JEE Main 2023 — Mathematics Probability & Statistics

Given,

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

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

So let $y=x(66-x)$y=x(66−x)

$$\Rightarrow y=66x-{x}^{2}$$⇒y=66x−x2

Now differentiating to find maxima and minima we get,

$$\frac{dy}{dx}=66-2x$$dxdy​=66−2x

Now equating with zero to find point of maxima as $y=66x-{x}^{2}$y=66x−x2 represents a downward parabola so it will give maxima,

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

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

Now solving $x(66-x)\geq \frac{5M}{9}$x(66−x)≥95M​

$$\Rightarrow x(66-x)\geq \frac{5\times 1089}{9}$$⇒x(66−x)≥95×1089​

$$\Rightarrow x(66-x)\geq 605$$⇒x(66−x)≥605

$$\Rightarrow {x}^{2}-66x+605\leq 0$$⇒x2−66x+605≤0

$$\Rightarrow (x-11)(x-55)\leq 0$$⇒(x−11)(x−55)≤0

So $x\in [11,55]\rightarrow$x∈[11,55]→total $45$45 numbers,

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

So probability will be $\frac{15}{45}=\frac{1}{3}$4515​=31​