Solveeit Logo
Login

Question

Mathematics Question on Probability

One hundred identical coins, each with probability p, of showing up heads are tossed once. If 0<p<10 < 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

A

12\frac{1}{2}

B

49101\frac{49}{101}

C

50101\frac{50}{101}

D

51101\frac{51}{101}

Answer

51101\frac{51}{101}

Explanation

Solution

Let X be the number of coins showing heads. Let X be a binomial variate with parameters n = 100 and p.
Since \hspace30mm P(X=50)=P(x=51)P(X= 50) = P(x=51)
⇒        100C50p50(1−p)50=100C51p51(1−p)49\Rightarrow \, \, \, \, \, \, \, \, ^100 C_50 p_50(1-p)_50 = ^100 C_51 p_51 (1-p)_49
⇒      (100)!(50!)(50!).(51!)×(49!)100!=p1−p\Rightarrow \, \, \, \, \, \, \frac{(100)!}{(50!)(50!)}.\frac{(51!)\times(49!)}{100!} = \frac{p}{1-p}
⇒              p1−p=5150⇒p=51101\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{p}{1-p} = \frac{51}{50} \Rightarrow p= \frac{51}{101}