Question

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

• A. $1/2$
• B. $1/4$
• C. $1/8$
• D. $1/16$

Answer 1

The correct option is B $1/4$

Let $p=$ probability of getting a tail in a single trial $=1/2$
$n=$ number of trails $=100$
and $X=$ number of tails in 100 trials.
We have
$P(X=r){=}^{100}{C}_r{p}^r{q}^{n-r}$
${=}^{100}{C}_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{100-r}{=}^{100}{C}_r{\left(\frac{1}{2}\right)}^{100}$
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,
$2{(}^{100}{C}_1{+}^{100}{C}_3+...{+}^{100}{C}_{49})=2^{99}$
${\Rightarrow }^{100}{C}_1{+}^{100}{C}_3+...{+}^{100}{C}_{49}=2^{98}$
$\therefore$ Probability of required event $=\frac{2^{98}}{2^{100}}=\frac{1}{4}$.