JEE Main 2023 — Probability Question with Solution

Given,

Sum of two integer is $66$, so one number will be $x$ and other will be $66-x$,

And given $M$ is maximum value of their product,

So let $y=x\left(66-x\right)$

$\Rightarrow y=66x-x^2$

Now differentiating to find maxima and minima we get, 

$\frac{dy}{dx}=66-2x$

Now equating with zero to find point of maxima as $y=66x-x^2$ represents a downward parabola so it will give maxima,

So $\frac{dy}{dx}=0\Rightarrow 66-2x=0\Rightarrow x=33$,

Hence, the value of $M=33\times 33=1089$

Now solving $x\left(66-x\right)\ge \frac{5M}{9}$

$\Rightarrow x\left(66-x\right)\ge \frac{5\times 1089}{9}$

$\Rightarrow x\left(66-x\right)\ge 605$

$\Rightarrow x^2-66x+605\le 0$

$\Rightarrow \left(x-11\right)\left(x-55\right)\le 0$

So $x\in \left[11,55\right]\to$total $45$ numbers,

Now for probability of $A$, favourable outcomes will be $x=3k\Rightarrow x=\left\{12,15,18,.....54\right\}\to$ total $15$ numbers,

So probability will be $\frac{15}{45}=\frac{1}{3}$