 3, 4, 6, 8}.

To find \( A \cup B \):
\[
A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8\}
\]
Counting the outcomes in \( A \cup B \) gives us 8 outcomes. The probability is:
\[
P(A \text{ or } B) = \frac{|A \cup B|}{|S|} = \frac{8}{10} = 0.8
\]

### b) \( P(A \text{ and } B) \)
This refers to the intersection of A and B, denoted \( A \cap B \).

- The common outcome in A and B is {3}.

So,
\[
A \cap B = \{3\}
\]
There is 1 outcome in \( A \cap B \). The probability is:
\[
P(A \text{ and } B) = \frac{|A \cap B|}{|S|} = \frac{1}{10} = 0.1
\]

### c) \( P(A \text{ and } B^c) \)
Here \( B^c \) (the complement of B) includes all outcomes in the sample space that are not in B.
\[
B^c = \{1, 5, 7, 9, 10\}
\]

Now, we find \( A \cap B^c \):
The outcomes in A are {1, 3, 5, 7} and in \( B^c \) are {1, 5, 7, 9, 10}.

So,
\[
A \cap B^c = \{1, 5, 7\}
\]
There are 3 outcomes in \( A \cap B^c \). The probability is:
\[
P(A \text{ and } B^c) = \frac{|A \cap B^c|}{|S|} = \frac{3}{10} = 0.3
\]

### d) \( P((A \text{ or } B)^c) \)
\( (A \text{ or } B)^c \) contains all outcomes that are not in \( A \cup B \).
The outcomes in \( A \cup B \) are {1, 2, 3, 4, 5, 6, 7, 8}, so:
\[
(A \text{ or } B)^c = \{9, 10\}
\]
There are 2 outcomes in \( (A \text{ or } B)^c \). The probability is:
\[
P((A \text{ or } B)^c) = \frac{|(A \text{ or } B)^c|}{|S|} = \frac{2}{10} = 0.2
\]

### Summary of the Probabilities
- \( P(A \text{ or } B) = \frac{8}{10} \)
- \( P(A \text{ and } B) = \frac{1}{10} \)
- \( P(A \text{ and } B^c) = \frac{3}{10} \)
- \( P((A \text{ or } B)^c) = \frac{2}{10} \)

Everything checks out with the provided probabilities!