Question

Let $A,B$, and $C$be events relating to the experiment of rolling a pair of dice.

(a) If localid="1647938016434" $P\left(A\right|C)>P(B\left|C\right)$and localid="1647938126689" $P\left(A\right|C^c)>P(B|C^c)$either prove that localid="1647938033174" $P\left(A\right)>P\left(B\right)$or give a counterexample by defining events $B$and $C$for which that relationship is not true.

(b) If localid="1647938162035" $P\left(A\right|C)>P(A|C^c)$and $P\left(B\right|C)>P(B|C^c)$either prove that $P\left(AB\right|C)>P(AB\left|C^c\right)$or give a counterexample by defining events $A,B$and $C$for which that relationship is not true. Hint: Let $C$be the event that the sum of a pair of dice is $10$; let $A$be the event that the first die lands on $6$; let $B$be the event that the second die lands on $6$.

Short answer

a)Use the law of total probability to prove that $P\left(A\right)>P\left(B\right)$.

b)The hint gives a counterexample. The events $A$, $B$, and $C$ for which that relationship is not true.

Step-by-step solution

Step 1:Given Information(part a)

Given that
$P(A\mid C)>P(B\mid C)\text{and}P\left(A\mid C^c\right)>P\left(B\mid C^c\right)$

The events $A$, $B$, and $C$ for which that relationship is not true.

Step 2:Explanation(part a)

Think,

$\left.P(A\mid C)>P(B\mid C),P\left(A\mid C^c\right)>P\left(B\mid C^c\right)\right\}$

The rules of total probability are used in $1$st and$3$rd row (dividing by$C$ )

$P\left(A\right)=P(A\mid C)P\left(C\right)+P\left(A\mid C^c\right)P\left(C^c\right)$

$>P(B\mid C)P\left(C\right)+P\left(B\mid C^c\right)P\left(C^c\right)\text{Assumptions}$

$=P\left(B\right)$

Consolidating the start and the end we demonstrate:

$P\left(A\right)>P\left(B\right)$

Step 3:Final Answer(part a)

Use the law of total probability to prove that$P\left(A\right)>P\left(B\right)$.

Given Information(part b)

Given that,

$P(A\mid C)>P\left(A\mid C^c\right)\text{and}P(B\mid C)>P\left(B\mid C^c\right)$

The events localid="1647939230660" $A$$B$, and $C$ for which that relationship is not true.

Step 5:Explanation(part b)

Think,

$\left.P(A\mid C)>P\left(A\mid C^c\right),P(B\mid C)>P\left(B\mid C^c\right)\right\}P(AB\mid C)>P\left(AB\mid C^c\right)$

can not be finalized.

Hint gives one possibility for a counterexample, with there expressed occasions:

$A$- the first of two dice arrived on six, $B$- the other one arrived on six, and localid="1647939355923" $C$- the amount of the numbers on the dice is $10$.

$P(A\mid C)=\frac{1}{3},P(B\mid C)=\frac{1}{3},P(AB\mid C)=0,\text{and}$

$P\left(A\mid C^c\right)=\frac{5}{33},P\left(B\mid C^c\right)=\frac{5}{33},P\left(AB\mid C^c\right)=\frac{1}{33}$

Step 6:Final Answer(part b)

The hint gives a counterexample. The events $A$, B, and localid="1647939910420" $C$for which that relationship is not true.