# Conditional Probability for a coin to be fair

A gambler has in his pocket a fair coin and a biased coin which will land heads with probability $\frac34$.
He selects one of the coins at random; when he tosses it, it lands heads. What is the probability it is the fair coin?
(II)If he tosses the same coin a second time, and again it lands heads. What now is the probability it is the fair coin?
(III) If he tosses the same coin for a third time, and this time it lands tails. What now is the probability it is the fair coin?

My solution using Bayes:

(I)$$P(\text{fair}|\text{heads})=\frac{P(\text{fair}\cap\text{heads})}{P(\text{heads})}=\frac{\frac12\cdot\frac12}{\frac12\cdot\frac12+\frac12\cdot\frac34}=\frac{\frac14}{\frac58}=\frac25$$

(II) $$P(\text{fair}|\text{heads,heads})=\frac{P(\text{fair}\cap\text{heads}\cap\text{heads})}{P(\text{heads}\cap\text{heads})}=\frac{\frac12\cdot\frac12\cdot\frac12}{\frac12\cdot\frac12\cdot\frac12+\frac12\cdot\frac34\cdot\frac34}=\frac{4}{13}$$

(III) $$P(\text{fair}|\text{heads,heads,Tails})=\frac{P(\text{fair}\cap\text{heads}\cap\text{heads}\cap\text{tails})}{P(\text{heads}\cap\text{heads}\cap\text{tails})}=\frac{\frac12\cdot\frac12\cdot\frac12\cdot\frac12}{\frac12\cdot\frac12\cdot\frac12\cdot\frac12+\frac12\cdot\frac34\cdot\frac34\cdot\frac14}=\frac{8}{17}$$ Please can someone help me if my understanding is correct.

• Looks good to me. Sep 4, 2016 at 5:44
• Yes, you are correct. Alternatively notice that if the probabilities of the two coins landing on heads are pooled, the fair coin has 2/5 of the "share". Sep 4, 2016 at 5:45

• @rowang Second verse, same as the first. There's no question; you've got this down okay. $\color{green}\checkmark$ (Assuming the repeat of $(1/4)/(5/8)$ to be a cut and paste typo.) Sep 5, 2016 at 0:00