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Mutally Exclusive

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
Let A and B be two events defi ned on a sample space S. Suppose that P(A) = 0:4 and
P(B) = 0:8. Is it possible to say whether or not A and B are mutually exclusive? Justify
your answer.

I know that mutually exclusive means that its impossible for them both to occur at the same time.

Therefore I think that the answer is no, because the ratio of P(A) 0:4 and P (B) 0:8 would not be able to occur in both events but, Im not really sure how they got the ratio of 0:4 and 0:8 I just kinda guess on this one! I know how to figure it out if they gave you an A B and C but since they only gave you an A and B Im fairly confused.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,688
quaidy4 said:
Let A and B be two events defi ned on a sample space S. Suppose that P(A) = 0:4 and P(B) = 0:8. Is it possible to say whether or not A and B are mutually exclusive? Justify
your answer.
This must be true: \(\displaystyle 1\ge\mathcal{P}(A\cup B)=\mathcal{P}(A)+\mathcal{P}(B)-\mathcal{P}(A\cap B).\)
Is that possible if \(\displaystyle \mathcal{P}(A\cap B)= 0~?\)
 

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
I don't really understand that though, I understand the formula and such but I don't understand how you would determine if it's possible or not when only given 2 numbers.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,688
quaidy4 said:
I don't really understand that though, I understand the formula and such but I don't understand how you would determine if it's possible or not when only given 2 numbers.
What does \(\displaystyle \mathcal{P}(A)+\mathcal{P}(B)=~?\)
Remember that \(\displaystyle 1\ge\mathcal{P}(A\cup B)\)
How can you make that happen unless \(\displaystyle \mathcal{P}(A\cap B)>0~?\)
 

quaidy4

New member
Joined
Jan 19, 2011
Messages
15
thanks I understand it now!
 
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