Question
One hundred identical coins, each with probability $$p$$ of showing up heads, are tossed. If $$0 < p < 1$$ and the probability of heads showing on $$50$$ coins is equal to that of heads showing on $$51$$ coins. The value of $$p$$ is :
A.
$$\frac{1}{2}$$
B.
$$\frac{{49}}{{101}}$$
C.
$$\frac{{50}}{{101}}$$
D.
$$\frac{{51}}{{101}}$$
Answer :
$$\frac{{51}}{{101}}$$
Solution :
Let $$X \sim B\left( {100,\,p} \right)$$ be the number of coins showing heads, and let $$q = 1 - p.$$
Then, since $$P\left( {X = 51} \right) = P\left( {X = 50} \right),$$ we have
$$\eqalign{
& {}^{100}{C_{51}}\left( {{p^{51}}} \right)\left( {{q^{49}}} \right) = {}^{100}{C_{50}}\left( {{p^{50}}} \right)\left( {{q^{50}}} \right) \cr
& \Rightarrow \frac{p}{q} = \left( {\frac{{100!}}{{50!\,50!}}} \right)\left( {\frac{{51!\,49!}}{{100!}}} \right) \cr
& \Rightarrow \frac{p}{{1 - p}} = \frac{{51}}{{50}} \cr
& \Rightarrow 50p = 51 - 51p \cr
& \Rightarrow p = \frac{{51}}{{101}} \cr} $$