A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let P(K) be the probability of scoring at least K points. The largest value of K such that `P(K)gt1//2` is -

Options

A. a. 14
B. b. 15
C. c. 16
D. d. 17

Explanation

To solve the problem, we need to find the largest value of \( K \) such that the probability \( P(K) \) of scoring at least \( K \) points in 10 tosses of a coin is greater than \( \frac{1}{2} \). The scoring system awards 1 point for each head and 2 points for each tail.

### Step-by-Step Solution:

1. **Understanding the Problem**:
Each coin toss can result in either a head (1 point) or a tail (2 points). If we toss the coin 10 times, let \( H \) be the number of heads and \( T \) be the number of tails. We know that \( H + T = 10 \).

2. **Points Calculation**:
The total score \( S \) can be expressed as:
\[ S = H \cdot 1 + T \cdot 2 = H + 2T \]
Since \( T = 10 - H \), we can rewrite the score as:
\[ S = H + 2(10 - H) = H + 20 - 2H = 20 - H \]

3. **Finding the Score**:
The score \( S \) can take values depending on the number of heads \( H \):
- If \( H = 0 \) (all tails), \( S = 20 \).
- If \( H = 1 \), \( S = 19 \).
- If \( H = 2 \), \( S = 18 \).
- ...
- If \( H = 10 \) (all heads), \( S = 10 \).

Thus, the possible scores range from 10 to 20.

4. **Probability Calculation**:
The probability \( P(K) \) of scoring at least \( K \) points can be calculated using the binomial distribution. The number of ways to get \( H \) heads in 10 tosses is given by \( \binom{10}{H} \).

We need to find the minimum number of heads \( H \) such that the score \( S \geq K \):
\[ 20 - H \geq K \implies H \leq 20 - K \]

5. **Finding the Probability**:
The probability \( P(K) \) can be expressed as:
\[ P(K) = \frac{\sum_{H=0}^{20-K} \binom{10}{H}}{2^{10}} \]
We want \( P(K) > \frac{1}{2} \).

6. **Finding the Largest \( K \)**:
To find the largest \( K \) such that \( P(K) > \frac{1}{2} \), we calculate:
- For \( K = 17 \):
\[ P(17) = \frac{\binom{10}{0} + \binom{10}{1} + \binom{10}{2}}{2^{10}} \]
- For \( K = 16 \):
\[ P(16) = \frac{\binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3}}{2^{10}} \]
- For \( K = 15 \):
\[ P(15) = \frac{\binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4}}{2^{10}} \]
- For \( K = 14 \):
\[ P(14) = \frac{\binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5}}{2^{10}} \]

We find that \( P(16) > \frac{1}{2} \) but \( P(17) < \frac{1}{2} \).

Thus, the largest value of \( K \) such that \( P(K) > \frac{1}{2} \) is:

\[ \boxed{16} \]