Statistics 1- Probability of both mutually exclusive and independent events. Watch

Michael Boom
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Hello guys, could you please tell me how to work out:
-P(A u B) of both mutually exclusive and independent events
-P(A n B) of both mutually exclusive and independent events
-P(A/B) of both mutually exclusive and independent events
-P(B/A) of both mutually exclusive and independent events
I wasn't at college the day my teacher covered this because i was sick, i feel a bit behind and confused because different forums say different stuff.
Thank you very much in advance
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BobBobson
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If A and B are both mutually exclusive and independent:
 

P(A \cup B) = P(A) + P(B)

P(A \cap B) = 0

P (A \mid B) = 0

P (B \mid A) = 0
If you need an explanation of any of these, feel free to ask.
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Michael Boom
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(Original post by BobBobson)
If A and B are both mutually exclusive and independent:
 

P(A \cup B) = P(A) + P(B)

P(A \cap B) = 0

P (A \mid B) = 0

P (B \mid A) = 0
If you need an explanation of any of these, feel free to ask.
Oh I feel so bad because i didn't explain myself properly, i wanted to know how to work them out first if they are independent, and secondly if they are mutually exclusive, not both at the same time.
I am sorry i made you waste time, but if you are kind enough could you please tell me how to work them out only if they are independent, then how to work them out if they are mutually exclusive.
Thanks
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BobBobson
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(Original post by Michael Boom)
Oh I feel so bad because i didn't explain myself properly, i wanted to know how to work them out first if they are independent, and secondly if they are mutually exclusive, not both at the same time.
I am sorry i made you waste time, but if you are kind enough could you please tell me how to work them out only if they are independent, then how to work them out if they are mutually exclusive.
Thanks
Oh, sure, no worries
Mutually Exclusive and Independent. These are the same equations as below, but just simplified for specifically Exclusive events:


P(A \cup B) = P(A) + P(B)

P(A \cap B) = 0

P(A \mid B) = 0

P(B \mid A) = 0
Not mutually exclusive but it is independent:


P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cap B) = P(A) + P(B) - P(A \cup B) \mathrm{\ OR\ } P(A \mid B) \times P(B) \mathrm{\ OR\ } P(B \mid A) \times P(A)

P(A \mid B) = P(A \cap B) \div P(B)

P(B \mid A) = P(A \cap B) \div P(A)

Dependent events will be the same as independent events, but usually, you would have to draw a tree diagram since each probabiblity won't be obvious from the question.
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Michael Boom
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(Original post by BobBobson)
Oh, sure, no worries
Mutually Exclusive and Independent. These are the same equations as below, but just simplified for specifically Exclusive events:


P(A \cup B) = P(A) + P(B)

P(A \cap B) = 0

P(A \mid B) = 0

P(B \mid A) = 0
Not mutually exclusive but it is independent:


P(A \cup B) = P(A) + P(B) - P(A \cap B)

P(A \cap B) = P(A) + P(B) - P(A \cup B) \mathrm{\ OR\ } P(A \mid B) \times P(B) \mathrm{\ OR\ } P(B \mid A) \times P(A)

P(A \mid B) = P(A \cap B) \div P(B)

P(B \mid A) = P(A \cap B) \div P(A)

Dependent events will be the same as independent events, but usually, you would have to draw a tree diagram since each probabiblity won't be obvious from the question.

Thank you very much
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