Answer: Our required probability is [tex]\dfrac{1}{2}[/tex]
Step-by-step explanation:
Since we have given that
Number of coins = 3
Number of coin has 2 heads = 1
Number of fair coins = 2
Probability of getting one of the coin among 3 = [tex]\dfrac{1}{3}[/tex]
So, Probability of getting head from fair coin = [tex]\dfrac{1}{2}[/tex]
Probability of getting head from baised coin = 1
Using "Bayes theorem" we will find the probability that it is the two headed coin is given by
[tex]\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}[/tex]
Hence, our required probability is [tex]\dfrac{1}{2}[/tex]
No, the answer is not [tex]\dfrac{1}{3}[/tex]