If A and B are arbitrary events, then

English

Gujarati

Hindi

14.Probability

normal

If $A$ and $B$ are arbitrary events, then

A

$P(A \cap B) \ge P(A) + P(B)$

B

$P(A \cup B) \le P(A) + P(B)$

C

$P(A \cap B) = P(A) + P(B)$

D

None of these

Solution

(b) $P(A \cup B) = P(A) + P(B) – P(A \cap B) \le P(A) + P(B)$,$(\because \,\,P(A \cap \,B) \geqslant \,0)$.

Standard 11

Mathematics

Share

Similar Questions

Fill in the blanks in following table :

$P(A)$ $P(B)$ $P(A \cap B)$ $P (A \cup B)$

$0.5$ $0.35$ ……… $0.7$

easy

The chances to fail in Physics are $20\%$ and the chances to fail in Mathematics are $10\%$. What are the chances to fail in at least one subject ………… $\%$

medium

A coin is tossed twice. If events $A$ and $B$ are defined as :$A =$ head on first toss, $B = $ head on second toss. Then the probability of $A \cup B = $

easy

Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $P(B)=p .$ Find $p$ if they are independent.

easy

If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are mutually exclusive, then $x = $

easy