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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!