Two events A and B will be independent, if

A and B are mutually exclusive

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P (A^{'} B^{'}) = [1 - P (A)] [1 - P (B)]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

P (A) = P (B)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

P (A) + P (B) = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $P (A^{'} B^{'}) = [1 - P (A)] [1 - P (B)]$
$P (A^{'} B^{'}) = P (A^{'} \cap B^{'})$
$= P A ¯ \cup B$
$= 1 - P (A \cup B)$
$= 1 - P (A) - P (B) - P (A) - P (B)$
$= 1 - P (A) - P (B) (1 - P (A)]$
$= [1 - P (A)] [1 - P (B)]$

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

Join BYJU'S Learning Program

Two events A and B will be independent, if

The correct option is A P(A′B′)=[1−P(A)][1−P(B)]P(A′B′)=P(A′∩B′)=PA¯∪B=1−P(A∪B)=1−P(A)−P(B)−P(A)−P(B)=1−P(A)−P(B)(1−P(A)]=[1−P(A)][1−P(B)]

The correct option is A $P (A^{'} B^{'}) = [1 - P (A)] [1 - P (B)]$
$P (A^{'} B^{'}) = P (A^{'} \cap B^{'})$
$= P A ¯ \cup B$
$= 1 - P (A \cup B)$
$= 1 - P (A) - P (B) - P (A) - P (B)$
$= 1 - P (A) - P (B) (1 - P (A)]$
$= [1 - P (A)] [1 - P (B)]$