<PRE><B>Question 628947</B>
Brian a landscape architect submitted a bid on each of three home landscaping projects. He estimates that the probabilities of winning the bid on Project A, Project B, and Project C are .7, .6, .5 respectively. Also the probability of winning a bid on one of the three projects is independent of winning or losing the bids on the other two projects. Find the probability that Brian will:
Win all 3 bids::: 0.7*0.6*0.5 = 0.21
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Win Exactly two of the bids::: 0.7*0.6*0.5 + 0.6*0.3*0.5 + 0.5*0.7*0.4 
= 0.21 +  0.09 + 0.14 = 0.44
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Win exactly one bid::: 0.7*0.4*0.5 + 0.6*0.3*0.5 + 0.5*0.3*04 
= 0.14 + 0.09 + 0.06 = 0.29
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Cheers,
Stan H.
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