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Q.
A fair coin is tossed 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
Let the coin be tossed $n$ times.
Let getting head is consider to be success.
$\therefore p=\frac{1}{2},\, q=1-p=1-\frac{1}{2}=\frac{1}{2}$
It is given that,
$P(X=3)=P(X=5)$
$\Rightarrow { }^{n} C_{3}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{n-3}={ }^{n} C_{5}\left(\frac{1}{2}\right)^{5}\left(\frac{1}{2}\right)^{n-5}$
$\Rightarrow { }^{n} C_{3}={ }^{n} C_{6}$
$\Rightarrow n=3+5 \left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x +y=n\right]$
$\Rightarrow n=8$
Now, $P(X=1)={ }^{8} C_{1}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{8-1}$
$={ }^{8} C_{1} \times\left(\frac{1}{2}\right)^{8}=\frac{1}{32}$