Name: Conditional Probability
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: What is the probability of occurrence of an event when another event associated with it has already occurred? Check out this video to learn about it.

Question

State and prove Baye's Theorem.

Accepted Answer

Baye's Theorem states that let $S$ be the sample space and let $E_{1}, E_{2}, . . ., E_{n}$ be $n$ mutually exclusive and exhaustive events associated with a random experiment. If $A$ is any event which occurs with $E_{1}$ or $E_{2}$ or .... or $E_{n}$ , then

$P (E_{i} / A) = \frac{P (E_{i}) P (A / E_{i})}{\sum_{i = 1}^{n} P (E_{i}) P (A / E_{i})}, i = 1, 2 . . . n$

Proof: Since $E_{1}, E_{2}, . . ., E_{n}$ are $n$ mutually exclusive and exhaustive events, we have

$S = E_{1} \cup E_{2} \cup \dots \cup E_{n}, where E_{i} \cap E_{j} = ϕ for i \neq j$

$\Rightarrow A = A \cap S$

$\Rightarrow A = (A \cap E_{1}) \cup (A \cap E_{2}) \cup \dots \cup (A \cap E_{n})$

$\Rightarrow P (A) = P (A \cap E_{1}) + P (A \cap E_{2}) + \dots + P (A \cap E_{n})$ [By addition theorem.]

$\Rightarrow P (A) = \sum_{i = 1}^{n} P (A \cap E_{i})$

$\Rightarrow P (A) = \sum_{i = 1}^{n} P (E_{i}) P (A / E_{i})$ $[∵ P (A \cap E_{i}) = P (E_{i}) P (A / E_{i})] . . . (1)$

Now, using multiplication theorem of probability, we have

$P (A \cap E_{i}) = P (A) P (E_{i} / A) for i = 1, 2, \dots, n$

$\Rightarrow P (E_{i} / A) = \frac{P (A \cap E_{i})}{P (A)}$

$\Rightarrow P (E_{i} / A) = \frac{P (E_{i}) P (A / E_{i})}{P (A)} [∵ P (A \cap E_{i}) = P (E_{i}) P (A / E_{i})]$

$\Rightarrow P (E_{i} / A) = \frac{P (E_{i}) P (A / E_{i})}{\sum_{i = 1}^{n} P (E_{i}) P (A / E_{i})}$ [Using $(1)$ ]

Hence, $P (E_{i} / A) = \frac{P (E_{i}) P (A / E_{i})}{\sum_{i = 1}^{n} P (E_{i}) P (A / E_{i})}, i = 1, 2, \dots, n$ .

State and prove Baye's Theorem.

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