If $A$ and $B$ are any two events such that $P (A) + P (B) - P (A a n d B) = P (A)$ , then

P (B | A) = 1

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P (A | B) = 1

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P (B | A) = 0

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P (A | B) = 0

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Solution

The correct option is B $P (A | B) = 1$
$P (A) + P (B) - P (A a n d B) = P (A)$
$\Rightarrow P (A) + P (B) - P (A \cap B) = P (A)$
$\Rightarrow P (B) - P (A \cap B) = 0$
$\Rightarrow P (A \cap B) = P (B)$
$∴ P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{P (B)}{P (B)} = 1$
Thus, the correct answer is B.

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Similar questions

Q. 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 (C) P (B|A) = 0 (D) P (A|B) = 0

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) $P (\frac{A}{B}) = 0$

Q. Let

A

and

B

be two events such that

P (A \cup B) = P (A) + P (B) - P (A) P (B)

. If

0 < P (A) < 1

and

0 < P (B) < 1

, then

P (A \cup B)^{'}

is equal to

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

Q. If

A

and

B

are two events such that

A \subset B

and

P (B) \neq 0

, then which of the following is correct?

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

The correct option is B P(A|B)=1P(A)+P(B)−P(AandB)=P(A)⇒P(A)+P(B)−P(A∩B)=P(A)⇒P(B)−P(A∩B)=0⇒P(A∩B)=P(B)∴P(A|B)=P(A∩B)P(B)=P(B)P(B)=1Thus, the correct answer is B.

If $A$ and $B$ are any two events such that $P (A) + P (B) - P (A a n d B) = P (A)$ , then