Question

Let the lines $x+y=20$ and $x+y=10$ meet the coordinates axes at $A,B$ and $C,D$ respectively. If a point $M(a,b)$ where $a,b\in N,$ is randomly selected from inside the triangle $AOB,$ then the probability that it lies outside the triangle $COD$ is
(Here, $O$ is the origin)

• A. $\frac{3}{4}$
• B. $\frac{27}{38}$
• C. $\frac{15}{19}$
• D. $\frac{14}{19}$

Answer 1

The correct option is D $\frac{14}{19}$


$x+y=20$
Number of points inside $△AOB$ is given by,
$x+y\le 19$
${}^{17+3-1}{C}_{3-1}={}^{19}{C}_2=171$

Similarly, number of points inside the $△COD={}^9{C}_2=36$
So, number of points outside $△COD$ and inside $△AOB$
$=171-36-9\left[9\text{points on line}CD\right]$
$=126$

Therefore, required probability is,
$P=\frac{126}{171}=\frac{14}{19}$