Probability

2014-03-06 4:22 am
Let A and B be events with probabilities P(A) = 3/4 and P(B) = 1/3.

(a) Show that 1/12 <= P(A∩B) <= 1/3 and give examples to show that both extremes are possible.
(b) Find corresponding bounds for P(A∪B).

回答 (3)

2014-03-06 6:48 am
✔ 最佳答案
a)P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 3/4 + 1/3 - P(A ∩ B)
13/12 - P(A ∩ B) = P(A ∪ B) ≤ 1
1/12 ≤ P(A ∩ B)Also P(A ∩ B) ≤ P(B) = 1/3 < P(A) = 3/4
∴ 1/12 ≤ P(A ∩ B) ≤ 1/3
For example :
Choosing an integer from 1 to 12 ,
A : Choosing 1 to 9 , then P(A) = 3/4.
B : Choosing 9 to 12 , then P(B) = 1/3.
We have P(A ∩ B) = P(Choosing 9) = 1/12.Choosing an integer from 1 to 12 ,
A : Choosing 1 to 9 , then P(A) = 3/4.
B : Choosing 1 to 4 , then P(B) = 1/3.
We have P(A ∩ B) = P(Choosing 1 to 4) = 1/3.
b)By example in a) , we can see that
if P(A ∩ B) = 1/12 , then P(A ∪ B) = 1 ; if P(A ∩ B) = 1/3 , then P(A ∪ B) = 3/4.Therefore 13/12 - 1/3 ≤ 13/12 - P(A ∩ B) ≤ 13/12 - 1/12
⇒ 3/4 ≤ P(A ∪ B) ≤ 1 .
2014-03-08 11:45 pm
好嘢呀,我係第18個呀~
2014-03-06 6:52 am
(◕‿◕✿)
好野呀,我係第一個呀~


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