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Given two coins; the first coin is fair and the second coin has two heads. One coin is picked at random and tossed two times. It shows heads both times. What is the probability that the coin picked is fair? | Given two coins; the first coin is fair and the second coin has two heads. One coin is picked at random and tossed two times. It shows heads both times. What is the probability that the coin picked is fair? | ||
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| + | ==Share and discuss your solutions below.== | ||
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| + | ==Solution 1== | ||
• <math class="inline">F=\left\{ \text{fair coin is selected}\right\}</math> | • <math class="inline">F=\left\{ \text{fair coin is selected}\right\}</math> | ||
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• By using Bayes' theorem,<math class="inline">P\left(F|H2\right)=\frac{P\left(H2|F\right)P\left(F\right)}{P\left(H2|F\right)P\left(F\right)+P\left(H2|S\right)P\left(S\right)}=\frac{P\left(H2|F\right)}{P\left(H2|F\right)+P\left(H2|S\right)}=\frac{\frac{1}{4}}{\frac{1}{4}+1}=\frac{1}{5}.</math> | • By using Bayes' theorem,<math class="inline">P\left(F|H2\right)=\frac{P\left(H2|F\right)P\left(F\right)}{P\left(H2|F\right)P\left(F\right)+P\left(H2|S\right)P\left(S\right)}=\frac{P\left(H2|F\right)}{P\left(H2|F\right)+P\left(H2|S\right)}=\frac{\frac{1}{4}}{\frac{1}{4}+1}=\frac{1}{5}.</math> | ||
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Latest revision as of 17:35, 13 March 2015
Communication, Networking, Signal and Image Processing (CS)
Question 1: Probability and Random Processes
January 2002
1. (20 pts)
Given two coins; the first coin is fair and the second coin has two heads. One coin is picked at random and tossed two times. It shows heads both times. What is the probability that the coin picked is fair?
Solution 1
• $ F=\left\{ \text{fair coin is selected}\right\} $
• $ S=\left\{ \text{the coin that has two heads is selected}\right\} $
• $ H2=\left\{ \text{heads are shown both times}\right\} $
• $ P\left(H2|F\right)=\frac{1}{4},\; P\left(H2|S\right)=1,\; P\left(F\right)=P\left(S\right)=\frac{1}{2}. $
• By using Bayes' theorem,$ P\left(F|H2\right)=\frac{P\left(H2|F\right)P\left(F\right)}{P\left(H2|F\right)P\left(F\right)+P\left(H2|S\right)P\left(S\right)}=\frac{P\left(H2|F\right)}{P\left(H2|F\right)+P\left(H2|S\right)}=\frac{\frac{1}{4}}{\frac{1}{4}+1}=\frac{1}{5}. $
