Bayes' theorem problem, Struggling with this the whole night, .Thank you.

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Two states of nature exist for a particular situation: a good economy and a poor economy. An economic study may be performed to obtain more information about which of these will actually occur in the coming year. The study may forecast either a good economy or a poor economy. Currently there is 60% chance that the economy will be good and a 40% chance that the economy will be poor. In the past, whenever the economy was good, the economic study predicted it would be good 80% of the time. (The other 20% of the time the prediction was wrong.) In the past, whenever the economy was poor, the economic study predicted it would be 90% of the time.(The other 10% of the time the prediction was wrong.)

a) Use Bayes' theorem and find the following:
P(good economy| prediction of good economy)
P(poor economy| prediction of good economy)
P(good economy| prediction of poor economy)
P(poor economy| prediction of poor economy)

b) Suppose the initial (prior) probability of a good economy is 70% (instead of 60%), and the probability of a poor economy is 30% (instead of 40%). Find the posterior probabilities in part a based on these new values.
 
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Hi hildanhk,

Can you show us what you have tried so our helpers can see where you are stuck and can then offer help?:)
 
I am try to use the formula and plug the numbers in. But I could not get the correct answers.
P(A|B)= P(B|A)P(A)/P(B|A)P(A)+P(B|A')P(A')

For example, I put
P(good economy| prediction of good economy)= 0.8*0.6/0.8*0.6+0.9*0.6= 0.51

But the answer should be 0.923
 
hildanhk said:
I am try to use the formula and plug the numbers in. But I could not get the correct answers.
P(A|B)= P(B|A)P(A)/P(B|A)P(A)+P(B|A')P(A')

For example, I put
P(good economy| prediction of good economy)= 0.8*0.6/0.8*0.6+0.9*0.6= 0.51

But the answer should be 0.923

Welcome to MHB, hildanhk! :)

You seem to have mixed up the last part (0.9 * 0.6).

EDIT: The phrasing is somewhat confusing.
We have P(predict good | good) = 80%.
The remainder is P(predict poor | good) = 20%
And from P(predict poor | poor) = 90%,
we get that P(predict good | poor) = 10%.

So it should be (edited):
$$P(B|A')\ P(A') = P(\text{predict good}|\text{poor})\ P(\text{poor}) = 0.1 \cdot 0.4$$

So you should have:
$$P(\text{good}|\text{predict good}) = \frac{0.8 \cdot 0.6}{0.8 \cdot 0.6 + 0.1 \cdot 0.4} \approx 0.923$$
 
Last edited:
I have updated my previous post, since I made a mistake with my interpretation of the wording in the problem statement.