Let 0

Question

(D) Given that $P(A \cup B) = P(A) + P(B) - P(A)P(B)$.We know that the general addition theorem is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Comparing

Vedclass · Accepted Answer

Both $(a)$ and $(c)$

Vedclass · Answer

(D) Given that $P(A \cup B) = P(A) + P(B) - P(A)P(B)$.We know that the general addition theorem is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Comparing these two,we get $P(A \cap B) = P(A)P(B)$,which implies that events $A$ and $B$ are independent.If $A$ and $B$ are independent,then $P(A/B) = P(A)$,so statement $(a)$ is correct.Also,if $A$ and $B$ are independent,then their complements $A^c$ and $B^c$ are also independent.By De Morgan's Law,$(A \cup B)^c = A^c \cap B^c$.Therefore,$P((A \cup B)^c) = P(A^c \cap B^c) = P(A^c)P(B^c)$ because $A^c$ and $B^c$ are independent. Thus,statement $(c)$ is also correct.Since both $(a)$ and $(c)$ are correct,the final answer is $(d)$.

Let $0 < P(A) < 1$,$0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ Then

Similar Questions

In a single throw of $2$ dice, what is the probability of a doublet?

There are $n$ letters and $n$ addressed envelopes. The probability that all the letters are not kept in the right envelope,is

If $A$ and $B$ are two independent events,then $P(A/B) = $

$A$ bag contains $4$ white,$5$ red,and $6$ green balls. Three balls are picked up randomly. The probability that a white,a red,and a green ball is drawn is

Five horses are in a race. $Mr. A$ selects two of the horses at random and bets on them. The probability that $Mr. A$ selected the winning horse is (in $/5$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module