If $A, B, C$ are three events associated with a random experiment, prove that
$P ( A \cup B \cup C ) $ $= P ( A )+ P ( B )+ P ( C )- $ $P ( A \cap B )- P ( A \cap C ) $ $- P ( B \cap C )+ $ $P ( A \cap B \cap C )$
Consider $E = B \cup C$ so that
$P ( A \cup B \cup C ) = P ( A \cup E )$
$= P ( A )+ P ( E )- P ( A \cap E )$ ...... $(1)$
Now
$P ( E )= P ( B \cup C )$
$= P ( B )+ P ( C )- P ( B \cap C )$ ......... $(2)$
Also $A \cap E=A \cap(B \cup C)$ $=(A \cap B) \cup(A \cap C)$ [using distribution property of intersection of sets over the union]. Thus
$P(A \cap E)=P(A \cap B)+P(A \cap C)$ $-P[(A \cap B) \cap(A \cap C)]$
$= P ( A \cap B )+ P ( A \cap C )- P [ A \cap B \cap C ] $ ......... $(3)$
Using $(2)$ and $( 3 )$ in $(1)$, we get
$P [ A \cup B \cup C ]= P ( A )+ P ( B )$ $+ P ( C )- P ( B \cap C )$ $- P ( A \cap B )- P ( A \cap C )$ $+ P ( A \cap B \cap C )$
If $A$ and $B$ are two independent events, then $A$ and $\bar B$ are
Given two mutually exclusive events $A$ and $B$ such that $P(A) = 0.45$ and $P(B) = 0.35,$ then $P (A$ or $B ) =$
Three persons $P, Q$ and $R$ independently try to hit a target . If the probabilities of their hitting the target are $\frac{3}{4},\frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by $P$ or $Q$ but not by $R$ is
The probability that $A$ speaks truth is $\frac{4}{5}$, while this probability for $B$ is $\frac{3}{4}$. The probability that they contradict each other when asked to speak on a fact
For the three events $A, B$ and $C, P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs)= $P$ (exactly one of the events $C$ or $A$ occurs)= $p$ and $P$ (all the three events occur simultaneously) $ = {p^2},$ where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is