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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.

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    $\begingroup$ Looks good to me. $\endgroup$ – barak manos Sep 4 '16 at 5:44
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    $\begingroup$ 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". $\endgroup$ – Parcly Taxel Sep 4 '16 at 5:45
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Please can someone help me if my understanding is correct.

No, you have correctly employed Bayes' Rule and the Law of Total Probability to arrive at the correct answer, so there is nothing left to help you with.

Good work.

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  • $\begingroup$ @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.) $\endgroup$ – Graham Kemp Sep 5 '16 at 0:00
  • $\begingroup$ edited the typo. Thanks. I was not sure if my application was right. $\endgroup$ – roang Sep 5 '16 at 0:22

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