$\begin{array}{1 1}(A)\;0.556\\(B)\;0.666\\(C)\;0.776\\(D)\;0.576\end{array} $

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- Required probability =$\large\frac{n(E)}{n(S)}$
- $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Step 1:

Given integers 1 to 1000

Total number of sample space n(S)=1000

Step 2:

Total integers multiple of 2 =$\large\frac{1000}{2}$

P(A)=500

Total integers multiple of 9 =$\large\frac{1000}{9}$

P(B)=111

Total integers multiple of both (2 and 9)=$\large\frac{1000}{2\times 9}$

$P(A \cap B)=55$

Step 3:

$\therefore$ Total integers multiple of (2 or 9)=$P(A)+P(B)-P(A \cap B)$

$\Rightarrow 500+111-55$

$n(E)=556$

Step 4:

$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$

$\Rightarrow \large\frac{556}{1000}$

$\Rightarrow 0.556$

Hence (A) is the correct answer.

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