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Question
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 `31/42`, determine the value of n.
Solution
Given that n coins are two-headed coins and the remaining (n + 1) coins are fair.
Let E1: the event that unfair coin is selected
E2: the event that the fair coin is selected
E: the event that the toss results in a head
∴ P(E1) = `"n"/(2"n" + 1)` and P(E2) = `("n" + 1)/(2"n" + 1)`
`"P"("E"/"E"_1)` = 1(sure event) and `"P"("E"/"E"_2) = 1/2`
∴ P(E) = `"P"("E"_1)*"P"("E"/"E"_1) + "P"("E"_2)*"P"("E"/"E"_2)`
= `"n"/(2"n" + 1)*1 + ("n" + 1)/(2"n" + 1)*1/2`
= `1/(2"n" + 1)("n" + ("n" + 1)/2)`
= `1/(2"n" + 1) ((2"n" + "n" + 1)/2)`
= `(3"n" + 1)/(2(2"n" + 1))`
But P(E) = `31/42` ....(Given)
∴ `(3"n" + 1)/(2(2"n" + 1)) = 31/42`
⇒ `(3"n" + 1)/(2"n" + 1) = 31/21`
⇒ 63n + 21 = 62n + 31
⇒ n = 10
Hence, the required value of n is 10.
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