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mathmath333
 one year ago
Probablity Question
mathmath333
 one year ago
Probablity Question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6 (ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i found an answer here but it doesn't making sense to me http://www.meritnation.com/askanswer/question/checkwhetherthefollowingprobabilitiespaandpbare/probability/6322947

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1in order to be consistently defined: P(A ∩ B) must be less than or equal to P(A) and P(B) P(A ∪ B) must be greater than or equal to P(A) and P(B)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0u mean both conditions or any one condition should be satisfied

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1so, for the first one (i) what do you think the answer is? (i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0consistentlyt defined

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1not quite, check the first condition again... P(A ∩ B) must be less than or equal to P(A) and P(B)

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.1notice how P(A ∩ B) is greater than P(B)?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1these "P(A ∩ B) must be less than or equal to P(A) and P(B) P(A ∪ B) must be greater than or equal to P(A) and P(B)" are necessary conditions but not sufficient

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439556261906:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The second one is possible, since they can overlap 0.2 to get A U B = 0.8 dw:1439556485375:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439556527719:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That makes the expression hold.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1for example you can't have P(A)=.8 , P(B)=.9 and P(A\(\cap\)B)=0 and you can't have P(A)=.2 , P(B)=.3 and P(A\(\cup\)B)=.8 they satisfy "P(A ∩ B) must be less than or equal to P(A) and P(B) P(A ∪ B) must be greater than or equal to P(A) and P(B)" but they are not consistently defined
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