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QJEEM23Ma3472: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3472 Solution: If the probability that the random variable takes values is…

If the probability that the random variable X\mathrm{X} takes values xx is given by P(X=x)=k(x+1)3−x,x=0,1,2,3,…\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 3^{-x}, x=0,1,2,3, \ldots , where k\mathrm{k} is a constant, then P(X≥2)\mathrm{P}(\mathrm{X} \geq 2) is equal to :

Multiple Choice Options

  • A 718\frac{7}{18}
  • B 2027\frac{20}{27}
  • C 727\frac{7}{27}
  • D 1118\frac{11}{18}

Step-by-Step Verified Solution

As, we know that sum of all the probabilities =1=1

amp; So, ∑x=1∞P(X=x)=1amp;⇒k[1+2⋅3−1+3⋅3−2+….∞]=1\begin{aligned}& \text { So, } \sum_{x=1}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\\\& \Rightarrow k\left[1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+\ldots . \infty\right]=1\end{aligned}

amp; Let S=1+23+332+….+∞amp;⇒S3=0+13+232+333+….+∞\begin{aligned}& \text { Let } S=1+\frac{2}{3}+\frac{3}{3^2}+\ldots .+\infty \\\\& \Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots .+\infty\end{aligned}

On subtracting, we get

amp;2S3=1+13+132+….+∞amp;⇒2S3=11−13=123amp;⇒2S3=32amp;⇒S=94\begin{aligned}& \frac{2 S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots .+\infty \\\\& \Rightarrow \frac{2 S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}} \\\\& \Rightarrow \frac{2 S}{3}=\frac{3}{2} \\\\& \Rightarrow S=\frac{9}{4}\end{aligned}

amp; So, k×94=1⇒k=49amp; Now, P(X≥2)=1−P(Xlt;2)amp;=1−P(X=0)−P(X=1)amp;=1−49(1)−49×23amp;=1−49−827=27−12−827=727\begin{aligned}& \text { So, } k \times \frac{9}{4}=1 \Rightarrow k=\frac{4}{9} \\\\& \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{X}<2) \\\\& =1-\mathrm{P}(\mathrm{X}=0)-\mathrm{P}(\mathrm{X}=1) \\\\& =1-\frac{4}{9}(1)-\frac{4}{9} \times \frac{2}{3} \\\\& =1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27}\end{aligned}

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