Probability Ques 122
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: 122.(d)
Solution:
Formula:
- 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) $
$\Rightarrow { }^{100} C _{50} p^{50}(1-p)^{50} ={ }^{100} C _{51}(p)^{51}(1-p)^{49} $
$\Rightarrow \frac{(100) !}{(50 !)(50 !)} \cdot \frac{(51 !) \times(49 !)}{100 !} =\frac{p}{1-p} \Rightarrow \frac{p}{1-p}=\frac{51}{50} $
$\Rightarrow p =\frac{51}{101}$
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