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

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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Two integers $\mathrm{x}$ and $\mathrm{y}$ are chosen with replacement from the set $\{0,1,2,3, \ldots, 10\}$. Then the probability that $|x-y|>5$, is:

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Total Probability Theorem and Bayes' Theorem

Theorem of Total Probability

Suppose A1 , A2 ,..., An  are n mutually exclusive and exhaustive set of events and suppose that each of the events A1 , A2 ,..., An 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(An ) P(A|An )

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

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

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

Now, for any event A

                               A = A ∩ S

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

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

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 ∩ An )]

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

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 (An )P(A|An )

or             

\mathrm{P}(\mathrm{A})=\sum_{i=1}^{n} \mathrm{P}\left(\mathrm{A}_{i}\right) \mathrm{P}\left(\mathrm{A} | \mathrm{A}_{i}\right)

 

Bayes’ Theorem

Suppose A1 , A2 ,..., An  are n mutually exclusive and exhaustive set of events. Then the conditional probability that Ai, happens (given that event A has happened) is given by

 

\\\mathrm{P}\left(\mathrm{A}_{i} | \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} | \mathrm{A}_{i}\right)}{\sum_{j=1}^{n} \mathrm{P}\left(\mathrm{A}_{j}\right) \mathrm{P}\left(\mathrm{A} | \mathrm{A}_{j}\right)} \quad \\\\\text {for any } i=1,2,3, \ldots, n

 

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Total Probability Theorem and Bayes' Theorem

Mathematics Part II Textbook for Class XII

Page No. : 549

Line : 12

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