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? - Mathematics

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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 E1 be the event of selecting the two-headed coin,
E2 be the event of selecting the biased coin that comes up heads 75% of the times,
E3 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/E1) = Probability of getting a head on the coin, given that the coin is two-headed.

`⇒ P(A/E_1)=1`

P(A/E2) = 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/E3) = 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(E1/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.

 
  Is there an error in this question or solution?
2013-2014 (March) All India Set 1

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