Is this true in probability? P(AUB)' = (P(A) + P(B)) '

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SUMMARY

The discussion centers on the probability expression P(A ∪ B)' = (P(A) + P(B))'. The user initially calculates P(B) using the provided probabilities P(A) = 0.4, P(A ∩ B) = 0.1, and P(A' ∩ B') = 0.2, leading to an incorrect conclusion of P(B) = -0.5. The correct interpretation reveals that the formula holds true only when events A and B do not occur simultaneously. The general formula for the union of two events is P{A ∪ B} = P{A} + P{B} - P{A ∩ B}.

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huan.conchito
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Please help me with Probability

is this true in probability? P(AUB)' = (P(A) + P(B)) '

The question is
a) Assume that P(A) = 0.4 P(AnB)=0.1 P(A'nB')=0.2
P(B) = ?
what i did is:
P(AUB)= P(A)+P(B)- P(AnB)
P(AUB)= 0.4 + P(B)-0.1
P(A'nB')= 0.2 = P(AUB)' :confused: = 0.2 = 1 - (0.4 + P(B)-0.1)
P(B)= -0.5

NVM I GOT IT MYSELF
 
Last edited:
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Only if A and B don't occur at the same time (simultaneously)

marlon
 
what is the formula to manipulate such an expression if they occur at the same time?
 
huan.conchito said:
is this true in probability? P(AUB)' = (P(A) + P(B)) '
Here is the general form:
P{A ∪ B} = P{A} + P{B} - P{A ∩ B}


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