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Mathematics 21 Online
OpenStudy (anonymous):

There is a 0.63 probability that Mobil stock will rise, a 0.75 probability Gulf stock will rise and a 0.4 chance that both stocks will rise. Find the probability that neither stock will rise. There is a 0.63 probability that Mobil stock will rise, a 0.75 probability Gulf stock will rise and a 0.4 chance that both stocks will rise. Find the probability that neither stock will rise. @Mathematics

OpenStudy (anonymous):

So .37*.25=.0925 But the answer key for a similar problem( only difference is .47 instead of .4), the answer is .2775. The answers shouldnt differ that much right?

OpenStudy (zarkon):

why are you multiplying those two numbers?

OpenStudy (anonymous):

isnt that how you get the probability for AND?

OpenStudy (zarkon):

not in this case

OpenStudy (anonymous):

Then Im completely lost.

OpenStudy (zarkon):

your events are NOT independent

OpenStudy (anonymous):

I think that's the concept that I dont quite understand

OpenStudy (anonymous):

would the formula be P(A U B)/P(A)?

OpenStudy (zarkon):

no

OpenStudy (zarkon):

use \[P(A^c\cap B^c)=1-P(A\cup B)\]

OpenStudy (anonymous):

so the 1-P(none) formula

OpenStudy (zarkon):

for any event E \[P(E)=1-P(E^c)\] or \[P(E^c)=1-P(E)\]

OpenStudy (anonymous):

so how exactly does not being independent effect the problem?

OpenStudy (zarkon):

events A and B are independent iff \[P(A\cap B)=P(A)P(B)\]

OpenStudy (zarkon):

you cant use this formula since A and B are not independent

OpenStudy (anonymous):

So in terms of a venn diagram, independent is two circles that dont intersect. and not independent has two circles that do intersect. and in this problem the intersection is .4? So I need to find the sum of each part of the venn diagram, subtract from one (to get the number outside the diagram) which would be the "neither". Correct?

OpenStudy (zarkon):

NO

OpenStudy (zarkon):

independent events intersect (most of the time)

OpenStudy (zarkon):

if A and B are independent and P(A)>0 and P(B)>0 then A and B have to intersect.

OpenStudy (zarkon):

use \[P(A^c\cap B^c)=1-P(A\cup B)=1-[P(A)+P(B)-P(A\cap B)]\]

OpenStudy (anonymous):

|dw:1321204843404:dw| Is what Im picturing Basically, I was doing P(Ac∩Bc)=1−P(A∪B)=1−[P(A)+P(B)−P(A∩B)] except that mine adds instead of subtracts the intersection.

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