NIMCET 2012 Question Paper with solutions

Question : 40 of 120

Marks: +1, -0

One hundred identical coins each withprobability P of showing up heads are tossed. If0 < P < 1 and the probability of heads showing on50 coins is equal to that of heads on 51 coins,then the value of P is

$1/2$
${49}/{101}$
${50}/{101}$
${51}/{101}$

Validate

Solution:

By condition, using binomial distribution,

↖{100}C_{50}P^{50}(1-P)^{50}

↖{100}C_{51}P^{51}(1-P)^{49}

⇒

{100!}/{50!50!}

(1 - P) =

{100!}/{51!49!}

. P
⇒

1/{50}

(1 - P) =

P/{51}

⇒ 51 - 51 P = 51 P
⇒ 101 P = 51
∴ P =

{51}/{101}

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