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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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An integer is chosen at random from first two hundred numbers. Then the probability that the integer chosen is divisible by 6 or 8 is

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


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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.
$A\cap B=\{24,48,......192\}$
$\Rightarrow\:n(A)=33$, $n(B)=25$ 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}$
Ans : (A)
answered Jan 15, 2014 by thanvigandhi_1
edited Mar 26, 2014 by rvidyagovindarajan_1

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