Question

A fair coin is tossed $100$ times. The probability of getting tails $1,3,....49$ times is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{1}{16}$

Answer 1

The correct option is B $\frac{1}{4}$

Let the probability of getting a tail in a single trial be $p=\frac{1}{2}$. The number of trials be $n=100$ and the number of trials in $100$ trials be $X$. We have,
$P(X=r)={}^{100}{C}_r{p}^r{q}^{100-r}$
$={}^{100}{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{100-r}$
$={}^{100}{C}_r{\left(\frac{1}{2}\right)}^{100}$
Now,
$P(X=1)+P(X=3)+...+P(X=49)$
$={}^{100}{C}_1{\left(\frac{1}{2}\right)}^{100}+{}^{100}{C}_3{\left(\frac{1}{2}\right)}^{100}+...+{}^{100}{C}_{49}{\left(\frac{1}{2}\right)}^{100}$
$=({}^{100}{C}_1+{}^{100}{C}_3+...+{}^{100}{C}_{49}){\left(\frac{1}{2}\right)}^{100}$
But
${}^{100}{C}_1+{}^{100}{C}_3+...+{}^{100}{C}_{99}=2^{99}$
Also,
${}^{100}{C}_{99}={}^{100}{C}_1$
${}^{100}{C}_{97}={}^{100}{C}_3,...{}^{100}{C}_{51}={}^{100}{C}_{49}$
Thus,
$2({}^{100}{C}_1+{}^{100}{C}_3+...+{}^{100}{C}_{49})=2^{99}$
or ${}^{100}{C}_1+{}^{100}{C}_3+...+{}^{100}{C}_{49}=2^{98}$
Therefore, probability of the required event $=\frac{2^{98}}{2^{100}}=\frac{1}{4}$