When a coin is tossed n times, if the probability for getting `6` heads is equal to the probability of getting `8 ` heads, then the value of `n` is

Options

A. `10`
B. `12`
C. `14`
D. `20`

Answer

To solve the problem where the probability of getting 6 heads is equal to the probability of getting 8 heads when a coin is tossed \( n \) times, we can use the binomial distribution formula. Let's go through the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Binomial Probability Formula**:
The probability of getting \( k \) heads in \( n \) tosses of a fair coin is given by:
\[ P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{n-k} = \binom{n}{k} \left(\frac{1}{2}\right)^n \]
where \( \binom{n}{k} \) is the binomial coefficient.

2. **Setting Up the Equation**:
According to the problem, the probability of getting 6 heads is equal to the probability of getting 8 heads:
\[ P(X = 6) = P(X = 8) \]
This can be expressed using the binomial formula:
\[ \binom{n}{6} \left(\frac{1}{2}\right)^n = \binom{n}{8} \left(\frac{1}{2}\right)^n \]

3. **Canceling Out Common Terms**:
Since \( \left(\frac{1}{2}\right)^n \) is common on both sides, we can cancel it out:
\[ \binom{n}{6} = \binom{n}{8} \]

4. **Using the Property of Binomial Coefficients**:
The property of binomial coefficients states that:
\[ \binom{n}{k} = \binom{n}{n-k} \]
Therefore, we can rewrite \( \binom{n}{8} \) as \( \binom{n}{n-8} \):
\[ \binom{n}{6} = \binom{n}{n-8} \]

5. **Setting the Indices Equal**:
For the two binomial coefficients to be equal, the indices must satisfy:
\[ 6 = n - 8 \quad \text{or} \quad 8 = n - 6 \]
Solving the first equation:
\[ n = 6 + 8 = 14 \]

6. **Conclusion**:
Thus, the value of \( n \) is:
\[ n = 14 \]

### Final Answer:
The value of \( n \) is \( 14 \).