Probability question (rewriting an expression?)

[nicholaskong100]

 

[blamocur]

The last transformation uses the distributive property of multiplication and addition/subtraction.

 

[Harry_the_cat]

-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]

You have the answer! Just one more line.

[ 1- p(A)][1 - p(B)] = p(A)^c*p(B)^c !!!!!!

 

[pka]

You have the answer! Just one more line.

[ 1- p(A)][1 - p(B)] = p(A)^c*p(B)^c !!!!!!

[imath] \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}) \\ [/imath]