A fair coin is tossed n times . If the probability of getting seven heads is equal to the probability of getting 9 heads , then the probability of getting two heads is

Options

A. `2/15`
B. `15/(2^(8))`
C. `15/(2^(3))`
D. none of these

Answer

To solve the problem step by step, we need to find the probability of getting 2 heads when a fair coin is tossed \( n \) times, given that the probability of getting 7 heads is equal to the probability of getting 9 heads.

### Step 1: Set up the probability equation
The probability of getting \( r \) heads when a fair coin is tossed \( n \) times can be expressed using the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( p \) is the probability of getting heads (which is \( \frac{1}{2} \)), and \( q \) is the probability of getting tails (also \( \frac{1}{2} \)).

### Step 2: Write the equations for 7 heads and 9 heads
Given that the probability of getting 7 heads is equal to the probability of getting 9 heads, we can write:
\[ \binom{n}{7} \left(\frac{1}{2}\right)^7 \left(\frac{1}{2}\right)^{n-7} = \binom{n}{9} \left(\frac{1}{2}\right)^9 \left(\frac{1}{2}\right)^{n-9} \]

### Step 3: Simplify the equation
This simplifies to:
\[ \binom{n}{7} \left(\frac{1}{2}\right)^{n} = \binom{n}{9} \left(\frac{1}{2}\right)^{n} \]
Since \( \left(\frac{1}{2}\right)^{n} \) is common on both sides, we can cancel it out:
\[ \binom{n}{7} = \binom{n}{9} \]

### Step 4: Use the property of binomial coefficients
From the property of binomial coefficients, we know that:
\[ \binom{n}{r} = \binom{n}{n-r} \]
Thus, we have:
\[ \binom{n}{7} = \binom{n}{9} \implies n - 7 = 9 \implies n = 16 \]

### Step 5: Find the probability of getting 2 heads
Now that we know \( n = 16 \), we can find the probability of getting 2 heads:
\[ P(X = 2) = \binom{16}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{16-2} \]
This simplifies to:
\[ P(X = 2) = \binom{16}{2} \left(\frac{1}{2}\right)^{16} \]

### Step 6: Calculate \( \binom{16}{2} \)
Calculating \( \binom{16}{2} \):
\[ \binom{16}{2} = \frac{16 \times 15}{2 \times 1} = 120 \]

### Step 7: Substitute back into the probability formula
Now we substitute back:
\[ P(X = 2) = 120 \cdot \left(\frac{1}{2}\right)^{16} = \frac{120}{2^{16}} = \frac{120}{65536} = \frac{15}{8192} \]

### Conclusion
The probability of getting 2 heads when a fair coin is tossed 16 times is:
\[ \frac{15}{8192} \]