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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 - Mathematics

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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

  • 1/2

  • 51/101

  • 49/101

  • None of these

     

Solution

51/101 

\[\text{ Let X denote the number of coins showing head .}  \]
\[\text{ Therefore, X follows a binomial distribution with p and n as parameters . }  \]
\[\text{ Given that } P(X = 50) = P(X = 51)\]
\[ \Rightarrow ^{100}{}{C}_{50} \ p^{50} q^{50} = ^{100}{}{C}_{51} \ p^{51}\  q^{49} \]
\[\text{ on simplifying we get } , \]
\[\frac{51}{50} = \frac{p}{q}\]
\[ \Rightarrow \frac{51}{50} = \frac{p}{1 - p} (\text{ since}  \ p + q = 1)\]
\[ \Rightarrow p = \frac{51}{101}\]

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Solution 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 Concept: Bernoulli Trials and Binomial Distribution.
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