Consider 3 boxes. Box A contains 2 pennies and 4 nickels; box B contains 6 penni

Question

Consider 3 boxes. Box A contains 2 pennies and 4 nickels; box B contains 6 pennies and 6 nickels; and box C contains 1 penny and 3 nickels. If one coin is selected at random from each box, what is the probability that a penny was selected from box A, given that exactly 2 pennies were selected? Prove (b) and (c) of Theorem 3.2. A pathologist is certain that the cause of death is due to one of three poisons. Up to this point she feels that the test results indicate probabilities of .6, .3, and .1 that the death was caused by poisons 1, 2, and 3, respectively. She conducts another test. She was startled by the result, because the probability of this result if the cause of death is poison 1, 2, or 3, is .1, .7, or .2, respectively. How might the pathologist use Bayes' Theorem to update her

Explanation / Answer

given

3 boxes

A contains 2 pennies and 4 nickels

B contains 6 Pennies and 6 Nickeles

C Contains 1 penny and 3 nickeles

given P(A)=P(B) =P(C) =1/3

let S is selected coin is Penny so

P(S|A) =1/3 P(S|B) =1/2 P(S|C)=1/4

let D is event that 2 penny selected

E is event that penny is selected from A

so we Have to find P(E|D) =P(E and D) /P(D)

now

P(D) =Penny from A and B ,nickele from C or Penny from A and C ,nickel from B or nickele from A ,Penny from B and C

=(1/3)(1/2) (3/4) +(1/3)(1/4)(1/2) +(2/3)(1/2)(1/4)=(1/8) +(1/24) +(1/12) =6/24 =1/4

P(E and D) =Penny from A and B ,nickele from C or Penny from A and C ,nickel from B

  =(1/3)(1/2) (3/4) +(1/3)(1/4)(1/2)

=(1/8) +(1/24) =4/24 =1/6

now

P(E|D) =(1/6)/(1/4) =2/3

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