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i HAVE ATTACHED THE PROBLEM STATEMENT IMAGE .CAN ANYONE SOLVE THIS PROBLEM WITH LOGIC AND NOT USING BAYES THEOREM.

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one year ago

```							Let E1 E2 E3 be events such that E1 represents the selection of Box I E2 represents Selection of Box II And E3 represents the Selection of Box III Let A denotes the event such that the coin drawn is of gold. ∵ there are 3 boxes- ∴ P(E1) = P(E2) = P(E3) = 1/3 Since, all boxes contains two coins. And box I have both the coins gold. Box II has only silver coins, and box III has one gold and one silver coin. ∴ P(A|E1) = P(drawing gold coin from box I) = 2/2 = 1 Similarly, P(A|E2) = 0 And P(A|E3) = 1/2 We need to find the probability of the event such that another coin in the box is also gold and this is only possible if the coin is drawn from the box I i.e. we need to find P(E1|A). From Baye’s theorem– ⇒ P(E1|A) = ∴ The probability that the other coin in the box is also of gold = 2/3
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one year ago
```							Dear student Let E1 , E2 nd E3 be the events that boxex I II and III are chosen respectively ∵ there are 3 boxes- ∴ P(E1) = P(E2) = P(E3) = 1/3 Since, all boxes contains two coins. And box I have both the coins gold. Box II has only silver coins, and box III has one gold and one silver coin. ∴ P(A|E1) = P(drawing gold coin from box I) = 2/2 = 1 Similarly, P(A|E2) = 0 And P(A|E3) = 1/2 We need to find the probability of the event such that another coin in the box is also gold and this is only possible if the coin is drawn from the box I i.e. we need to find P(E1|A). From Baye’s theorem– ⇒ P(E1|A) =[P(E1)P(A/E1)] / [P(E1) P(A/E1) + P(E2)P(A/E2)+ P(E3)P(A/E3)][This part was missing in above ans] ∴ The probability that the other coin in the box is also of gold = 2/3
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one year ago
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