anonymous
  • 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
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • 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?
Zarkon
  • Zarkon
why are you multiplying those two numbers?
anonymous
  • anonymous
isnt that how you get the probability for AND?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Zarkon
  • Zarkon
not in this case
anonymous
  • anonymous
Then Im completely lost.
Zarkon
  • Zarkon
your events are NOT independent
anonymous
  • anonymous
I think that's the concept that I dont quite understand
anonymous
  • anonymous
would the formula be P(A U B)/P(A)?
Zarkon
  • Zarkon
no
Zarkon
  • Zarkon
use \[P(A^c\cap B^c)=1-P(A\cup B)\]
anonymous
  • anonymous
so the 1-P(none) formula
Zarkon
  • Zarkon
for any event E \[P(E)=1-P(E^c)\] or \[P(E^c)=1-P(E)\]
anonymous
  • anonymous
so how exactly does not being independent effect the problem?
Zarkon
  • Zarkon
events A and B are independent iff \[P(A\cap B)=P(A)P(B)\]
Zarkon
  • Zarkon
you cant use this formula since A and B are not independent
anonymous
  • 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?
Zarkon
  • Zarkon
NO
Zarkon
  • Zarkon
independent events intersect (most of the time)
Zarkon
  • Zarkon
if A and B are independent and P(A)>0 and P(B)>0 then A and B have to intersect.
Zarkon
  • Zarkon
use \[P(A^c\cap B^c)=1-P(A\cup B)=1-[P(A)+P(B)-P(A\cap B)]\]
anonymous
  • 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.
anonymous
  • anonymous
so to get neither dont you need to get rid of all that is in the circles? including the intersection?
Zarkon
  • Zarkon
yes
anonymous
  • anonymous
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)]
anonymous
  • anonymous
oh wait
anonymous
  • anonymous
I see nevermind
anonymous
  • anonymous
|dw:1321205205257:dw| So the answer is .02? that's seems really small.(compared to the other answer)
Zarkon
  • Zarkon
it is .02
anonymous
  • anonymous
ok thanks
Zarkon
  • Zarkon
I would check your key again...if you just change from .4 to .47 then .2775 is not the answer

Looking for something else?

Not the answer you are looking for? Search for more explanations.