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

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

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Grade:12th pass

2 Answers

Arun
25763 Points
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

Vikas TU
14149 Points
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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