


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 ( D_{1} D_{2} D_{3} ) in the previous version of this problem, using the multiplication rule twice (Rolf Pg 380):
We seek P(D_{1} D_{2} Z_{3}) + P(D_{1} Z_{2} D_{3}) + P(Z_{1} D_{2} D_{3}) 
= P(D_{1})^{.} P(D_{2}D_{1})^{.} P(Z_{3}D_{1}D_{2}) + P(D_{1})^{.} P(Z_{2}D_{1})^{.} P(D_{3}D_{1}Z_{2}) + P(Z_{1})^{.} P(D_{2}Z_{1}) ^{.} P(D_{3}Z_{1}D_{2}) 
= (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.
