$\begin {array} {1 1} (A)\;\large\frac{1}{4} & \quad (B)\;\large\frac{2}{4} \\ (C)\;\large\frac{3}{4} & \quad (D)\;None \: of \: these \end {array}$

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- $ P( A \cup B ) = P(A)+P(B)-P( A \cap B)$

One integer can be chosen out of 200 integers in $^{200}C_1=200$ ways.

Let A be the event that an integer selected is divisible by 6 and

B be the even that the integer is divisible by 8.

$i.e.,A=\{6,12,18,...........198\}$

$B=\{8,16,24,.......200\}$

$A\cap B=\{24,48,......192\}$

$\Rightarrow\:n(A)=33$, $n(B)=25$ and $n(A\cap B)=8$

Then $P(A) = \large\frac{33}{200}$ ,

$ P(B)= \large\frac{25}{200}$ and

$ P( A \cap B ) = \large\frac{8}{200}$

$P($getting a number divisible by $6$ or $8)=P(A\cup B)$

We know that $ P( A \cup B ) = P(A)+P(B)-P( A \cap B)$

$\Rightarrow$ Required probability $ = \large\frac{33}{200}+\large\frac{25}{200}-\large\frac{8}{200}$

$ = \large\frac{1}{4}$

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