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  • Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

Answers (1)

Let E1 be the event that a fair coin is drawn
E2 be the event that two headed coin is drawn
E be the event that tossed coin get a head

\begin{aligned} & \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)=\frac{1}{2}, \text { And } \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)=1\\ &\text { Using Bayes' theorem, we have }\\ &\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{E}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{2}\right)}\\ &=\frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times 1}\\ &=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}\\ &=\frac{\frac{1}{4}}{\frac{3}{4}}\\ &=\frac{1}{3} \end{aligned}

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