Q.3.8 - A First Course in Probability

Question

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

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

(b) If $P\left(A\right|C) > P(A|C^c )$ 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 $6$.

Step-by-Step Solution

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 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 $P\left(A\right)>P\left(B\right)$.

Step 4: 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 $A$ $B$, 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 $C$ - 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 $C$ for which that relationship is not true.