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Probability question

2025 · 3 Apr · Shift 2 · Q40
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Probability question

2025 · 3 Apr · Shift 2 · Q40

JEE MainMathematicsProbabilityMCQ+4 / −1
If the probability that the random variable XXX takes the value xxx is given by P(X=x)=k(x+1)3−x,x=0,1,2,3…P(X=x)=k(x+1) 3^{-x}, x=0,1,2,3 \ldotsP(X=x)=k(x+1)3−x,x=0,1,2,3…, where kkk is a constant, then P(X≥3)P(X \geq 3)P(X≥3) is equal to
  1. A
    19\frac{1}{9}91​
  2. B
    827\frac{8}{27}278​
  3. C
    727\frac{7}{27}277​
  4. D
    49\frac{4}{9}94​
View written solutionFree

Correct answer: A

  1. Given probability mass function

    P(X=x)=k(x+1)3−x,x=0,1,2,3,…P(X=x)=k(x+1)3^{-x}, \quad x=0,1,2,3,\dotsP(X=x)=k(x+1)3−x,x=0,1,2,3,…

    Since total probability is 111, we use

    ∑x=0∞P(X=x)=1\sum_{x=0}^{\infty} P(X=x)=1∑x=0∞​P(X=x)=1

    So,

    k∑x=0∞(x+1)(13)x=1k\sum_{x=0}^{\infty}(x+1)\left(\frac13\right)^x=1k∑x=0∞​(x+1)(31​)x=1

  2. Evaluate the series

    We use the standard result:

    ∑x=0∞(x+1)rx=1(1−r)2,∣r∣<1\sum_{x=0}^{\infty}(x+1)r^x=\frac{1}{(1-r)^2}, \quad |r|<1∑x=0∞​(x+1)rx=(1−r)21​,∣r∣<1

    Here r=13r=\frac13r=31​. Therefore,

    =\frac{1}{\left(\frac23\right)^2} =\frac{9}{4}$$ Hence, $$k\cdot \frac94=1 \implies k=\frac49$$

  3. Find P(X≥3)P(X\ge 3)P(X≥3)

    P(X≥3)=1−P(X=0)−P(X=1)−P(X=2)P(X\ge 3)=1-P(X=0)-P(X=1)-P(X=2)P(X≥3)=1−P(X=0)−P(X=1)−P(X=2)

    Now compute each term:

    P(X=0)=k(0+1)30=49P(X=0)=k(0+1)3^0=\frac49P(X=0)=k(0+1)30=94​

    P(X=1)=k(1+1)3−1=49⋅2⋅13=827P(X=1)=k(1+1)3^{-1}=\frac49\cdot 2\cdot \frac13=\frac{8}{27}P(X=1)=k(1+1)3−1=94​⋅2⋅31​=278​

    P(X=2)=k(2+1)3−2=49⋅3⋅19=427P(X=2)=k(2+1)3^{-2}=\frac49\cdot 3\cdot \frac19=\frac{4}{27}P(X=2)=k(2+1)3−2=94​⋅3⋅91​=274​

    Therefore,

    P(X≥3)=1−49−827−427P(X\ge 3)=1-\frac49-\frac{8}{27}-\frac{4}{27}P(X≥3)=1−94​−278​−274​

    Convert to denominator 272727:

    49=1227\frac49=\frac{12}{27}94​=2712​

    So,

    =1-\frac{24}{27} =\frac{3}{27} =\frac19$$

  4. Check options

    P(X≥3)=19P(X\ge 3)=\frac19P(X≥3)=91​

    So the correct option is A.

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