R raybies New member Joined Jan 7, 2021 Messages 9 Feb 21, 2021 #1 P(A or B)=P(A)+P(B). Is this an accurate way to find the probability? Why or why not?
D Deleted member 4993 Guest Feb 21, 2021 #2 raybies said: P(A or B)=P(A)+P(B). Is this an accurate way to find the probability? Why or why not? Click to expand... You tell us .. then we will critique your logic and if needed we will guide you to correct answer. Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
raybies said: P(A or B)=P(A)+P(B). Is this an accurate way to find the probability? Why or why not? Click to expand... You tell us .. then we will critique your logic and if needed we will guide you to correct answer. Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Feb 21, 2021 #3 raybies said: P(A or B)=P(A)+P(B). Is this an accurate way to find the probability? Why or why not? Click to expand... NO! it is the same a counting: \(\mathcal{P}(A \cup B) = \mathcal{P}(A) + \mathcal{P}(B) - \mathcal{P}(A \cap B)\)
raybies said: P(A or B)=P(A)+P(B). Is this an accurate way to find the probability? Why or why not? Click to expand... NO! it is the same a counting: \(\mathcal{P}(A \cup B) = \mathcal{P}(A) + \mathcal{P}(B) - \mathcal{P}(A \cap B)\)