A bag contains `n+1` coins. If is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is 7/12, then find the value of `ndot`

Explanation

Let `E_(1)` denote an event when a coin with two heads is selected and `E_(2)` an event when a fair coin is selected. Let A be the event when the toss results in head. Then,
`P(E_(1))=1//(n+1),P(E_(2))=n//(n+1),P(A//E_(1))=1,andP(A//E_(2))=1//2.`
Using total probability theorem, we have
`P(A)=P(E_(1))P(A//E_(1))+P(A//E_(2))P(A//E_(2)),`
`or(7)/(12)=(1)/(nxx1)xx1+(n)/(n+1)xx1/2`
`or12+6n=7n+7 or n=5`