35 votes 35 votes Let $A$ and $B$ be any two arbitrary events, then, which one of the following is TRUE? $P (A \cap B) = P(A)P(B)$ $P (A \cup B) = P(A)+P(B)$ $P (A \mid B) = P(A \cap B)P(B)$ $P (A \cup B) \leq P(A) + P(B)$ Probability gate1994 probability conditional-probability normal isro2017 + – Kathleen asked Oct 4, 2014 recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 13.8k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply cseBiswajit commented May 7, 2017 i moved by Satbir Oct 28, 2019 reply Follow Share (d)P(A∪B)≤P(A)+P(B) 2 votes 2 votes adactive18 commented Jan 27, 2018 reply Follow Share Why should it be? The operator is multiplication not division. 0 votes 0 votes Lakshman Bhaiya commented Sep 29, 2020 reply Follow Share For option $(D)$ $P(A \cup B) \leq P(A) + P(B)$ $\implies P(A) + P(B) – P(A \cap B) \leq P(A) + P(B)$ $\implies -P(A \cap B)\leq 0$ $\implies P(A \cap B) \geq 0\;;\text{Always true.}$ 1 votes 1 votes Please log in or register to add a comment.
Best answer 53 votes 53 votes Is true only if events are independent. Is true only if events are mutually exclusive i.e. $P (A \cap B) = 0$ Is false everywhere. Is always true as $P (A \cup B) = P(A)+P(B)-P (A \cap B)$ Since, $P (A \cap B) >=0$, $P (A \cup B) \leq P(A)+P(B)$ Correct Answer: D. Happy Mittal answered Oct 6, 2014 edited Apr 23, 2021 by Lakshman Bhaiya Happy Mittal comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Peeyush Pandey commented Jan 8, 2019 reply Follow Share There must be a division sign after P(A∩B) that is why c is incorrect correct formula is P(A|B)=P(A∩B)/P(B) 4 votes 4 votes Tmajestical commented Jan 4, 2023 reply Follow Share Just to add, option D is a famous inequality called as the Union bound aka the Boole’s Inequality. 0 votes 0 votes tanktopblackhole commented Jun 18, 2023 reply Follow Share Just want to add, it seems option C isn’t false $\textbf{everywhere}$ if $B = \Omega$(sample space) then the equality holds! 0 votes 0 votes Please log in or register to add a comment.
9 votes 9 votes If A and B are mutually exclusive events then: A$\cap$B = $\Phi$ and A$\cup$B = A+B If A and B are non mutually exclusive events then: A$\cap$B != $\Phi$ and A$\cup$B = A+B-A$\cap$B (principle of inclusion-exclusion) therefore A$\cup$B <= A + B so D is the answer... Nirmal Gaur answered May 7, 2017 edited Jan 18, 2018 by Nirmal Gaur Nirmal Gaur comment Share Follow See all 0 reply Please log in or register to add a comment.
7 votes 7 votes Option D is Ans If A and B be two arbitrary events, then P (A ∪ B) = P (A) +P (B) - P(A $\cap$ B) Let say 10=7+8-5 So now 10 <= 7+8 So option D ....P (A ∪ B) <= P (A) +P (B) is correct Ans. Rajesh Pradhan answered May 7, 2017 Rajesh Pradhan comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes What are arbitrary events?If they are independent then Option A and if disjoint then Option C. Purvi Agrawal answered May 7, 2017 Purvi Agrawal comment Share Follow See all 2 Comments See all 2 2 Comments reply shraddha_gami commented May 7, 2017 reply Follow Share Option A is correct –1 votes –1 votes Dheeraj Pant commented May 7, 2017 reply Follow Share Option D should be correct. 0 votes 0 votes Please log in or register to add a comment.