Probability Example: A vs B | Competing Software Companies

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SUMMARY

The discussion centers on the probability of two competing software companies, A and B, winning a contract, where Company A is twice as likely to win as Company B. The established probabilities are P(A) = 2/3 and P(B) = 1/3, derived from the relationship P(A) + P(B) = 1 and the condition P(B) = 2P(A). The confusion regarding the denominator of 3 arises from the need to normalize the probabilities based on their relative likelihoods, confirming that the events are mutually exclusive.

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r0bHadz
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Homework Statement


I was reading this example in a book...

"
Two competing software companies are after an important contract. Company
A is twice as likely to win this competition as company B. Hence, the probability to
win the contract equals 2/3 for A and 1/3 for B."

Homework Equations

The Attempt at a Solution


What is the significance of the word "twice as likely?" It seems to me like if company 1 had a probability of .5, then there is no way company 2 had a probability of 1

And where did they get the 3 in the denominator of the fraction? How can we know that its not 2/7 and 4/7 for the two companies?
 
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r0bHadz said:
It seems to me like if company 1 had a probability of .5, then there is no way company 2 had a probability of 1
Yes, the events "company 1 is twice as likely to win" and "company 1 has 0.5 probability" exclude each other.
r0bHadz said:
And where did they get the 3 in the denominator of the fraction? How can we know that its not 2/7 and 4/7 for the two companies?
The problem makes the assumption that exactly one of them will win the contract. P(A)+P(B)=1. Together with P(B)=2P(A) you can calculate the probabilities.
 
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