Question

Three integers are chosen at random from the first 20 integers. The probability that their product is even is
(a) 2/19
(b) 3/29
(c) 17/19
(d) 4/19

Answer 1

(c) 17/19

Number of ways in which we can choose three distinct integers from 20 integers = ²⁰C₃ = 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= ¹⁰C₃ = 120
P(product is even) = 1 $-$ P(product is odd)

= $1-\frac{120}{1140}=\frac{1140-120}{1140}=\frac{1020}{1140}=\frac{17}{19}$