SOLUTION: Please help me solve this of simple way. Thank You. Three contractors (call them A, B, and C) bid on a project to build an addition to the UVA Rotunda. Suppose that you believe

Algebra ->  Probability-and-statistics -> SOLUTION: Please help me solve this of simple way. Thank You. Three contractors (call them A, B, and C) bid on a project to build an addition to the UVA Rotunda. Suppose that you believe       Log On


   



Question 1119886: Please help me solve this of simple way. Thank You.
Three contractors (call them A, B, and C) bid on a project to build an addition to the UVA Rotunda. Suppose that you believe that Contractor A is 4 times more likely to win than Contractor B, who in turn is 7 times more likely to win than Contractor C. What are each of their probabilities of winning?
P(A Wins) =
P(B Wins) =
P(C Wins) =

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


Contractor C is the least likely to win. So let x be the probability that C wins.

Then "contractor B is 7 times MORE likely to win than contractor C" means the probability that contractor B wins is x + 7(x) = 8x.

Then "contractor A is 4 times MORE likely to win than contractor B" means the probability that contractor A wins is 8x + 4(8x) = 40x.

The sum of the probabilities must be equal to 1:

x+%2B+8x+%2B+40x+=+1
49x+=+1
x+=+1%2F49

The probability that C wins is x = 1/49.
The probability that B wins is 8x = 8/49.
The probability that A wins is 40x = 40/49.