Question

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 5 times more likely to win than Contractor B, who in turn is 6 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 1

Answers:

• P(A wins) = 30/37
• P(B wins) = 6/37
• P(C wins) = 1/37

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Step-by-step explanation:

Let

• x = P(A wins)
• y = P(B wins)
• z = P(C wins)

Contractor A is 5 times more likely to get the job compared to contractor B. We can say x = 5y based on that given info.

We can also say that y = 6z since "B is 6 times more likely to win than C"

Using x = 5y and y = 6z, we get....

x = 5y

x = 5(y)

x = 5(6z) .. replace y with 6z

x = 30z

Showing contractor A is 30 times more likely to win compared to contractor C. We'll use this later.

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From here we add the probabilities x,y,z and set this sum equal to 1. The sum of all probabilities in any probability distribution is always 1 to represent 100%. Keep in mind that
`0 \le x \le 1` and the same applies to y and z as well. Basically the x,y,z values must be between 0 and 1.

So,

x+y+z = 1

30z+6z+z = 1 .... replace x with 30z; replace y with 6z

37z = 1

z = 1/37

Meaning that,

x = 30z

x = 30(1/37)

x = 30/37

and,

y = 6z

y = 6(1/37)

y = 6/37

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As a check, we add x,y,z to get

x+y+z = (30/37)+(6/37)+(1/37)

x+y+z = (30+6+1)/37

x+y+z = 37/37

x+y+z = 1

which helps confirm our answer.