For two events A and B with PB 1 prove that PA-B-PA-PA- B1-PB-

Now, P(A|B')=P(A∩ B')P(B') [ According to the definition of conditional probability]or, #160; P(A|B')=P(A) P(A∩ B)1 P(B) [ Since P(B')=1 P(B) ...

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Standard XII

Mathematics

Conditional Probability

For two event...

Question

For two events $A$ and $B$ with $P (B) \neq 1$ prove that $P (A | B^{'}) = \frac{P (A) - P (A \cap B)}{1 - P (B)}$ .

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Solution

Now,
$P (A | B^{'}) = \frac{P (A \cap B^{'})}{P (B^{'})}$ [ According to the definition of conditional probability]
or, $P (A | B^{'}) = \frac{P (A) - P (A \cap B)}{1 - P (B)}$ [ Since $P (B^{'}) = 1 - P (B)$ and $P (A \cap B^{'}) = P (A) - P (A \cap B)$ ] [Given $P (B) \neq 1$ ]

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Similar questions

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

Q. Assertion :If

P (A / B) = P (B / A)

. A, B are two non mutually exclusive events then

P (A) = P (B)

. Reason: For non mutually exclusive events

(A \cap B) \neq ϕ

and

P (A / B) = \frac{P (A \cap B)}{P (B)}

P (B / A) = \frac{P (A \cap B)}{P (A)}

Q. If A and B are two events, then P (

A

∩ B) =
(a) P

(A)

(B)

(b) 1 − P (A) − P (B)
(c) P (A) + P (B) − P (A ∩ B)
(d) P (B) − P (A ∩ B)

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

. Then,
Statement 1:

P (A \cap ¯ ¯¯ ¯ B) = P (¯ ¯¯ ¯ A \cap B) = 0

Statement 2: If

P (A) + P (B) = 1

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For two events A and B with P(B)≠1 prove that P(A|B′)=P(A)−P(A∩B)1−P(B).

Now,P(A|B′)=P(A∩B′)P(B′) [ According to the definition of conditional probability]or, P(A|B′)=P(A)−P(A∩B)1−P(B) [ Since P(B′)=1−P(B) and P(A∩B′)=P(A)−P(A∩B)] [Given P(B)≠1]

For two events $A$ and $B$ with $P (B) \neq 1$ prove that $P (A | B^{'}) = \frac{P (A) - P (A \cap B)}{1 - P (B)}$ .

Now,
$P (A | B^{'}) = \frac{P (A \cap B^{'})}{P (B^{'})}$ [ According to the definition of conditional probability]
or, $P (A | B^{'}) = \frac{P (A) - P (A \cap B)}{1 - P (B)}$ [ Since $P (B^{'}) = 1 - P (B)$ and $P (A \cap B^{'}) = P (A) - P (A \cap B)$ ] [Given $P (B) \neq 1$ ]