Question

A bag contains `(2n+1)` coins. It is known that `n` of these coins have a head on both sides whereas 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` .

Answer 1

Given , n coins have head on both sides and (n+ 1) coins are fair coins .
Let `E_(1)` = Event that an unfair coin is selected
`E_(2)` = Event that a fair coin is selected
E = Event that the toss results in a head
`P (E_(1)) = (n)/(2n+1)` and `P(E_(2)) = (n+ 1)/(2n+1)`
Also , `" " P((E)/(E_(1)))= 1 and P((E)/(E_(2))) = (1)/(2)`
`therefore " " P(E) = P(E_(1)) . P((E)/(E_(1))) + P(E_(2)) . P((E)/(E_(2))) = (n)/(2n+1)* 1 + (n+1)/(2n+1) * (1)/(2)`
`implies " " (31)/(42) = (2n + n + 1)/(2(2n+1)) implies (31)/(42) = (3n+1)/(4n+1)`
`implies " " 124 n + 62 = 126n + 42`
`implies " " 2n = 20 implies n = 10`