A coin is tossed twice. If E and F denote occurrence of head on first toss and second toss respectively, then what is `P(E cup F)` equal to ?

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. `(3)/(4)`
D. `(1)/(3)`

Answer

To solve the problem, we need to find the probability of the union of two events E and F, where:

- E is the event of getting a head on the first toss.
- F is the event of getting a head on the second toss.

### Step-by-Step Solution:

1. **Identify the Sample Space**:
When a coin is tossed twice, the sample space \( S \) consists of the following outcomes:
\[ S = \{ \text{HH, HT, TH, TT} \} \]
Here, HH = Head on first toss and Head on second toss, HT = Head on first toss and Tail on second toss, TH = Tail on first toss and Head on second toss, TT = Tail on first toss and Tail on second toss.

2. **Determine the Events**:
- Event E (Head on the first toss):
\[ E = \{ \text{HH, HT} \} \]
- Event F (Head on the second toss):
\[ F = \{ \text{HH, TH} \} \]

3. **Calculate the Probability of Event E**:
The number of favorable outcomes for event E is 2 (HH, HT), and the total number of outcomes in the sample space is 4. Thus, the probability of E is:
\[ P(E) = \frac{\text{Number of favorable outcomes for E}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} \]

4. **Calculate the Probability of Event F**:
The number of favorable outcomes for event F is also 2 (HH, TH). Thus, the probability of F is:
\[ P(F) = \frac{\text{Number of favorable outcomes for F}}{\text{Total outcomes}} = \frac{2}{4} = \frac{1}{2} \]

5. **Calculate the Probability of Intersection of E and F**:
The intersection of E and F, denoted as \( E \cap F \), is the event where both E and F occur, which corresponds to the outcome HH. Thus:
\[ E \cap F = \{ \text{HH} \} \]
The number of favorable outcomes for \( E \cap F \) is 1. Therefore, the probability of \( E \cap F \) is:
\[ P(E \cap F) = \frac{1}{4} \]

6. **Use the Formula for Union of Two Events**:
The probability of the union of two events is given by:
\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
Substituting the values we found:
\[ P(E \cup F) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} \]

7. **Simplify the Expression**:
\[ P(E \cup F) = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]

### Final Answer:
\[ P(E \cup F) = \frac{3}{4} \]