Calculating Probabilities for Job Bidding: V and W Firms

[phy]

Two firms V and W consider bidding on a road-building job, which may or may not be awarded depending on the amounts of the bids. Firm V submits a bid and the probability is 3/4 that it will get the job provided firm W does not bid. The probability is 3/4 that W will bid and if it does, the probability that V will get the job is only 1/3.
1. What is the probability that V will get the job?
2. If V gets the job, what is the probability that W did not bid?

I spoke to my professor on how I should do this question and all she would tell me was that when we're given a lot of information in the question, apply Bayes' Theorem but I'm really not sure how to do that. I've never taken any courses in probability or stats before so any suggestions or ideas on how I could start off the question would be greatly appreciated. Thanks =)

 

[Justin Lazear]

Problems like these aren't completely dissimilar to the basic counting approach to probability, i.e. # of ways to succeed / # of ways total.

Company W has a 3/4 probability of bidding. We have two cases.

Case 1: Company W bids. Chance of this happening: 3/4

Case 2: Company W does not bid. Chance of this happening : 1/4

Within Case 1, there are again two cases.

Case 1: Company W bids (3/4):
----Case a: Company V wins the bid (1/3)
----Case b: Company V does not win the bid (2/3)

Within case 2, there are also two cases.

Case 2: Company W does not bid (1/4):
----Case a: Company V wins the bid (3/4)
----Case b: Company V does not win the bid (1/4)

So in what cases does Company V get the bid? Cases 1.a and 2.a. So we add the probabilities of these cases happening.

P(1.a) = P(1)*P(1.a) = (3/4)*(1/3)
P(2.a) = P(2)*P(2.a) = (1/4)*(3/4)

P(V wins the bid) = P(1.a) + P(2.a) = 3/4*1/3 + 1/4*3/4 = 3/4(7/12) = 7/16

For part b, it is assumed that V wins the bid. So Cases 1.b and 2.b clearly did not happen. This changes the total probability, analagous to reducing the (total # of ways). Now the only two viable options are Cases 1.a and 2.a, so the total probability is P(1.a) + P(2.a). The probability of Company W not bidding is exhibited by Case 2.a, so the probability that Company W does not bid given that Company V wins the bid will be P(2.1)/(P(1.a) + P(2.a)).

--J

 

[wais]

Hi there! It's great that you reached out to your professor for help with this question. Bayes' Theorem is definitely a useful tool for calculating probabilities in these types of scenarios.

First, let's define some variables to make things easier to understand:

V = probability that firm V gets the job
W = probability that firm W gets the job
B = probability that W bids on the job

Now, let's use Bayes' Theorem to calculate the probability that V gets the job:

P(V) = P(V|B') * P(B') + P(V|B) * P(B)

Where B' represents the event that W does not bid on the job.

We know that P(V|B') = 3/4, meaning that if W does not bid, there is a 3/4 chance that V will get the job. We also know that P(B) = 3/4, so the probability of W bidding is 3/4.

Now, we need to find P(V|B) - the probability that V gets the job given that W does bid. We are given that if W bids, there is only a 1/3 chance that V will get the job. So, P(V|B) = 1/3.

Putting it all together, we have:

P(V) = (3/4) * (1 - 3/4) + (1/3) * (3/4) = 3/16 + 1/4 = 11/16

Therefore, the probability that V will get the job is 11/16.

Now, for the second part of the question - if V gets the job, what is the probability that W did not bid?

Using Bayes' Theorem again, we have:

P(B'|V) = P(B'|V) * P(V) / P(B)

We know that P(V) = 11/16 from the previous calculation. We also know that P(B) = 3/4. And we can calculate P(B'|V) by subtracting P(B|V) from 1, since these two events are complementary. P(B|V) can be calculated as P(V|B) * P(B) / P(V).

So, we have:

P(B'|V) = (1 - P(V|B) * P(B) / P(V)) * P(V