SOLUTION: THE PROBABILITY THAT AN AMERICAN INDUSTRY WILL LOCATE IN MUNICH IS 0.7, THE PROBABILITY THAT IT WILL LOCATE IN BRUSSELS IS 0.4, AND THE PROBABILITY THAT IT WILL LOCATE IN EITHER MU

Question 643396: THE PROBABILITY THAT AN AMERICAN INDUSTRY WILL LOCATE IN MUNICH IS 0.7, THE PROBABILITY THAT IT WILL LOCATE IN BRUSSELS IS 0.4, AND THE PROBABILITY THAT IT WILL LOCATE IN EITHER MUNICH OR BRUSSELS OR BOTH IS 0.8. WHAT IS THE PROBABILITY THAT THE INDUSTRY WILL LOCATE
A. IN BOTH CITIES?
B. IN NEITHER CITY?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let m represent munich and b represent brussels.
equations become:
p(m) = .7
p(b) = .4
p(m or b or both) = .8
the general formula is:
p(m or b or both) = p(m) + p(b) - p(m and b)
replacing variables with their values, the formula becomes:
.8 = .7 + .4 - p(m and b)
we can solve for p(m and b) to get:
p(m and b) = .7 + .4 - .8 which becomes:
p(m and b) = .3
p(both cities) = p(m and b) = .3
p(neither city) is equal to the probability that the location will not be in munich nor will it be in brussels nor will it be in both.
this is equal to 1 - p(m or b or both).
since p(m or b or both) is equal to .8, then p(neither city) becomes 1 - .8 which is equal to .2.
understanding why p(m or b or both) is equal to p(m) + p(b) - p(m and b) helps to understand why the formula is the way it is.
p(m) includes p(m and b)
p(b) includes p(m and b) as well.
when you add p(m) and p(b) together, you are adding p(m and b) twice.
that is why you have to subtract p(m and b) once.
it removes the double counting.
i'll try to show you what the numbers mean.
probabilities and percents go together.
.8 probability means that, if you could do it all over again many times ad nauseum, the event will occur 80% of the time.
with that interpretation, we get:
80% of the time the location will be munich or brussels or both.
this breaks down as follows:
30% of the time the location will be munich and brussels.
40% of the time the location will be munich only.
10% of the time the location will be brussels only.
30% + 40% + 10% = 80% which is the percent of the time that the location will be either munich or brussels or both.
the percent of the time it will be munich is equal to 70% which is equal to 40% of the time that it will be munich only plus 30% of the time that it will be munich and brussels.
the percent of the time it will be brussels is equal to 40% which is equal to 10% of the time that it will be brussels only plus 30% of the time that it will be munich and brussels.
to summarize your answers:
p(m) = .7
p(b) = .4
p(m or b or both) = .8
p(m and b) = .3
p(not (m or b or both)) = 1 - .8 = .2

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