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 3 years ago
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
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

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INT
 3 years ago
Best ResponseYou've already chosen the best response.0So .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?

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

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

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

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

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

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

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

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

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

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

INT
 3 years ago
Best ResponseYou've already chosen the best response.0So 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?

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

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

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

INT
 3 years ago
Best ResponseYou've already chosen the best response.0dw: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.

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

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

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

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