Two events $A$ and $B$ will be independent, if

A

and

B

are mutually exclusive

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P (A' B') = [1 - P (A)] [1 - P (B)]

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P (A) = P (B)

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P (A) + P (B) = 1

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Solution

The correct option is A $P (A' B') = [1 - P (A)] [1 - P (B)]$
Two events $A$ and $B$ are said to be independent, if $P (A B) = P (A) \times P (B)$
Consider the result given in alternative $B$ .
$P (A' B') = [1 - P (A)] [1 - P (B)]$
$\Rightarrow P (A' \cap B') = 1 - P (A) - P (B) + P (A) \cdot P (B)$
$\Rightarrow 1 - P (A \cup B) = 1 - P (A) - P (B) + P (A) \cdot P (B)$
$\Rightarrow P (A \cup B) = P (A) + P (B) - P (A) \cdot P (B)$
$\Rightarrow P (A) + P (B) - P (A B) = P (A) + P (B) - P (A) \cdot P (B)$
$\Rightarrow P (A B) = P (A) \cdot P (B)$
This implies that $A$ and $B$ are independent, if $P (A' B') = [1 - P (A)] [1 - P (B)]$

Suggest Corrections

Similar questions

Two events A and B will be independent, if

(A) A and B are mutually exclusive

(B)

(D) P(A) + P(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)}

Choose the correct answer,
Two events A and B are said to be independent, if
(a) A and B are mutually exclusive
(b) $P (A^{'} \cap B^{'}) = [I - P (A)] [1 - P (B)]$
(c) $P (A) = P (B)$
(d) $P (A) + P (B) = 1$

If A and B are mutually exclusive events then
(a) PA≤PB
(b) PA≥PB
(c) PA<PB
(d) None of these

Q. If

A

and

B

are mutually exclusive events, then

P (A) > 0

and

P (B) \neq 1,

P (¯ ¯¯¯¯¯¯¯¯¯ ¯ A / B)

is equal to

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