I’m not sure if this changes the math but A-->G and B-->G aren’t given. P(A) conjoined to P(A-->G) should = 1⁄3. Same for P(B) and P (B-->G). No?
I’m not sure if this changes the math but A-->G and B-->G aren’t given. P(A) conjoined to P(A-->G) should = 1⁄3. Same for P(B) and P (B-->G). No?