mathmath333
  • mathmath333
Probablity Question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathmath333
  • mathmath333
Check 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
  • mathmath333
i found an answer here but it doesn't making sense to me http://www.meritnation.com/ask-answer/question/check-whether-the-following-probabilities-p-a-and-p-b-are/probability/6322947
Vocaloid
  • Vocaloid
in 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)

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mathmath333
  • mathmath333
u mean both conditions or any one condition should be satisfied
Vocaloid
  • Vocaloid
both
Vocaloid
  • Vocaloid
so, 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
  • mathmath333
consistentlyt defined
Vocaloid
  • Vocaloid
not quite, check the first condition again... P(A ∩ B) must be less than or equal to P(A) and P(B)
Vocaloid
  • Vocaloid
notice how P(A ∩ B) is greater than P(B)?
Zarkon
  • Zarkon
these "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
mathmath333
  • mathmath333
i am confused
anonymous
  • anonymous
|dw:1439556261906:dw|
anonymous
  • anonymous
The second one is possible, since they can overlap 0.2 to get A U B = 0.8 |dw:1439556485375:dw|
anonymous
  • anonymous
|dw:1439556527719:dw|
anonymous
  • anonymous
That makes the expression hold.
Zarkon
  • Zarkon
for 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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