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

$A$ problem in Statistics is given to four students $A, B, C$ and $D$. Their chances of solving it are $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}$ and $\frac{1}{6}$ respectively. What is the probability that the problem will be solved?

$10$ coins are tossed. What is the probability that exactly $5$ heads appear?

Three identical dice are rolled. The probability that the same number will appear on each of them is:

What is the probability that a leap year selected at random will contain $53$ Sundays?

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is

Vedclass Test Series

Exam Paper Generator

Online Exam Module