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 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?

Answer 1

Answer:  The probabilities of winning a contract are

$P(A) = \frac{28}{36}$  

$P(B) = \frac{7}{36}$  

$P(C) = \frac{1}{36}$

Let the Probability of C winning the contract - P(C) be 'X'

Then,

Probability of B winning the contract - P(B) will be '7X'     and

Probability of A winning the contract - P(A) will be $\mathbf{P(A) = 4 * P(B) = 4*7X = 28X}$

Since the total of all the probabilities is 1,

$\mathbf{P(A) + P(B) + P(C) =1}$

$\mathbf{28X + 7X + X =1}$

$\mathbf{36X =1}$

$\mathbf{X =\frac{1}{36}}$

So,

$P(A) = \frac{28}{36}$

$P(B) = \frac{7}{36}$

$P(C) = \frac{1}{36}$