MCQ
Three integers are chosen at random from the first $20$ integers. The probability that their product is even is:
  • A
    $\frac{2}{19}$
  • B
    $\frac{3}{29}$
  • $\frac{17}{19}$
  • D
    $\frac{4}{19}$

Answer

Correct option: C.
$\frac{17}{19}$
Number of ways in which we can choose three distinct integers from $20$ integers $\ ^{20}\text{C}_3=1140$
We know that, if we take three odd numbers, there product will always be an odd number.
Out of $20$ consecutive integers, $10$ are even and $10$ are odd integers.
Number of ways in which we can choose three distinct odd integers from $10$ odd integers $=\ ^{10}\text{C} _3=120$
$P($product is even$) = 1 - P($product is odd$),$
$=1-\frac{120}{1140}$
$=\frac{1140-120}{1140}$
$=\frac{1020}{1140}$
$=\frac{17}{19}$

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