Answer:
The (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)
Step-by-step explanation:
Given
Total unbiased coin = n
Normal coins =n - 1
Fake = 1
The (conditional) probability that the coin you chose is the fake coin is represented by
P(Fake | Head)
And it's calculated as follows;
P(Fake | Head) = P(Fake, Head) ÷ P(Head) ----- (1)
Where P(Fake, Head) = P(Fake) * P(Head | Fake)
P(Fake) = 1/n --- because only one is fake
P(Head | Fake) = n/n because all coins (including the fake) have head
So, P(Fake, Head) = P(Fake) * P(Head | Fake) becomes
P(Fake, Head) = 1/n * n/n
P(Fake, Head) = 1/n
P(Head) is calculated by
P(Fake) * P(Head | Fake) + P(Normal) * P(Head | Normal)
P(Fake) * P(Head | Fake) = P(Fake, Head) = 1/n (as calculated above)
P(Normal) * P(Head | Normal) = ½ * (n - 1)/n ----- considering that the coin also has a tail with equal probability as that of the head.
Going back to (1)
P(Fake | Head) = P(Fake, Head) ÷ P(Head) becomes
P(Fake | Head) = (1/n) ÷ ((1/n) + (½(n-1)/n))
= (1/n) ÷ ((1/n) + (½(n-1)/n))
= (1/n) ÷ (1/n + (n - 1)/2n)
= (1/n) ÷ (2 + n - 1)/(2n)
= (1/n) ÷ (1 + n)/(2n)
= (1/n) * (2n)/(1 + n)
= 2/(1 + n)
Hence, the (conditional) probability that the coin you chose is the fake coin is 2/(1 + n)
