A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is

Let the coin be tossed $$n$$ times. The probability of getting exactly $$r$$ heads is $$P(r) = \binom{n}{r}\left(\frac{1}{2}\right)^n$$.

Given that $$P(7) = P(9)$$, we have $$\binom{n}{7} = \binom{n}{9}$$.

Since $$\binom{n}{r} = \binom{n}{n-r}$$, this equation holds when $$7 = n - 9$$, giving $$n = 16$$.

Now we find the probability of getting exactly 2 heads when the coin is tossed 16 times:

$$P(2) = \binom{16}{2}\left(\frac{1}{2}\right)^{16} = \frac{120}{2^{16}}$$

We compute $$\binom{16}{2} = \frac{16 \times 15}{2} = 120$$, and simplify $$\frac{120}{2^{16}} = \frac{15 \times 8}{8 \times 2^{13}} = \frac{15}{2^{13}}$$.

The answer is Option 1: $$\frac{15}{2^{13}}$$.