There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what i

Answer

`4//9`

Explanation

Let `E_(1),E_(2) and E_(3)` be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin. Then
`P(E_(1))=P(E_(2))=P(E_(3))=1/3`
Let A be the evetn that the coin shows heads.
A two-headed coin will always show heads. Therefore,
`P(A//E_(1))=P` (coin showing heads, given that it is a two headed coun)=1
Probability of heads coming up, given that it is a baised coin) `(75)/(100)=(3)/(4)`
Since the third coin is unbiased, the probability that it shows heads is always 1/2
`therefore P(A//E_(3))=P` (coin showing heads, given that it is an unbised coin) `=1/2`
The probability that the coin is two-headed, given that it shows heads, is given by `P(E_(1)//A)`
By using Bayes' theorem, we batain
`P(E_(1)//A)=(P(E_(1)).P(A//E_(1)))/(P(E_(1)).P(A//E_(1))+P(A//E_(2))+P(E_(2)).P(A//E_(3)))`
`=(1/3xx1)/(1/3xx1+1/3xx3/4+1/3xx1/2)=4/9`