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  • If A and B are two events and A ≠ θ, B ≠ θ, then A. P(A | B) = P(A).P(B)

If A and B are two events and A ≠ θ, B ≠ θ, then
A. P(A | B) = P(A).P(B)
B. P(A \mid B)=\frac{P(A \cap B)}{P(B)}

C. P(A | B).P(B | A)=1
D. P(A | B) = P(A) | P(B)

Answers (1)

Given-

A \neq \phi, B \neq \phi$
CASE 1 : If we take option (\mathrm{A})$ i.e. $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A) \times P(B)}{P(B)}=P(A) \neq R H S$
CASE 2 : If we take option (B) i.e.
P(A \mid B)=\frac{P(A \cap B)}{P(B)}$
this is true, knowing that this is conditional probability.
CASE 3: If we take option \mathrm{C}$.i.e. $\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B} \mid \mathrm{A})=1$
\\\mathrm{LHS}=\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B} \mid \mathrm{A})$ \\$=\frac{P(A \cap B)}{P(B)} \times \frac{P(B \cap A)}{P(A)}$ \\$=\frac{[P(A \cap B)]^{2}}{P(B) P(A)}$ \\$\neq \mathrm{RHS}$

Hence, the correct option is B

Posted by

infoexpert22

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