Q63 / 160 · WBJEE 2015 · Mathematics
A fair coin is tossed at a fixed number of times. If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then the probability of getting exactly one head is
- A$1 / 64$
- Correct answerB$1 / 32$
- C$1 / 16$
- D$1 / 8$
Solution
Let the coin be tossed $n$ times
Let getting head is consider to be success $\therefore \quad p=\frac{1}{2}, q=1-p=1-\frac{1}{2}=\frac{1}{2}$
It is given that,
$P(X=3)=P(X=5)$
$\Rightarrow \quad{ }^{n} C_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{n-3}={ }^{n} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{n-5}$
$\Rightarrow \quad{ }^{n} C_{3}={ }^{n} C_{5}$
$\Rightarrow \quad n=3+5 \quad\left[\because^{n} C_{x}={ }^{n} C_{y} \Rightarrow x+y=n \mid\right.$
$\Rightarrow \quad n=8$
Now, $P(X=1)={ }^{8} C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{8-1}$
$={ }^{8} C_{1} \times\left(\frac{1}{2}\right)^{8}=\frac{1}{32}$
Hence, the correct answer is option: 'B'.