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

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) (15/28) (B) (15/213) (C) (2/15) (D) (2/45)

Tardigrade Admin · Accepted Answer

Let the coin be tossed n times. ∴ n C 7((1/2)) n = n C 9((1/2)) n ⇒ n =7+9 ⇒ n =16 So, required probability = 16 C 2((1/2))16=((16 × 15/2)) × (1/216)=(15/213).

Q. 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

$\frac{15}{2 ^{8}}$

$\frac{15}{2 ^{13}}$

$\frac{2}{15}$

$\frac{2}{45}$

Let the coin be tossed $n$ times.
$∴^{n} C_{7} (\frac{1}{2})^{n} =^{n} C_{9} (\frac{1}{2})^{n}$
$\Rightarrow n = 7 + 9 \Rightarrow n = 16$
So, required probability $=^{16} C_{2} (\frac{1}{2})^{16} = (\frac{16 \times 15}{2}) \times \frac{1}{2 ^{16}} = \frac{15}{2 ^{13}}$ .

Q. 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

2815​

21315​

152​

452​

Let the coin be tossed n times. ∴nC7​(21​)n=nC9​(21​)n ⇒n=7+9⇒n=16 So, required probability =16C2​(21​)16=(216×15​)×2161​=21315​.

$\frac{15}{2 ^{8}}$

$\frac{15}{2 ^{13}}$

$\frac{2}{15}$

$\frac{2}{45}$

Let the coin be tossed $n$ times.
$∴^{n} C_{7} (\frac{1}{2})^{n} =^{n} C_{9} (\frac{1}{2})^{n}$
$\Rightarrow n = 7 + 9 \Rightarrow n = 16$
So, required probability $=^{16} C_{2} (\frac{1}{2})^{16} = (\frac{16 \times 15}{2}) \times \frac{1}{2 ^{16}} = \frac{15}{2 ^{13}}$ .