A unbaised coin is tossed n times. If the probability of getting 5 heads is equal to the probability of getting 6 heads then probability of getting 3 heads is

Options

A. `""^(11)C_(5)(1/2)^(5)`
B. `""^(11)C_(6)(1/2)^(6)`
C. `""^(11)C_(3)(1/2)^(11)`
D. `(11)/(1024)`

Answer

To solve the problem, we need to find the probability of getting 3 heads when an unbiased coin is tossed \( n \) times, given that the probability of getting 5 heads is equal to the probability of getting 6 heads.

### Step-by-Step Solution:

1. **Understanding the Problem**: We know that the probability of getting \( k \) heads when tossing a coin \( n \) times follows the binomial distribution formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here, \( p = \frac{1}{2} \) (the probability of getting heads) and \( 1-p = \frac{1}{2} \) (the probability of getting tails).

2. **Setting Up the Equation**: We are given that:
\[ P(X = 5) = P(X = 6) \]
Using the binomial formula, we can write:
\[ \binom{n}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{n-5} = \binom{n}{6} \left(\frac{1}{2}\right)^6 \left(\frac{1}{2}\right)^{n-6} \]
Simplifying this, we have:
\[ \binom{n}{5} \left(\frac{1}{2}\right)^n = \binom{n}{6} \left(\frac{1}{2}\right)^n \]
Since \( \left(\frac{1}{2}\right)^n \) is common on both sides, we can cancel it out:
\[ \binom{n}{5} = \binom{n}{6} \]

3. **Using the Binomial Coefficient Property**: The relationship between binomial coefficients gives us:
\[ \binom{n}{5} = \frac{n!}{5!(n-5)!} \quad \text{and} \quad \binom{n}{6} = \frac{n!}{6!(n-6)!} \]
Setting these equal, we have:
\[ \frac{n!}{5!(n-5)!} = \frac{n!}{6!(n-6)!} \]
Cancelling \( n! \) from both sides:
\[ \frac{1}{5!(n-5)!} = \frac{1}{6!(n-6)!} \]

4. **Cross Multiplying**: This leads to:
\[ 6!(n-6)! = 5!(n-5)! \]
Expanding \( 6! \) and \( (n-5)! \):
\[ 6(n-6)! = (n-5)(n-6)! \]
Cancelling \( (n-6)! \) from both sides (assuming \( n \neq 6 \)):
\[ 6 = n - 5 \]
Thus, we find:
\[ n = 11 \]

5. **Finding the Probability of Getting 3 Heads**: Now that we have \( n = 11 \), we can find the probability of getting 3 heads:
\[ P(X = 3) = \binom{11}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{11-3} \]
Simplifying this:
\[ P(X = 3) = \binom{11}{3} \left(\frac{1}{2}\right)^{11} \]

6. **Calculating \( \binom{11}{3} \)**:
\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]
Therefore, the probability becomes:
\[ P(X = 3) = 165 \left(\frac{1}{2}\right)^{11} = \frac{165}{2048} \]

### Final Answer:
The probability of getting 3 heads is:
\[ \frac{165}{2048} \]