Conditional Probability - Practice Questions & MCQ

Conditional Probability

Conditional probability is a measure of the probability of an event given that another event has already occurred. If A and B are two events associated with the same sample space of a random experiment, the conditional probability of the event A given that B

has already occurred is written as $P(A \mid B), P(A / B)$ or

$
P\left(\frac{A}{B}\right)
$

The formula to calculate $P(A \mid B)$ is
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$ where $P(B)$ is greater than zero.
For example, suppose we toss one fair, six-sided die. The sample space $S=\{1,2,3,4,5,6\}$. Let $A=$ face is 2 or 3 and $B=$ face is even number $(2,4,6)$.
Here, $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$ means that the probability of occurrence of face 2 or 3 when an even number has occurred which means that one of 2, 4 and 6 has occurred.
To calculate $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$, we count the number of outcomes 2 or 3 in the modified sample space $B=\{2,4,6\}$ : meaning the common part in $A$ and $B$. Then we divide that by the number of outcomes in $B$ (rather than $S$ ).

$
\begin{aligned}
\mathrm{P}(\mathrm{~A} \mid \mathrm{B}) & =\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\frac{\mathrm{n}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{n}(\mathrm{~S})}}{\frac{\mathrm{n}(\mathrm{~B})}{\mathrm{n}(\mathrm{~S})}} \\
& =\frac{\frac{\text { (the number of outcomes that are } 2 \text { or } 3 \text { and even in } \mathrm{S})}{6}}{\frac{\text { (the number of outcomes that are even in } \mathrm{S})}{6}} \\
& =\frac{\frac{1}{6}}{\frac{3}{6}}=\frac{1}{3}
\end{aligned}
$
 

Properties of Conditional Probability
Let $A$ and $B$ are events of a sample space $S$ of an experiment, then
Property $1 P(S \mid A)=P(A \mid A)=1$
Proof:

Also,

$
\begin{aligned}
& P(S \mid A)=\frac{P(S \cap A)}{P(A)}=\frac{P(A)}{P(A)}=1 \\
& P(A \mid A)=\frac{P(A \cap A)}{P(A)}=\frac{P(A)}{P(A)}=1
\end{aligned}
$

Thus,

$
\mathrm{P}(\mathrm{~S} \mid \mathrm{A})=\mathrm{P}(\mathrm{~A} \mid \mathrm{A})=1
$
 

Property 2 If $A$ and $B$ are any two events of a sample space $S$ and $C$ is an event of $S$ such that $P(C) \neq 0$, then

$
P((A \cup B) \mid C)=P(A \mid C)+P(B \mid C)-P((A \cap B) \mid C)
$

In particular, if A and B are disjoint events, then

$
P((A \cup B) \mid C)=P(A \mid C)+P(B \mid C)
$

Proof:

$
\begin{aligned}
\mathrm{P}((\mathrm{~A} \cup \mathrm{~B}) \mid \mathrm{C}) & =\frac{\mathrm{P}[(\mathrm{~A} \cup \mathrm{~B}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})} \\
& =\frac{\mathrm{P}[(\mathrm{~A} \cap \mathrm{C}) \cup(\mathrm{B} \cap \mathrm{C})]}{\mathrm{P}(\mathrm{C})}
\end{aligned}
$

(by distributive law of union of sets over intersection)

$
\begin{aligned}
& =\frac{P(A \cap C)+P(B \cap C)-P((A \cap B) \cap C)}{P(C)} \\
& =\frac{P(A \cap C)}{P(C)}+\frac{P(B \cap C)}{P(C)}-\frac{P[(A \cap B) \cap C]}{P(C)} \\
& =P(A \mid C)+P(B \mid C)-P((A \cap B) \mid C)
\end{aligned}
$

When A and B are disjoint events, then

$
\begin{aligned}
& \mathrm{P}((\mathrm{~A} \cap \mathrm{~B}) \mid \mathrm{C}) \\
& \Rightarrow \mathrm{P}((\mathrm{~A} \cup \mathrm{~B}) \mid \mathrm{F}) \\
&=\mathrm{P}(\mathrm{~A} \mid \mathrm{F})+\mathrm{P}(\mathrm{~B} \mid \mathrm{F})
\end{aligned}
$

Property $3 P\left(A^{\prime} \mid B\right)=1-P(A \mid B)$, if $P(B) \neq 0$
Proof:
From Property 1, we know that $\mathrm{P}(\mathrm{S} \mid \mathrm{B})=1$

$
\begin{array}{lll}
\Rightarrow & P\left(\left(A \cup A^{\prime}\right) \mid B\right)=1 & \text { (as } A \cup A^{\prime}=S \text { ) } \\
\Rightarrow & P(A \mid B)+P\left(A^{\prime} \mid B\right)=1 & \text { (as } A \text { and } A^{\prime} \text { are disjoint event) } \\
\Rightarrow & P\left(A^{\prime} \mid B\right)=1-P(A \mid B) &
\end{array}
$