A coin is tossed repeatedly. A and B call alternately for winning a prize of Rs 30. One who calls correctly first wins the prize. A starts the call. Then the expectationof

Options

A. A is Rs. 10
B. B is Rs. 10
C. A is Rs. 20
D. B is Rs. 20

Answer

To solve the problem, we need to calculate the expected winnings for both A and B when they call alternately for a prize of Rs 30, starting with A.

### Step-by-step Solution:

1. **Understanding the Game**:
- A and B take turns calling heads or tails on a coin toss.
- A starts first. The game continues until one of them calls correctly.

2. **Probability of Winning**:
- The probability of A winning on his first call is \( \frac{1}{2} \) (if he calls correctly).
- If A does not win on his first call (which happens with probability \( \frac{1}{2} \)), then B gets a chance to call.

3. **Calculating A's Expected Winnings**:
- If A wins on his first call, he wins Rs 30. The expected value for this event is:
\[ E_A = \left(\frac{1}{2} \times 30\right) \]
- If A does not win on the first call (probability \( \frac{1}{2} \)), then B calls. If B wins, A gets nothing. If B does not win, it goes back to A. The probability that A wins in the second round is:
\[ E_A = \frac{1}{2} \times 30 + \frac{1}{2} \left(\frac{1}{2} \times 30 + \frac{1}{2} E_A\right) \]
- This can be simplified to:
\[ E_A = \frac{30}{2} + \frac{30}{4} + \frac{1}{4}E_A \]

4. **Setting Up the Equation**:
- Rearranging gives us:
\[ E_A - \frac{1}{4}E_A = 15 + 7.5 \]
\[ \frac{3}{4}E_A = 22.5 \]
\[ E_A = \frac{22.5 \times 4}{3} = 30 \]

5. **Finding B's Expected Winnings**:
- Since the total prize is Rs 30, and A's expected winnings are Rs 20, B's expected winnings can be calculated as:
\[ E_B = 30 - E_A = 30 - 20 = 10 \]

6. **Final Results**:
- The expected winnings for A is Rs 20 and for B is Rs 10.

### Conclusion:
- A's expectation is Rs 20, and B's expectation is Rs 10.