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A bag contains $2n+1$ coins. It is known that $n$ of these coins have a head on both sides. Whereas the remaining $n+1$ coins are fair. A coins is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\large\frac{31}{42}.$ then $n$=

\[(a)\;10 \quad (b)\;11 \quad (c)\;12 \quad (d)\;13 \]
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(a) 10
answered Nov 7, 2013 by pady_1
 

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