Question
One hundred identical coins, each with probability, $$p,$$ of showing up heads are tossed once. If 0 < $$p$$ < 1 and the probabilitity 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 :
$$\frac{51}{101}$$
Solution :
Prob. of one coin showing head = $$p$$
∴ Prob. of one coin showing tail = $$1 - p$$
ATQ coin is tossed 100 times and prob. of 50 coins showing head = prob. of 51 coins showing head.
Using binomial prob. distribution
$$\eqalign{
& P\left( {X = r} \right) = \,{\,^n}{C_r}{p^r}{q^{n - r}}, \cr
& {\text{we get,}}{{\text{ }}^{100}}{C_{50}}{p^{50}}{\left( {1 - p} \right)^{50}} = {\,^{100}}{C_{51}}{p^{51}}{\left( {1 - p} \right)^{49}} \cr
& \Rightarrow \,\,\frac{{1 - p}}{p} = \frac{{^{100}{C_{51}}}}{{^{100}{C_{50}}}} = \frac{{50!50!}}{{51!49!}} = \frac{{50}}{{51}} \cr
& \Rightarrow \,\,51 - 51p = 50p \cr
& \Rightarrow \,\,101p = 51 \cr
& \Rightarrow \,\,p = \frac{{51}}{{101}} \cr} $$