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The $12$ houses on one side of a street are numbered with even numbers starting at $2$ and going up to $24$. A free newspaper is delivered on Monday to $3$ different houses chosen at random from these $12$. Find the probability that at least $2$ of these newspapers are delivered to houses with numbers strictly greater than $14$.

  1. $\frac{7}{11}$
  2. $\frac{5}{12}$
  3. $\frac{4}{11}$
  4. $\frac{5}{22}$
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Answer is (C)

There are  $12$ houses on one side of a street are numbered with even numbers.

In which $5$ houses are strictly greater than Number $14.$

And remaining $7$ houses are numbered smaller than $14$ (i.e. including $14$)

No of way of choosing at least $2$ of these newspapers are delivered to houses with numbers strictly greater than $14.$

$^{5}C_{3}  +^{5}C_{2}\times  ^{7}C_{1} =80$

Total way of choosing $3$ houses$=^{12}C_{3} =220$

So Required probability=$\dfrac{80}{220}=\dfrac{4}{11}.$

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