Company X has submitted bids on two construction contracts, successively.

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Zpoutof

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Company X has submitted bids on two contracts, successively.
The company believes that there is a 40% probability of winning the first contract. If the company wins the first contract, the probability of winning the second contract is 70%.
If the company loses the first contract, the probability of winning the second contract is 50%.

Question:What is the probability that the would win either contract?

I am unsure if the correct answer will be 0.42 or do we need to use the addition rule to solve this? Which method is correct?
 
Last edited:
Zpoutof said:
Company X has submitted bids on two construction contracts, successively.
The company believes that there is a 40% probability of winning the first contract. If the company wins the first contract, the probability of winning the second contract is 70%.
If the company loses the first contract, the probability of winning the second contract is 50%.
Question:What is the probability that the would win either contract?
I am unsure if the correct answer will be 0.42 or do we need to use the addition rule to solve this? Which method is correct?
Click to expand...
Here is what you need to find the final answer:
\(\displaystyle \mathcal{P}(I\cup II)=\mathcal{P}(I)+\mathcal{P}(II)-\mathcal{P}(I\cap II)=\mathcal{P}(I)+\mathcal{P}(II)-\mathcal{P}(II|I)\mathcal{P}(I) \).

Now the are parts of that you must find from what is known. So please reply showing your work.
 
Using the formula, it will be 0.4+0.5 -(0.4 X 0.5) =0.7, however I do not understand why the answer is not (0.4X0.3=0.12) + (0.6 X 0.5= 0.3) =0.42 from the tree diagram.
 
Zpoutof said:
Company X has submitted bids on two construction contracts, successively.
The company believes that there is a 40% probability of winning the first contract. If the company wins the first contract, the probability of winning the second contract is 70%.
If the company loses the first contract, the probability of winning the second contract is 50%.

Question:What is the probability that the would win either contract?

I am unsure if the correct answer will be 0.42 or do we need to use the addition rule to solve this? Which method is correct?
Click to expand...
You have four distinct cases on which others are built: (W, W), (W, F), (F, W), (F, F) and they have to add to 1 [or 100% if you prefer]. So what are you told
(1) (W, W) + (W, F) = 0.4 [the probability of winning the first is 40%]
(2) (W, W) = [(W, W) + (W, F)] * 0.7 [If we win the first, the probability of winning the second is 70%]
(3) ...
What can we deduce
(a) (F, W) + (F, F) = 1 - [(W, W) + (W, F)] = 0.6
(b) ...

Can you continue from there?

Note what this implies in the way of standard probability formulas such as what pka posted.
 
Zpoutof said:
Using the formula, it will be 0.4+0.5 -(0.4 X 0.5) =0.7, however I do not understand why the answer is not (0.4X0.3=0.12) + (0.6 X 0.5= 0.3) =0.42 from the tree diagram.
Click to expand...
We cannot see your tree diagram, so no comment.
You are given three facts: P(I)=0.4, P(II|I)=0.7, P(II|I^c)=0.5.

From those you can get: P(II & I)=0.28, P(II & I^c)=0.30. From which you get P(II)=0.58.

Can you finish?
 
Zpoutof said:
Using the formula, it will be 0.4+0.5 -(0.4 X 0.5) =0.7, however I do not understand why the answer is not (0.4X0.3=0.12) + (0.6 X 0.5= 0.3) =0.42 from the tree diagram.
Click to expand...
One [the 0.70] is the probability of winning at least one contract. The other [0.42] is the probability of winning only one contract. The way the question is worded is, IMO, ambiguous but I believe that the 0.7 answer is correct as the problem was stated.
 
Sorry, my bad. The question did ask us to use tree diagram to answer this question. Here is the diagram I drew. So I was wondering what it means if I add up Win first lose second (WFLS) and Lose first win second (LFWS) =0.42 means? Is the meaning like what Ishuda has stated? Or addition rule is still the correct method to answer this question?
 
Last edited:
Zpoutof said:
Sorry, my bad. The question did ask us to use tree diagram to answer this question. Here is the diagram I drew. So I was wondering what it means if I add up Win first lose second (WFLS) and Lose first win second (LFWS) =0.42 means? Is the meaning like what Ishuda has stated? Or addition rule is still the correct method to answer this question?
Click to expand...

The OP said
Zpoutof said:
Question:What is the probability that the would win either contract?
Click to expand...
In mathematics, "either" means one or both. So add the first three numbers on the tree diagram.
 
Zpoutof said:
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Sorry, my bad. The question did ask us to use tree diagram to answer this question. Here is the diagram I drew. So I was wondering what it means if I add up Win first lose second (WFLS) and Lose first win second (LFWS) =0.42 means? Is the meaning like what Ishuda has stated? Or addition rule is still the correct method to answer this question?
Click to expand...
Yes. My (W, L) and (L,W) are equivalent to your WFLS and LFWS respectively.
 
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