Aspire Faculty ID #16296 · Topic: NIMCET 2010 · Just now
NIMCET 2010

One hundred identical coins, each with probability $p$ of showing up a head, are tossed. If $0

Solution

Condition: $\binom{100}{50} p^{50} (1-p)^{50} = \binom{100}{51} p^{51} (1-p)^{49}$ Divide both sides: $\frac{1}{\frac{100-50}{51}} \cdot \frac{1-p}{p} = 1$ $\frac{51}{50} \cdot \frac{1-p}{p} = 1$ $\frac{1-p}{p} = \frac{50}{51}$ $51 - 51p = 50p$ $51 = 101p$ $p = \frac{51}{101}$

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