A company has three consulting firms named as A, B and C with probabilities 0.40

Question

A company has three consulting firms named as A, B and C with probabilities 0.40, 0.25 and 0.35. From past experience, it is known that the probability of cost overruns for the firms is 0.05, 0.03 and 0.15, respectively. Suppose a cost overrun is experienced by the company. a) What is the probability that consulting firms involved is firm C? b) What is the probability that it was firm A? c) What is the probability of a cost overrun by the company? d) What is the probability that overrun cost is experienced by firm A or B? e) What is the probability that cost is overrun is not experienced by firm B?

Explanation / Answer

Here, we are given that:

P(A) = 0.40,
P(B) = 0.25,
P(C) = 0.35

Also we are given here that:

P( cost overrun | A) = 0.05,
P( cost overrun | B) = 0.03,
P( cost overrun | C) = 0.15

Using law of total probability, we get here:

P( cost overrun ) = P( cost overrun | A)P(A) + P( cost overrun | B)P(B) + P( cost overrun | C)P(C)

P( cost overrun ) = 0.05*0.40 + 0.03*0.25 + 0.15*0.35 = 0.08

a) Now given that there is cost overrun, probability that firm C was involved is computed using bayes theorem here as:

P( C | cost overrun ) = P( cost overrun | C)P(C) / P( cost overrun )

P( C | cost overrun ) = 0.15*0.35 / 0.08 = 0.6563

Therefore 0.6563 is the required probability here.

b) Now similarly, probability that it was firm A is computed here as

P( A | cost overrun ) = 0.05*0.40 / 0.08 = 0.25

Therefore 0.25 is the required probability here.

c) The probability of a cost overrun by the company as computed using the law of total probability is given to be 0.08

Therefore 0.08 is the required probability here.

d) Here the required probability is computed as:

P( A | cost overrun ) + P( B | cost overrun ) = (0.05*0.40 + 0.03*0.25) / 0.08 = 0.3438

Therefore 0.3438 is the required probability here.

e) The required probability here is computed as:

= 1 - P( B | cost overrun )

= 1 - (0.03*0.25)/ 0.08

= 0.9063

Therefore 0.9063 is the required probability here.

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