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There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the times and the third is also a biased coin that comes up tails 40% of the time. One of the three coins is chosen at random and tossed and it shows heads. What is the probability that it was the two-headed coin?
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Solution
Let E_{1} be the event of selecting the two-headed coin,
E_{2} be the event of selecting the biased coin that comes up heads 75% of the times,
E_{3} be the event of selecting the biased coin that comes up tails 40% of the times
and A be the event of getting head on the coin.
Then,
`P(E_1)=P(E_2)=P(E_3)=1/3`
P(A/E_{1}) = Probability of getting a head on the coin, given that the coin is two-headed.
`⇒ P(A/E_1)=1`
P(A/E_{2}) = Probability of getting a head on the coin, given that the coin is a biased coin that comes up heads 75% of the time.
`⇒ P(A/E_2)=75/100=3/4`
Also, P(A/E_{3}) = Probability of getting a head on the coin, given that the coin is a biased coin that comes up tails 40% of the time.
`⇒ P(A/E_3)=60/100=3/5`
By Baye's theorem,
required probability = P(E_{1}/A)
`=( P(E_1) P(A/E_1))/(P(E_1) P(A/E_1)+P(E_2) P(A/E_2)+P(E_3) P(A/E_3))`
`= (1/3xx1)/(1/3xx1 )+ (1/3xx3/4) + (1/3xx3/5)`
`= 20/47`
Thus, the probability that it was the two-headed coin is 20/47.
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