Question

If $A,B$ and $C$ are independent events such that $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{3}{4}$ and $P\left(C\right)=\frac{2}{3}$
Consider the following statements:
$\left(A\right)$ : Probability that none of them occur is $p$
$\left(B\right)$ : Probability that at least one of them occur is $q.$
Which of the following is/are $TRUE$ ?

• A. $p=\frac{1}{24}$
• B. $q=\frac{7}{24}$
• C. $p+q=1$
• D. $p+q=\frac{1}{3}$

Answer 1

Answer: (A), (C)

$p+q=1$

Given : $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{3}{4}$ and $P\left(C\right)=\frac{2}{3}$

$\left(A\right)$Probability that none of them occur $=P(\overline{A}\cap \overline{B}\cap \overline{C})=(1-P(A\left)\right)(1-P(B\left)\right)(1-P(C)$
$=P(\overline{A}\cap \overline{B}\cap \overline{C})=\frac{1}{2}\times \frac{1}{4}\times \frac{1}{3}=\frac{1}{24}$
$p=\frac{1}{24}$

$\left(B\right)$ Probability that at least one of them occur is
$P(A\cup B\cup C)=1-(1-P(A\left)\right)(1-P(B\left)\right)(1-P(C\left)\right)=\frac{23}{24}$
$\therefore q=\frac{23}{24}$
$\Rightarrow p+q=1$