Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

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

Options:

$\frac{1}{2}$

$\frac{51}{101}$

$\frac{49}{101}$

none of these

Correct Answer:

$\frac{51}{101}$

Explanation:

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}$