91影视

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

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

Think,

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

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

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

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

$=P\left(B\right)$

Consolidating the start and the end we demonstrate:

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

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

Given that,

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

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

Think,

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

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 $10$.

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

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

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