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A bag contains (2n+1) coins.It is known that n of these coins have a head on both sides where as the rest of the coins are fair.A coin is picked up at random from the bag and is tossed.If the probability that the toss results in a head is $\Large \frac{31}{42}$,determine the value of n.

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• $E_1=chosing\; a \;baised \;coin(two\;headed)$
• $E_2=chosing\; a \;un\;baised \;coin$
• $A\;getting\;a\;head$
• $there\;are\;'n'\;baised\;coin\;and\;(n+1)\;unbaised\;coin\;in\;a\;total\;of\;(2n+1)\;coin$
• =$P(E_1)P(a/E_1)+P(E_2)P(A/E_2)$
$P(E_1)=P(chosing\;a\;baised\;coin)$
=$\Large\frac{n}{2n+1}$
$P(E_2)=P(chosing\;a\;unbaised\;coin)$
=$\Large\frac{n+1}{2n+1}$
$P(A/E_1)=P(getting\;a\;head/it\;is\;baised\;(two\;headed\;coin))$
=1
$P(A/E_2)=P(getting\;a\;head/it\;is\;unbaised\;coin)$
=$\Large\frac{1}{2}$
$\Large\;P(A)=1\times\frac{n}{2n+1}+\frac{1}{2}\times\frac{(n+1)}{(2n+1)}=\frac{31}{42}$[given]
=$\Large\;\frac{2n+n+1}{2(2n+1)}=\frac{3n+1}{2(2n+1)}=\frac{31}{42}$
=$42(3n+1)=2\times31(2n+1)$
=$126n+42=124n+62$
=$2n=20$
$n=10$

answered Mar 3, 2013
edited Jun 4, 2013

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