Probability question (rewriting an expression?)
blamocur Elite Member Joined Oct 30, 2021 Messages 3,205 Oct 30, 2021 #2 The last transformation uses the distributive property of multiplication and addition/subtraction.
Harry_the_cat Elite Member Joined Mar 16, 2016 Messages 3,782 Oct 30, 2021 #3 -Pr(B)[1 - Pr(a)] doesn't "become" [1 - Pr(B)]. To make it easier to see what is happening here , let Pr(A) = a and Pr(B) = b. What you have on the second last line is [1-a] - b[1-a]. This is an expression with 2 terms separated by the middle minus sign. Factorising out the common factor [1-a] gives [1-a][1-b].
-Pr(B)[1 - Pr(a)] doesn't "become" [1 - Pr(B)]. To make it easier to see what is happening here , let Pr(A) = a and Pr(B) = b. What you have on the second last line is [1-a] - b[1-a]. This is an expression with 2 terms separated by the middle minus sign. Factorising out the common factor [1-a] gives [1-a][1-b].
Steven G Elite Member Joined Dec 30, 2014 Messages 14,591 Oct 31, 2021 #4 You have the answer! Just one more line. [ 1- p(A)][1 - p(B)] = p(A)c*p(B)c !!!!!!
pka Elite Member Joined Jan 29, 2005 Messages 11,988 Oct 31, 2021 #5 Jomo said: You have the answer! Just one more line. [ 1- p(A)][1 - p(B)] = p(A)c*p(B)c !!!!!! Click to expand... P(Ac∩Bc)=P[(A∪B)c]=1−P(A∪B)=1−P(A)−P(B)+P(A∩B)=1−P(A)−P(B)+P(A)P(B)=[1−P(A)][1−P(B)]=P(Ac)P(Bc) \mathcal{P}({A^c} \cap {B^c}) \\ = \mathcal{P}[{(A \cup B)^c}] \\ = 1 - \mathcal{P}(A \cup B) \\ = 1 - \mathcal{P}(A) - \mathcal{P}(B) + \mathcal{P}(A \cap B) \\ = 1 - \mathcal{P}(A) - \mathcal{P}(B) + \mathcal{P}(A)\mathcal{P}(B) \\ = \left[ {1 - \mathcal{P}(A)} \right]\left[ {1 - \mathcal{P}(B)} \right] \\ = \mathcal{P}({A^c})\mathcal{P}({B^c}) \\ P(Ac∩Bc)=P[(A∪B)c]=1−P(A∪B)=1−P(A)−P(B)+P(A∩B)=1−P(A)−P(B)+P(A)P(B)=[1−P(A)][1−P(B)]=P(Ac)P(Bc)
Jomo said: You have the answer! Just one more line. [ 1- p(A)][1 - p(B)] = p(A)c*p(B)c !!!!!! Click to expand... P(Ac∩Bc)=P[(A∪B)c]=1−P(A∪B)=1−P(A)−P(B)+P(A∩B)=1−P(A)−P(B)+P(A)P(B)=[1−P(A)][1−P(B)]=P(Ac)P(Bc) \mathcal{P}({A^c} \cap {B^c}) \\ = \mathcal{P}[{(A \cup B)^c}] \\ = 1 - \mathcal{P}(A \cup B) \\ = 1 - \mathcal{P}(A) - \mathcal{P}(B) + \mathcal{P}(A \cap B) \\ = 1 - \mathcal{P}(A) - \mathcal{P}(B) + \mathcal{P}(A)\mathcal{P}(B) \\ = \left[ {1 - \mathcal{P}(A)} \right]\left[ {1 - \mathcal{P}(B)} \right] \\ = \mathcal{P}({A^c})\mathcal{P}({B^c}) \\ P(Ac∩Bc)=P[(A∪B)c]=1−P(A∪B)=1−P(A)−P(B)+P(A∩B)=1−P(A)−P(B)+P(A)P(B)=[1−P(A)][1−P(B)]=P(Ac)P(Bc)