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INT
Group Title
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
 3 years ago
 3 years ago
INT Group Title
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
 3 years ago
 3 years ago

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INT Group TitleBest ResponseYou've already chosen the best response.0
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?
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
why are you multiplying those two numbers?
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
isnt that how you get the probability for AND?
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
not in this case
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
Then Im completely lost.
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
your events are NOT independent
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
I think that's the concept that I dont quite understand
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
would the formula be P(A U B)/P(A)?
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
use \[P(A^c\cap B^c)=1P(A\cup B)\]
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
so the 1P(none) formula
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
for any event E \[P(E)=1P(E^c)\] or \[P(E^c)=1P(E)\]
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
so how exactly does not being independent effect the problem?
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
events A and B are independent iff \[P(A\cap B)=P(A)P(B)\]
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
you cant use this formula since A and B are not independent
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
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?
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
independent events intersect (most of the time)
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
if A and B are independent and P(A)>0 and P(B)>0 then A and B have to intersect.
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
use \[P(A^c\cap B^c)=1P(A\cup B)=1[P(A)+P(B)P(A\cap B)]\]
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
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.
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
so to get neither dont you need to get rid of all that is in the circles? including the intersection?
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
wouldnt that be adding the intersection? so 1−[P(A)+P(B)+P(A∩B)] instead of 1−[P(A)+P(B)−P(A∩B)]
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
I see nevermind
 3 years ago

INT Group TitleBest ResponseYou've already chosen the best response.0
dw:1321205205257:dw So the answer is .02? that's seems really small.(compared to the other answer)
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
I would check your key again...if you just change from .4 to .47 then .2775 is not the answer
 3 years ago
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