Probability 5 Question 11
11. 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
(1988, 2M)
(a) $1 / 2$
(b) $49 / 101$
(c) $50 / 101$
(d) $51 / 101$
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Answer:
Correct Answer: 11. (d)
Solution:
- Let $X$ be the number of coins showing heads. Let $X$ be a binomial variate with parameters $n=100$ and $p$.
Since,
$$ P(X=50)=P(X=51) $$
$$ \begin{aligned} \Rightarrow & { }^{100} C _{50} p^{50}(1-p)^{50} & ={ }^{100} C _{51}(p)^{51}(1-p)^{49} \\ \Rightarrow \quad & \frac{(100) !}{(50 !)(50 !)} \cdot \frac{(51 !) \times(49 !)}{100 !} & =\frac{p}{1-p} \Rightarrow \frac{p}{1-p}=\frac{51}{50} \\ \Rightarrow & p & =\frac{51}{101} \end{aligned} $$
