JEE Main 2023 — Mathematics Probability & Statistics

Given,

The probability that the random variable $X$X takes values $x$x is given by $P(X=x)=k(x+1){3}^{-x}$P(X=x)=k(x+1)3−x, $x=0,1,2,3,\ldots \ldots$x=0,1,2,3,……, where $k$k is a constant,

Now we know that,

$$P(X=0)+P(x=1)+P(x=2)+\ldots \ldots \ldots =1$$P(X=0)+P(x=1)+P(x=2)+………=1

$$\Rightarrow \frac{k}{{3}^{0}}+\frac{2k}{{3}^{1}}+\frac{3k}{{3}^{2}}+\ldots ..=1$$⇒30k​+312k​+323k​+…..=1

$$\Rightarrow k(1+\frac{2}{3}+\frac{3}{{3}^{2}}+\ldots .)=1$$⇒k(1+32​+323​+….)=1

Now finding,

$$S=1+\frac{2}{3}+\frac{3}{{3}^{2}}+\frac{4}{{3}^{3}}+\ldots ....(1)$$S=1+32​+323​+334​+…....(1)

$$\Rightarrow \frac{S}{3}=\frac{1}{3}+\frac{2}{{3}^{2}}+\frac{3}{{3}^{2}}+\ldots \ldots .....(2)$$⇒3S​=31​+322​+323​+…….....(2)

Now on subtracting above two equation we get,

$$\Rightarrow \frac{2S}{3}=1+\frac{1}{3}+\frac{1}{{3}^{2}}+\ldots \ldots$$⇒32S​=1+31​+321​+……

$$\Rightarrow S=\frac{9}{4}$$⇒S=49​

Hence, $k(1+\frac{2}{3}+\frac{3}{{3}^{2}}+\ldots .)=1$k(1+32​+323​+….)=1

$$\Rightarrow k=\frac{4}{9}$$⇒k=94​

Now finding,

$$P(X\geq 2)=P(2)+P(3)+\ldots \ldots$$P(X≥2)=P(2)+P(3)+……

$$\Rightarrow P(X\geq 2)=1-P(0)-P(1)$$⇒P(X≥2)=1−P(0)−P(1)

$$\Rightarrow P(X\geq 2)=1-(\frac{k}{1}+\frac{2k}{3})=1-\frac{20}{27}=\frac{7}{27}$$⇒P(X≥2)=1−(1k​+32k​)=1−2720​=277​