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 $50$ coins is equal to that of heads showing on $51$ coins, then the value of $p$ is:

• A. $\frac{1}{2}$
• B. $\frac{49}{101}$
• C. $\frac{50}{101}$
• D. $\frac{51}{101}$

Answer 1

The correct option is D $\frac{51}{101}$

Let X be the number of coins showing heads
Then $X\sim B(100,p)$

We have $P(X=51)=P(X=50)$

$\Rightarrow {100}_{C_{51}}{p}^{51}{q}^{49}={100}_{C_{50}}{p}^{50}{q}^{50}$

$\Rightarrow \frac{p}{q}=\frac{100!}{50!.50!}\times \frac{51!49!}{100!}$

$\Rightarrow \frac{p}{1-p}=\frac{51}{50}$ Where $q=1-p$
$\Rightarrow 50p=51-51p\Rightarrow 101p=51$

$\Rightarrow P=\frac{51}{101}$