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Total Probability Theorem and Bayes' Theorem is considered one of the most asked concept.
43 Questions around this concept.
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:
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
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
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