THE DIAMOND MERCHANT
(solution to extra question)
 Re-stated extra 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 after 3 examinations, the merchant will find exactly 2 of the diamonds mixed with zircons?
 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
The conditional probabilities assigned to the above tree branches are the same as were assigned in the previous version of this problem. Notice that the red colored events in the tree above are those for which exactly 2 of the 3 diamonds are found after 3 examinations. We seek the probability of the UNION of these three red events.

However, no 2 of these 3 red events can happen at the same time: it is impossible to pick both a diamond AND a zircon on the first examination, for example. Thus, the three red events are mutually exclusive, and to find the probability of their UNION, we add the separate probabilities of the three red events. Indeed, each of these red events is found exactly as we found P ( D1 D2 D3 ) in the previous version of this problem, using the multiplication rule twice (Rolf Pg 380):

 We seek P(D1 D2 Z3) + P(D1 Z2 D3) + P(Z1 D2 D3) = P(D1). P(D2|D1). P(Z3|D1D2) + P(D1). P(Z2|D1). P(D3|D1Z2) + P(Z1). P(D2|Z1) . P(D3|Z1D2) = (3/10) . (2/9) . (7/8)      +      (3/10) . (7/9) . (2/8)      +      (7/10) . (3/9) . (2/8) = (7/120) + (7/120) + (7/120) = 7/40

Note that as it turned out, all three red events had the same probability (7/120), but this is a coincidence: other events in the tree above have different probabilities.