Q51 / 150 · MHT CET 2021 (24 Sep Shift 2) · Mathematics
A fair coin is tossed for a fixed number of times. If probability of getting 7 heads is equal to probability of getting 9 heads, then probability of getting 2 heads is
- A$\frac{1}{15}$
- Correct answerB$\frac{15}{2^{13}}$
- C$\frac{15}{2^8}$
- D$\frac{2}{15}$
Solution
$\mathrm{p}=$ probability of getting head $=\frac{1}{2}$ and
$\mathrm{q}=$ probability of not getting head $\frac{1}{2}$
$\mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{n}-\mathrm{x}}$
As per data given, we write
$\begin{aligned}
& \mathrm{P}(\mathrm{x}=7)=\mathrm{P}(\mathrm{x}=9) \\
& \therefore{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{\mathrm{n}-7}={ }^{\mathrm{n}} \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{\mathrm{n}-9} \\
& \therefore \quad{ }^{\mathrm{n}} \mathrm{C}_7={ }^{\mathrm{n}} \mathrm{C}_9 \\
& \therefore \quad \mathrm{n}=16
\end{aligned}$
when $x=2$, we get
$\mathrm{P}(\mathrm{x}=2)={ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{14}=\frac{16 \times 15}{2}\left(\frac{1}{2}\right)^{16}=\frac{15}{(2)^{13}}$
Hence, the correct answer is option: 'B'.