Question

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, then the probability of getting two heads is

• A. $\frac{15}{2^8}$
• B. $\frac{15}{2^{13}}$
• C. $\frac{2}{15}$
• D. $\frac{2}{45}$

Answer 1

Answer: (B)

Let the coin be tossed $n$ times.
$\therefore{ }^{ n } C _7\left(\frac{1}{2}\right)^{ n }={ }^{ n } C _9\left(\frac{1}{2}\right)^{ n } $
$\Rightarrow n =7+9 \Rightarrow n =16$
So, required probability $={ }^{16} C _2\left(\frac{1}{2}\right)^{16}=\left(\frac{16 \times 15}{2}\right) \times \frac{1}{2^{16}}=\frac{15}{2^{13}}$.