Probability (altered die problem)

  • Thread starter bobsz
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Hello everyone,

I need some help with the following prob:

A bag contains 3 dice, 2 fair and 1 altered with all odd numbers replace with "1". One die is randomly selected and rolled independently twice. If the outcomes of both rolls were "1" and "1", what is the prob that the selected die is the altered die?
Here's what I've done:

Let A= Altered die and B=outcome is 1

then P(A|B) = P(A∩B)/ P(B)
P(A∩B) = 1/12 ( outcome of 3 11's)
P(B)= 5/9 ( total outcome of 1's)

therefore P(A|B) = (1/12)/(5/9) = 3/20

Is this correct? Is there another way to approach this problem?

Thanks!
 

Answers and Replies

  • #2
Office_Shredder
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First, it should be obvious that the odds of the altered die being picked is greater than 1/3.

In your attempt, the P(B) isn't 5/9, since you can only get 5 1s out of 18 total numbers. And the probability of it being the altered die and the outcome is one is actually 1/3*1/2, which is 1/6, not 1/12. That gives your method a much better answer.
 

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