A coin is tossed thrice. If event E denotes the number of heads is odd and event F denotes the number of tails is odd, then find the cases favourable to the event `EnnF`.

Answer

To solve the problem, we need to determine the cases that are favorable to the event \( E \cap F \), where:

- Event \( E \): The number of heads is odd.
- Event \( F \): The number of tails is odd.

Since a coin is tossed thrice, we can have the following outcomes for the number of heads (H) and tails (T):

1. **Total Outcomes**: When a coin is tossed three times, the total number of outcomes is \( 2^3 = 8 \). The possible outcomes are:
- HHH
- HHT
- HTH
- HTT
- THH
- THT
- TTH
- TTT

2. **Count Heads and Tails**: We will count the number of heads and tails for each outcome:
- HHH: 3 heads, 0 tails
- HHT: 2 heads, 1 tail
- HTH: 2 heads, 1 tail
- HTT: 1 head, 2 tails
- THH: 2 heads, 1 tail
- THT: 1 head, 2 tails
- TTH: 1 head, 2 tails
- TTT: 0 heads, 3 tails

3. **Identify Odd Counts**:
- For event \( E \) (number of heads is odd): The outcomes with an odd number of heads are HTT, THT, TTH (1 head).
- For event \( F \) (number of tails is odd): The outcomes with an odd number of tails are HHT, HTH, HTT, THT, TTH (1 or 3 tails).

4. **Find Intersection \( E \cap F \)**: We need to find the outcomes that satisfy both conditions (odd heads and odd tails):
- From the outcomes with odd heads (HTT, THT, TTH), we can check which of these have an odd number of tails:
- HTT: 1 head, 2 tails (not odd tails)
- THT: 1 head, 2 tails (not odd tails)
- TTH: 1 head, 2 tails (not odd tails)

None of the outcomes with an odd number of heads have an odd number of tails.

5. **Conclusion**: Therefore, there are no cases favorable to the event \( E \cap F \).

### Final Answer:
The number of cases favorable to the event \( E \cap F \) is **0**.