Total Probability Theorem and Bayes' Theorem - Practice Questions & MCQ

Theorem of Total Probability

Suppose $A_1, A_2, \ldots, A_n$ are $n$ mutually exclusive and exhaustive set of events and suppose that each of the events $A_1, A_2, \ldots, A_n$ has nonzero probability of occurrence. Let A be any event associated with S , then

$
P(A)=P\left(A_1\right) P\left(A \mid A_1\right)+P\left(A_2\right) P\left(A \mid A_2\right)+\ldots+P\left(A_n\right) P\left(A \mid A_n\right)
$

As from the image, $A_1, A_2, \ldots, A_n$ are $n$ mutually exclusive and exhaustive set of events
Therefore, $S=A_1 \cup A_2 \cup \ldots \cup A_n$
And $A i \cap A j=\varphi, i \neq j, i, j=1,2, \ldots, n$
Now, for any event A

$
\begin{aligned}
& A=A \cap S \\
& =A \cap\left(A_1 \cup A_2 \cup \ldots \cup A_n\right) \\
& =\left(A \cap A_1\right) \cup\left(A \cap A_2\right) \cup \ldots \cup\left(A \cap A_n\right)
\end{aligned}
$
Also $A \cap A i$ and $A \cap A j$ are respectively the subsets of $A i$ and $A j$
Since, $A i$ and $A j$ are disjoint, for $i \neq j$, therefore, $A \cap A i$ and $A \cap A j$ are also disjoint for all $i \neq j, i, j=1,2, \ldots, n$.
Thus, $P(A)=P\left[\left(A \cap A_1\right) \cup\left(A \cap A_2\right) \cup \ldots . . \cup\left(A \cap A_n\right)\right]$

$
=P\left(A \cap A_1\right)+P\left(A \cap A_2\right)+\ldots+P\left(A \cap A_n\right)
$
Using multiplication rule of probability

$
P(A \cap A i)=P(A i) P(A \mid A i) \text { as } P(A i) \neq 0 \forall i=1,2, \ldots, n
$
Therefore $\mathrm{P}(\mathrm{A})=\mathrm{P}\left(\mathrm{A}_1\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{A}_2\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{A}_2\right)+\ldots+\mathrm{P}\left(\mathrm{A}_n\right) \mathrm{P}\left(\mathrm{A} \mid \mathrm{A}_n\right)$

or             

$
\mathrm{P}(\mathrm{~A})=\sum_{i=1}^n \mathrm{P}\left(\mathrm{~A}_i\right) \mathrm{P}\left(\mathrm{~A} \mid \mathrm{A}_i\right)
$
Bayes' Theorem
Suppose $A_1, A_2, \ldots, A_n$ are $n$ mutually exclusive and exhaustive set of events. Then the conditional probability that $A_i$, happens (given that event $A$ has happened) is given by

$
\begin{aligned}
& \mathrm{P}\left(\mathrm{~A}_i \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{~A}_{\mathrm{i}} \cap \mathrm{~A}\right)}{\mathrm{P}(\mathrm{~A})}=\frac{\mathrm{P}\left(\mathrm{~A}_i\right) \mathrm{P}\left(\mathrm{~A} \mid \mathrm{A}_i\right)}{\sum_{j=1}^n \mathrm{P}\left(\mathrm{~A}_j\right) \mathrm{P}\left(\mathrm{~A} \mid \mathrm{A}_j\right)} \\
& \text { for any } i=1,2,3, \ldots, n
\end{aligned}
$