#### Question

One hundred identical coins, each with probability p of showing 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, the value of p is

1/2

51/101

49/101

None of these

#### Solution

51/101

\[\text{ Let X denote the number of coins showing head .} \]

\[\text{ Therefore, X follows a binomial distribution with p and n as parameters . } \]

\[\text{ Given that } P(X = 50) = P(X = 51)\]

\[ \Rightarrow ^{100}{}{C}_{50} \ p^{50} q^{50} = ^{100}{}{C}_{51} \ p^{51}\ q^{49} \]

\[\text{ on simplifying we get } , \]

\[\frac{51}{50} = \frac{p}{q}\]

\[ \Rightarrow \frac{51}{50} = \frac{p}{1 - p} (\text{ since} \ p + q = 1)\]

\[ \Rightarrow p = \frac{51}{101}\]

Is there an error in this question or solution?

Solution One Hundred Identical Coins, Each with Probability P of Showing 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 Concept: Bernoulli Trials and Binomial Distribution.