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 find the value of $p$ .

Question

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Answer 1

We have $^{100} C_{50} p^{50} (1 - p)^{50} =^{100} C_{51} p^{51} (1 - p)^{49}$
or $\frac{1 - p}{p} = \frac{1 0 0 !}{5 1 ! 4 9 !} \times \frac{5 0 ! 5 0 !}{1 0 0 !} = \frac{5 0}{5 1}$
$\Rightarrow 51 - 51 p = 50 p$
$\Rightarrow 51 = 50 p + 51 p$
$\Rightarrow 101 p = 51$
$∴ p = \frac{5 1}{1 0 1}$

Answer 2

We have

^{100} C_{50} p^{50} (1 - p)^{50} =^{100} C_{51} p^{51} (1 - p)^{49}

or

\frac{1 - p}{p} = \frac{1 0 0 !}{5 1 ! 4 9 !} \times \frac{5 0 ! 5 0 !}{1 0 0 !} = \frac{5 0}{5 1}

\Rightarrow 51 - 51 p = 50 p

\Rightarrow 51 = 50 p + 51 p

\Rightarrow 101 p = 51

∴ p = \frac{5 1}{1 0 1}