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anonymous
 5 years ago
find the indicated probablity if p(a and b)=0.5 p(a)=0.7 and p(b)=0.5 find p(a or b)
anonymous
 5 years ago
find the indicated probablity if p(a and b)=0.5 p(a)=0.7 and p(b)=0.5 find p(a or b)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If p(a)=.7, then p=.7/a if p(b)=.5, then p=.5/a right so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you need to find p of a&b? so p(ab)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it only says p( a or b)

M
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure p(a and b) = 0.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah that's what it says if P(a and B)=0.5

M
 5 years ago
Best ResponseYou've already chosen the best response.0do you have the answer? is it 0.7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am trying to figure it out still

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think for 'or' you add the probibility

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0.5/a+.7/b then? So p=.5/a and p=.7/b Would that be the final answer? Trying to understand this one too.

M
 5 years ago
Best ResponseYou've already chosen the best response.0if A = .5 and if B = .7 and P(A and B) = .5 that means A overlaps completely over B P(A or B) = P(A) + P(B) + P(A and B) = 0.7

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ignore my answer then, not bad for a first shot at a probability problem though ;)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0wait typo should be a "" P(A or B) = P(A) + P(B)  P(A and B) = 0.7

M
 5 years ago
Best ResponseYou've already chosen the best response.0last equation i wrote is wrong sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok I have a new single die one if u guys wanna look at it
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