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Question

Two distinct numbers are selected at random from the first twelve natural numbers. The probability that the sum will be divisible by 3 is


A
13
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B
2366
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C
12
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D
none of these
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Solution

The correct option is B $$\displaystyle \frac {1}{3}$$
$$First\quad 12\quad natural\quad numbers\quad are\quad 1,2,3,........,12\\ Number\quad of\quad ways\quad in\quad which\quad two\quad distinct\quad numbers\quad \\ can\quad be\quad selected\quad out\quad of\quad these\quad 12\quad natural\quad numbers= \scriptstyle\binom{ 12 }{ 2 }\\ Let\quad P\left( T \right) = \scriptstyle\binom{12}{2} \\ The\quad maximum\quad sum\quad of\quad two\quad selected\quad distinct\quad numbers=12+11=23\\ The\quad minimum\quad sum\quad of\quad two\quad selected\quad distinct\quad numbers=1+2=3\\ For\quad the\quad sum\quad being\quad divisible\quad by\quad 3\quad it\quad must\quad be\quad one\quad of\quad \\ 3,6,9,12,15,18,21\\ Let\quad S\quad be\quad the\quad event\quad of\quad sum\quad of\quad two\quad selected\quad number.\\ P\left( S=3 \right) =\dfrac { 1 }{ P\left( T \right)  } \\ P\left( S=6 \right) =\dfrac { 2 }{ P\left( T \right)  } \\ P\left( S=9 \right) =\dfrac { 4 }{ P\left( T \right)  } \\ P\left( S=12 \right) =\dfrac { 5 }{ P\left( T \right)  } \\ P\left( S=15 \right) =\dfrac { 5 }{ P\left( T \right)  } \\ P\left( S=18 \right) =\dfrac { 3 }{ P\left( T \right)  } \\ P\left( S=21 \right) =\dfrac { 2 }{ P\left( T \right)  } \\ Let\quad D\quad be\quad the\quad event\quad that\quad sum\quad is\quad divisible\quad by\quad 3\\ Then\quad P\left( D \right) =P\left( S=3 \right) +P\left( S=6 \right) +P\left( S=9 \right) +P\left( S=12 \right) +P\left( S=15 \right) +P\left( S=18 \right) +P\left( S=21 \right) \\ \Rightarrow P\left( D \right) =\dfrac { 22 }{ P\left( T \right)  } =\dfrac { 22 }{ 66 } \\ \Rightarrow P\left( D \right) =\dfrac { 1 }{ 3 } $$

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