THE DIAMOND MERCHANT
(completed solution)
Restated 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) 

As before, we define a few useful events
Let D_{1}= "a diamond is found on the 1st examination" 
Let D_{2}= "a diamond is found on the 2nd examination" 
Let D_{3}= "a diamond is found on the 3rd examination" 
Let Z_{1}= "a zircon is found on the 1st examination" 
Let Z_{2}= "a zircon is found on the 2nd examination" 
Let Z_{3}= "a zircon is found on the 3rd examination" 
 
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(D_{1}
D_{2}
D_{3}) 
= P(D_{1}) ^{.}
P(D_{2}  D_{1}) ^{.}
P(D_{3}  D_{1}D_{2}) 

 = (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: 

