There are three coins, One is two headed coin, another is a biased coin that comes up head 75% of the time and thrid is an unbaised coin. One of the three coins is chosen at random and tossed. If its hows head, what is the probability that it was the two headed coin?

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`