Question

One hundred identical coins, each with probability, $p$, of showing up heads are tossed once. If $0<p<1$ and the probability of heads showing on fifty coins is equal to that of heads showing on $51$ coins, then the value of $p$ is :

• A. $1/2$
• B. $49/101$
• C. $50/101$
• D. $51/101$

Answer 1

The correct option is D $51/101$

Applying binomial distribution for 100 trials we get

${}^{100}{C}_{50}{p}^{50}(1-p{)}^{50}={}^{100}{C}_{51}{p}^{51}(1-p{)}^{49}$

$\frac{(1-p)100!}{50!50!}=\frac{\left(p\right)100!}{51!49!}$

$\frac{1-p}{50}=\frac{p}{51}$

$51-51p=50p$

$101p=51$

$p=\frac{51}{101}$

Hence the required probability is $\frac{51}{101}$