Question 1122898
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It's unusual to talk about "weights" in a problem like this; so I assume you mean the same thing as the probabilities....<br>
The sum of all the probabilities is 5k+8k+2k+5k = 20k; since the sum of all the probabilities is 1, k = 1/20.<br>
Pr({o1,o4}) = 5k+5k = 10k = 10/20 = 1/2.