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Home  >>  CBSE XI  >>  Math  >>  Probability
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Suppose an integer from 1 through 1000 is chosen at random,find the probability that the integer is a multiple of 2 or a multiple of 9.

$\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 =$\large\frac{1000}{2}$
Total integers multiple of 9 =$\large\frac{1000}{9}$
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)=$P(A)+P(B)-P(A \cap B)$
$\Rightarrow 500+111-55$
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.
answered Jul 7, 2014 by sreemathi.v

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