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

Theorem of Total Probability

Suppose A₁ , A₂ ,..., A_n  are n mutually exclusive and exhaustive set of events and suppose that each of the events A₁ , A₂ ,..., A_n has nonzero probability of occurrence. Let A be any event associated with S, then

            P(A) = P(A₁ ) P(A|A₁ ) + P(A₂ ) P(A|A₂ ) + ... + P(A_n ) P(A|A_n )

As from the image, A₁ , A₂ ,..., A_n are n mutually exclusive and exhaustive set of events

Therefore,               S = A₁ ∪ A₂ ∪ ... ∪ A_n

And              Ai ∩ Aj = φ, i ≠ j, i, j = 1, 2, ..., n

Now, for any event A

                               A = A ∩ S

                                  = A ∩ (A₁ ∪ A₂ ∪ ... ∪ A_n )

                                  = (A ∩ A₁ ) ∪ (A ∩ A₂ ) ∪ ...∪ (A ∩ A_n) 

Also A ∩ Ai and A ∩ Aj are respectively the subsets of Ai and Aj  

Since, Ai and Aj are disjoint, for i ≠ j , therefore, A ∩ Ai and A ∩ Aj are also disjoint for all i ≠ j, i, j = 1, 2, ..., n.

Thus,                 P(A) = P [(A ∩ A₁ ) ∪ (A ∩ A₂ )∪ .....∪ (A ∩ A_n )]

                                  = P (A ∩ A₁ ) + P (A ∩ A₂ ) + ... + P (A ∩ A_n)

Using multiplication rule of probability

                 P(A ∩ Ai ) = P(Ai ) P(A|Ai ) as P (Ai ) ≠ 0 ∀ i = 1,2,..., n

Therefore,

P (A) = P (A₁) P (A|A₁ ) + P (A₂ ) P (A|A₂ ) + ... + P (A_n )P(A|A_n )

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}
$