If A and B be two mutually exclusive events in a sample space such that , `P(A) = 2/5 and P(B) = 1/2 ` then
find ` P(barA nn barB ) `:

Options

A. `4/5`
B. `1/10`
C. `8/9`
D. `13/20`

Answer

To solve the problem step by step, we will find \( P(\bar{A} \cap \bar{B}) \) given that \( P(A) = \frac{2}{5} \) and \( P(B) = \frac{1}{2} \).

### Step 1: Understand the concept of mutually exclusive events
Mutually exclusive events are events that cannot occur at the same time. Therefore, the intersection of two mutually exclusive events \( A \) and \( B \) is empty:
\[ P(A \cap B) = 0 \]

**Hint:** Remember that for mutually exclusive events, the occurrence of one event excludes the occurrence of the other.

### Step 2: Use the formula for the probability of the union of two events
The probability of the union of two events can be calculated using the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Since \( P(A \cap B) = 0 \) for mutually exclusive events, we can simplify this to:
\[ P(A \cup B) = P(A) + P(B) \]

**Hint:** Always check if the events are mutually exclusive before applying this formula.

### Step 3: Substitute the given probabilities into the formula
Substituting the values of \( P(A) \) and \( P(B) \):
\[ P(A \cup B) = \frac{2}{5} + \frac{1}{2} \]

**Hint:** When adding fractions, make sure to find a common denominator.

### Step 4: Find a common denominator and calculate
The common denominator for \( \frac{2}{5} \) and \( \frac{1}{2} \) is 10. Therefore:
\[ P(A) = \frac{2}{5} = \frac{4}{10} \]
\[ P(B) = \frac{1}{2} = \frac{5}{10} \]
Now, adding these:
\[ P(A \cup B) = \frac{4}{10} + \frac{5}{10} = \frac{9}{10} \]

**Hint:** Converting fractions to a common denominator makes addition straightforward.

### Step 5: Calculate the probability of the complement of the union
To find \( P(\bar{A} \cap \bar{B}) \), we can use the complement rule:
\[ P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) \]
Substituting the value we found:
\[ P(\bar{A} \cap \bar{B}) = 1 - \frac{9}{10} = \frac{1}{10} \]

**Hint:** The complement of an event gives you the probability of the event not occurring.

### Final Answer
Thus, the probability \( P(\bar{A} \cap \bar{B}) \) is:
\[ \boxed{\frac{1}{10}} \]