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
Answer & explanation
Correct answer: option 2
Let X be the number of coins showing heads. Then, X is a binomial variate with parameters n = 100 and p.
Now,
$P(X=50) = P(X =51)$
$ ⇒{^{100}C}_{50} \, p^{50} (1-p)^{50}= {^{100}C}_{51}(1-p)^{49}$
$ ⇒\frac{100!}{50!50!}×\frac{51!49!}{100!}=\frac{p}{1-p}⇒\frac{51}{1-p}⇒p=\frac{51}{101}$