Question

An integer is chosen at random from the first two hundred digits.What is the probability that the integer chosen is divisible by 6 or 8 ?

• A. $\frac{29}{100}.$
• B. $\frac{1}{4}.$
• C. $\frac{1}{8}.$
• D. $\frac{21}{100}.$

Answer 1

Answer: (B)

$\frac{1}{4}.$

Let A=the integer is divisible by 6
$A=6,12,18,....198$
$198=6+(n-1)6$
$\Rightarrow n=33$
So, $P\left(A\right)=\frac{33}{200}$
B=the integer is divisible by 8
$B=8,16,24,....200$
$200=8+(n-1)8$
$\Rightarrow n=25$
So, $P\left(B\right)=\frac{25}{200}$
Both are divisible by 6 and 8 both
$A\cap B=24,48,.....192$
$192=24+(n-1)24$
$\Rightarrow n=8$
So, $P(A\cap B)=\frac{8}{200}$
$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(AB\right)$
$=\frac{33}{200}+\frac{25}{200}-\frac{8}{200}=\frac{1}{4}.$