If A and B are any two events in a sample space S then $P (A \cup B)$ is

\geq P (A) + P (B)

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

P (A) + P (B)

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

\leq P (A) + P (B)

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

P (A \cap B)

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

Open in App

Solution

The correct option is C $\leq P (A) + P (B)$
From De-morgan's laws,
$P (A \cup B) = P (A) + P (B) - P (A \cap B) \leq P (A) + P (B)$
so, we conclude that $P (A \cup B) \leq P (A) + P (B)$

Suggest Corrections

Similar questions

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

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. 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

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 in a sample space S such that

P (A) \neq 0

and

P (B) \neq 0

, then

P (A \cap B)

Join BYJU'S Learning Program

If A and B are any two events in a sample space S then P(A∪B) is

The correct option is C ≤P(A)+P(B)From De-morgan's laws, P(A∪B)=P(A)+P(B)−P(A∩B)≤P(A)+P(B)so, we conclude that P(A∪B)≤P(A)+P(B)

If A and B are any two events in a sample space S then $P (A \cup B)$ is

The correct option is C $\leq P (A) + P (B)$
From De-morgan's laws,
$P (A \cup B) = P (A) + P (B) - P (A \cap B) \leq P (A) + P (B)$
so, we conclude that $P (A \cup B) \leq P (A) + P (B)$