Practicing Success
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 |
$\frac{1}{2}$ $\frac{49}{101}$ $\frac{50}{101}$ $\frac{51}{101}$ |
$\frac{51}{101}$ |
Let X denotes the number of coins showing heads up. Then, X is a binomial variate with n = 100 and probability of success p. We have, $P(X=51)=P(X=50)$ $⇒ {^{100}C}_{51}p^{51}q^{49}= {^{100}C}_{50}p^{50}q^{50},$ where q = 1- p $⇒ \frac{p}{q}=\frac{^{100}C_{50}}{^{100}C_{51}}=\frac{51}{50}⇒\frac{p}{1-p}=\frac{51}{50}⇒ p = \frac{51}{101}$ |