THE DIAMOND MERCHANT
(completed solution)
 Re-stated problem A diamond merchant accidentally dropped 3 valuable diamonds and 7 less valuable but similar zircons together into a bowl. To separate the two kinds of gems, the merchant must examine each with a hand lens. What is the probability that the merchant will find the 3 diamonds after 3 examinations? (obviously, two could never be sufficient)
 Let D1= "a diamond is found on the 1st examination" Let D2= "a diamond is found on the 2nd examination" Let D3= "a diamond is found on the 3rd examination" Let Z1= "a zircon is found on the 1st examination" Let Z2= "a zircon is found on the 2nd examination" Let Z3= "a zircon is found on the 3rd examination"

 1st examabove

 2nd examabove

 3rd examabove

Conditional probabilities have now been assigned to all branches of the tree which represents this problem. Using the Multiplication rule twice (Rolf Pg 537):

We seek P(D1 D2 D3) = P(D1) . P(D2 | D1) . P(D3 | D1D2)
= (3/10) . (2/9) . (1/8) = 1/120

And now that you are aware of the methods in this solution, try answering the following question about the same merchant and his/her bowl of ten mixed gems:

 What is the probability that after 3 examinations, the merchant will have located exactly 2 of his 3 diamonds?
Hint: all needed info is in the tree above
 Give up? Try the: