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 the distributive property of intersection over 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
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)$.
$P(A \cup B \cup C) = P(A \cup E)$
$= P(A) + P(E) - P(A \cap E)$ ...... $(1)$
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 the distributive property of intersection over 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
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)$.
Explore More
Similar Questions
Let $A$ and $B$ be two events such that $P(A) = 0.3$ and $P(A \cup B) = 0.8$. If $A$ and $B$ are independent events,then $P(B) = $
Two candidates $A$ and $B$ have attended an interview conducted by a recruitment board for two jobs. If the probability that candidate $A$ will get the job is $0.8$ and the probability that candidate $B$ will get the job is $0.7$,then the probability that at least one of them will get the job is
There are $100$ lottery tickets numbered from $1$ to $100$. If a ticket is drawn at random,find the probability that the number on it is a multiple of $3$ or $5$.
Medium
View SolutionIf $A$ and $B$ are two events such that $P(A) = 0.4$,$P(A \cup B) = 0.7$ and $P(A \cap B) = 0.2$,then $P(B) = $
Easy
View SolutionThe probability of occurrence of an event $A$ is $1/2$ and that of $B$ is $3/10$. If $A$ and $B$ are mutually exclusive, then the probability of occurrence of neither $A$ nor $B$ is (in $/5$)
MediumKCET 2026
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo