Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5), then `barE` is

Answer

To find \( \bar{E} \) (the complement of event \( E \)), we follow these steps:

### Step 1: Identify the Sample Space \( S \)
The sample space \( S \) is given as:
\[ S = \{1, 2, 3, 4, 5, 6\} \]

### Step 2: Identify the Event \( E \)
The event \( E \) is given as:
\[ E = \{1, 3, 5\} \]

### Step 3: Determine the Complement of \( E \)
The complement of \( E \), denoted as \( \bar{E} \), consists of all the elements in the sample space \( S \) that are not in \( E \).

To find \( \bar{E} \), we subtract the elements of \( E \) from \( S \):
\[ \bar{E} = S - E \]

### Step 4: Perform the Set Subtraction
Now we will remove the elements of \( E \) from \( S \):
- From \( S \), we remove \( 1, 3, \) and \( 5 \).

Thus, we are left with:
\[ \bar{E} = \{2, 4, 6\} \]

### Final Answer
The complement of event \( E \) is:
\[ \bar{E} = \{2, 4, 6\} \]

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