2.5.2 problem 58

(a)

If $A$ and $B$ are independent, then $$P\left(A\right|B,C)=P(A|B^c,C)=P\left(A\right|C).$$

$$P\left(A\right|B,C^c)=P(A|B^c,C^c)=P\left(A\right|C^c).$$

Thus, Simpson’s Paradox does not hold.

(b)

If $A$ and $C$ are independent, then $P\left(A\right|B,C)<P(A|B^c,C)\Rightarrow P\left(A\right|B)<P(A\left|B^c\right).$ Thus, Simpson’s Paradox does not hold.

(c)

If $B$ and $C$ are independent, then $$P\left(A\right|B)=P(C\left)P\right(A|B,C)+P\left(C^c\right)P\left(A\right|B,C^c).$$

$$P\left(A\right|B^c)=P(C\left)P\right(A|B^c,C)+P\left(C^c\right)P\left(A\right|B^c,C^c).$$

Since $P\left(A\right|B,C)>P(A|B^c,C)$ and $P\left(A\right|B,C^c)>P(A|B^c,C^c)$, $P\left(A\right|B)>P(A\left|B^c\right)$, so Simpson’s Paradox does not hold.