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JEE Mathematics 2025 Question with Solution

If the probability that the random variable XX takes the value xx is given by P(X=x)=k(x+1)3−xP(X = x) = k(x + 1)3^{-x}, x=0,1,2,3,…x = 0, 1, 2, 3, \dots, where kk is a constant, then P(X≥3)P(X \ge 3) is equal to

  • A

    727\frac{7}{27}

  • B

    49\frac{4}{9}

  • C

    827\frac{8}{27}

  • D

    19\frac{1}{9}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: P(X=x)=k(x+1)3−xP(X = x) = k(x + 1)3^{-x} for x=0,1,2,3,…x = 0, 1, 2, 3, \dots.

Find: P(X≥3)P(X \ge 3).

Since P(X=x)P(X = x) defines a probability distribution, the sum of probabilities over all possible values of xx must be equal to 11:

∑x=0∞P(X=x)=1\sum_{x=0}^{\infty} P(X = x) = 1 ∑x=0∞k(x+1)3−x=1\sum_{x=0}^{\infty} k(x + 1)3^{-x} = 1 k∑x=0∞(x+1)(13)x=1k \sum_{x=0}^{\infty} (x + 1)\left(\frac{1}{3}\right)^x = 1

Let

S=∑x=0∞(x+1)(13)x=1⋅(13)0+2⋅(13)1+3⋅(13)2+4⋅(13)3+…S = \sum_{x=0}^{\infty} (x + 1)\left(\frac{1}{3}\right)^x = 1 \cdot \left(\frac{1}{3}\right)^0 + 2 \cdot \left(\frac{1}{3}\right)^1 + 3 \cdot \left(\frac{1}{3}\right)^2 + 4 \cdot \left(\frac{1}{3}\right)^3 + \dots

So,

S=1+23+39+427+…S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots

Multiply by 13\frac{1}{3}:

13S=13+29+327+…\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots

Subtract the second equation from the first:

S−13S=(1+23+39+427+… )−(13+29+327+… )S - \frac{1}{3}S = \left(1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \dots\right) - \left(\frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \dots\right) 23S=1+(23−13)+(39−29)+(427−327)+…\frac{2}{3}S = 1 + \left(\frac{2}{3} - \frac{1}{3}\right) + \left(\frac{3}{9} - \frac{2}{9}\right) + \left(\frac{4}{27} - \frac{3}{27}\right) + \dots 23S=1+13+19+127+…\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots

The right side is a geometric series with first term a=1a = 1 and common ratio r=13r = \frac{1}{3}. Therefore,

23S=11−13=32\frac{2}{3}S = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} S=32×32=94S = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}

Hence,

kS=1  ⟹  k⋅94=1  ⟹  k=49kS = 1 \implies k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}

Now,

P(X≥3)=1−P(X<3)=1−[P(X=0)+P(X=1)+P(X=2)]P(X \ge 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

Compute each term:

P(X=0)=k(0+1)30=49P(X = 0) = k(0 + 1)3^{0} = \frac{4}{9} P(X=1)=k(1+1)3−1=49⋅2⋅13=827P(X = 1) = k(1 + 1)3^{-1} = \frac{4}{9} \cdot 2 \cdot \frac{1}{3} = \frac{8}{27} P(X=2)=k(2+1)3−2=49⋅3⋅19=1281=427P(X = 2) = k(2 + 1)3^{-2} = \frac{4}{9} \cdot 3 \cdot \frac{1}{9} = \frac{12}{81} = \frac{4}{27}

Therefore,

P(X≥3)=1−(49+827+427)=1−(1227+827+427)=1−2427=1−89=19P(X \ge 3) = 1 - \left( \frac{4}{9} + \frac{8}{27} + \frac{4}{27} \right) = 1 - \left( \frac{12}{27} + \frac{8}{27} + \frac{4}{27} \right) = 1 - \frac{24}{27} = 1 - \frac{8}{9} = \frac{1}{9}

Therefore, the correct option is D.

Using geometric series derivative

Given: P(X=x)=k(x+1)3−xP(X = x) = k(x + 1)3^{-x} for x=0,1,2,3,…x = 0, 1, 2, 3, \ldots.

Find: P(X≥3)P(X \ge 3).

Since this is a probability mass function,

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

So,

k∑x=0∞(x+1)3−x=1k \sum_{x=0}^{\infty} (x + 1)3^{-x} = 1

Use the standard geometric series:

∑x=0∞tx=11−t\sum_{x=0}^{\infty} t^x = \frac{1}{1-t}

Differentiate with respect to tt:

∑x=0∞xtx−1=1(1−t)2\sum_{x=0}^{\infty} x t^{x-1} = \frac{1}{(1-t)^2}

Multiplying by tt gives

∑x=0∞xtx=t(1−t)2\sum_{x=0}^{\infty} x t^x = \frac{t}{(1-t)^2}

Now substitute t=13t = \frac{1}{3}:

∑x=0∞x(13)x=13(1−13)2=13(23)2=34\sum_{x=0}^{\infty} x\left(\frac{1}{3}\right)^x = \frac{\frac{1}{3}}{\left(1 - \frac{1}{3}\right)^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{3}{4}

Also,

∑x=0∞(13)x=11−13=32\sum_{x=0}^{\infty} \left(\frac{1}{3}\right)^x = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}

Hence,

∑x=0∞(x+1)(13)x=34+32=94\sum_{x=0}^{\infty} (x+1)\left(\frac{1}{3}\right)^x = \frac{3}{4} + \frac{3}{2} = \frac{9}{4}

Therefore,

k⋅94=1  ⟹  k=49k \cdot \frac{9}{4} = 1 \implies k = \frac{4}{9}

Now calculate:

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=0)=49,P(X=1)=827,P(X=2)=427P(X=0) = \frac{4}{9}, \qquad P(X=1) = \frac{8}{27}, \qquad P(X=2) = \frac{4}{27}

So,

P(X<3)=49+827+427=2427=89P(X < 3) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{24}{27} = \frac{8}{9}

Thus,

P(X≥3)=1−89=19P(X \ge 3) = 1 - \frac{8}{9} = \frac{1}{9}

Therefore, the required probability is 19\frac{1}{9}, so the correct option is D.

Common mistakes

  • Students may forget to use the normalization condition for a probability distribution. This is wrong because kk is determined by ∑P(X=x)=1\sum P(X=x)=1. First find kk from the infinite series, then compute the required probability.

  • A common mistake is evaluating P(X≥3)P(X \ge 3) as only P(X=3)P(X=3). This is incorrect because X≥3X \ge 3 includes all values 3,4,5,…3,4,5,\ldots. Use the complement 1−P(X<3)1-P(X<3) to include the entire tail.

  • Students may mishandle the series ∑(x+1)(13)x\sum (x+1)\left(\frac{1}{3}\right)^x by treating it as an ordinary geometric series. This is wrong because of the extra factor (x+1)(x+1). Use the derivative of the geometric series or the subtraction method shown in the solution.

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