Two persons, A and B, have respectively n+1 and n coins, which they toss simultaneously. Then the probability that A will have more heads that B is

Options

A. `1/2`
B. `gt(1)/(2)`
C. `lt(1)/(2)`
D. none

Answer

To solve the problem, we need to find the probability that person A, who has \( n + 1 \) coins, will have more heads than person B, who has \( n \) coins, after tossing their coins simultaneously.

### Step-by-Step Solution:

1. **Understanding the Coin Tosses**:
- Person A has \( n + 1 \) coins.
- Person B has \( n \) coins.
- Both toss their coins simultaneously.

2. **Defining the Random Variables**:
- Let \( X_A \) be the number of heads obtained by A.
- Let \( X_B \) be the number of heads obtained by B.
- We want to find \( P(X_A > X_B) \).

3. **Possible Outcomes**:
- The number of heads \( X_A \) can take values from \( 0 \) to \( n + 1 \).
- The number of heads \( X_B \) can take values from \( 0 \) to \( n \).

4. **Condition for A to Have More Heads**:
- For A to have more heads than B, the possible scenarios are:
- If \( X_B = k \) (where \( k \) can be \( 0, 1, 2, \ldots, n \)), then A must have \( X_A \) equal to \( k + 1 \) or more.
- Thus, the condition can be summarized as:
- \( P(X_A > X_B) = \sum_{k=0}^{n} P(X_A > k) P(X_B = k) \)

5. **Calculating Probabilities**:
- The probability that A gets more than \( k \) heads is given by:
\[ P(X_A > k) = \frac{(n + 1) - k}{2^{n + 1}} \]
- The probability that B gets exactly \( k \) heads is given by:
\[ P(X_B = k) = \binom{n}{k} \left(\frac{1}{2}\right)^n \]

6. **Combining the Probabilities**:
- Therefore, we can express the overall probability as:
\[ P(X_A > X_B) = \sum_{k=0}^{n} P(X_A > k) P(X_B = k) = \sum_{k=0}^{n} \frac{(n + 1 - k)}{2^{n + 1}} \cdot \binom{n}{k} \left(\frac{1}{2}\right)^n \]

7. **Final Calculation**:
- This simplifies to:
\[ P(X_A > X_B) = \frac{1}{2} \text{ (since A has one extra coin)} \]

Thus, the probability that A will have more heads than B is \( \frac{1}{2} \).