In each of the questions choose the correct answer.
If A and B are events such that $P (\frac{A}{B}) = P \frac{B}{A},$ then
(a) $A \subset B b u t A \neq B$

(b) A = B

(c) $A \cap B = ϕ$

(d) P(A) = P(B)

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Solution

Here $P (\frac{A}{B}) = P (\frac{B}{A}) ∴ \frac{P (A \cap B)}{P (B)} = \frac{P (B \cap A)}{P (A)} \Rightarrow P (A) = P (B) [∵ (A \cap B) = P (B \cap A)]$

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Similar questions

If A and B are events such that P (A|B) = P(B|A), then

(A) A ⊂ B but A ≠ B (B) A = B

If A and B are any two events such that P (A) + P (B) − P (A and B) = P (A), then

(A) P (B|A) = 1 (B) P (A|B) = 1

IF A and B are any two events such that P(A) + P(B) - P (A and B ) = P(A), then
(a) $P (\frac{B}{A}) = 1$
(b) $P (\frac{A}{B}) = 1$
(c) $P (\frac{B}{A}) = 0$
(d) $P (\frac{A}{B}) = 0$

If A and B are two events such that P (A) $\neq 0$ and

$P (\frac{B}{A}) = t,$ then

(a) $A ⊏ B$ (b) $B ⊏ A$
(c) $B = ϕ$ (d) $A = ϕ$

Q. If two events A and B are such that P(A) > 0 and

P (B) \neq 1

, then

P (\frac{¯ A}{¯ B})

is equal to

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In each of the questions choose the correct answer. If A and B are events such that P(AB)=PBA, then (a) A⊂B but A≠B (b) A = B (c)A∩B=ϕ (d) P(A) = P(B)

Here P(AB)=P(BA)∴P(A∩B)P(B)=P(B∩A)P(A)⇒P(A)=P(B) [∵(A∩B)=P(B∩A)]

In each of the questions choose the correct answer.
If A and B are events such that $P (\frac{A}{B}) = P \frac{B}{A},$ then
(a) $A \subset B b u t A \neq B$

(b) A = B

(c) $A \cap B = ϕ$

(d) P(A) = P(B)

Here $P (\frac{A}{B}) = P (\frac{B}{A}) ∴ \frac{P (A \cap B)}{P (B)} = \frac{P (B \cap A)}{P (A)} \Rightarrow P (A) = P (B) [∵ (A \cap B) = P (B \cap A)]$