Question

Three integers are chosen at random from the first $20$ integers. The probability that their product is even is

• A. $\frac{2}{19}$
• B. $\frac{3}{29}$
• C. $\frac{17}{19}$
• D. $\frac{4}{19}$

Answer 1

The correct option is C

$\frac{17}{19}$

Explanation for the correct answer.

Given: $20$ integers, in which $10$ are even and $10$ are odd

The product of three integers is even if at least one of the integers is even.

Total cases of choosing three integers are ${}^{20}{\mathrm{C}}_3$

Probability $=\frac{Totalnumberoffavourableevents}{Totalpossibleevents}$

Since 10 integers are odd. Therefore we can select three odd numbers as ${}^{10}{C}_3$

If we find the probability of getting all odds, then after subtracting it from $1$we can get the required probability.

Required probability $=1-$Probability of integers such that all are odd

$$\begin{gathered} =1-\frac{{}^{10}{C}_3}{{}^{20}{C}_3} \\ =1-\frac{{\displaystyle \frac{10!}{3!\left(10-3\right)!}}}{{\displaystyle \frac{20!}{3!\left(20-3\right)!}}}\left({}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right) \\ =1-\frac{{\displaystyle \frac{10\times 9\times 8\times 7!}{6\times 7!}}}{{\displaystyle \frac{20\times 19\times 18\times 17!}{6\times 17!}}} \\ =1-\frac{120}{1140} \\ =1-\frac{2}{19} \\ =\frac{19-2}{19} \\ =\frac{17}{19} \end{gathered}$$

Hence option C is the correct answer.