Total Probability Theorem and Bayes' Theorem MCQ

Solve by difficulty

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:

$\frac{30}{121}$

$\frac{62}{121}$

$\frac{60}{121}$

$\frac{31}{121}$

View Solution

Concepts Covered - 1

Total Probability Theorem and Bayes' Theorem

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 )

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

Bayes’ Theorem

Suppose A₁ , A₂ ,..., 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

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