A bag contains 2n coins,out of which n-1 are | vedclass

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(C) Let $n-1$ be the number of unfair coins (heads on both sides) and $2n - (n-1) = n+1$ be the number of fair coins.Total coins = $2n$.The probabilit

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$13$

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(C) Let $n-1$ be the number of unfair coins (heads on both sides) and $2n - (n-1) = n+1$ be the number of fair coins.Total coins = $2n$.The probability of picking an unfair coin is $\frac{n-1}{2n}$ and a fair coin is $\frac{n+1}{2n}$.The probability of getting a head is given by:$P(H) = P(H|	ext{unfair})P(	ext{unfair}) + P(H|	ext{fair})P(	ext{fair})$$\frac{41}{56} = (1) 	imes \frac{n-1}{2n} + (\frac{1}{2}) 	imes \frac{n+1}{2n}$$\frac{41}{56} = \frac{2(n-1) + (n+1)}{4n}$$\frac{41}{56} = \frac{3n-1}{4n}$$41 	imes 4n = 56 	imes (3n-1)$$164n = 168n - 56$$4n = 56 \Rightarrow n = 14$.The number of unfair coins is $n-1 = 14-1 = 13$.

$A$ bag contains $2n$ coins,out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that a head appears in the toss is $\frac{41}{56}$,then the number of unfair coins in the bag is:

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