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Probability question

2023 · 8 Apr · Shift 2 · Q25
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  5. /2023 · 8 Apr · Shift 2 · Q25

Probability question

2023 · 8 Apr · Shift 2 · Q25

JEE MainMathematicsProbabilityMCQ+4 / −1
If the probability that the random variable X\mathrm{X}X takes values xxx 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, \ldotsP(X=x)=k(x+1)3−x,x=0,1,2,3,…, where k\mathrm{k}k is a constant, then P(X≥2)\mathrm{P}(\mathrm{X} \geq 2)P(X≥2) is equal to :
  1. A
    718\frac{7}{18}187​
  2. B
    2027\frac{20}{27}2720​
  3. C
    727\frac{7}{27}277​
  4. D
    1118\frac{11}{18}1811​
View written solutionFree

Correct answer: C

  1. Given probability mass function

We have 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 this is a probability distribution, ∑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)3−x=1.k\sum_{x=0}^{\infty}(x+1)3^{-x}=1.k∑x=0∞​(x+1)3−x=1.

  1. Evaluate the series

Let r=13.r=\frac13.r=31​. Then we need ∑x=0∞(x+1)rx.\sum_{x=0}^{\infty}(x+1)r^x.∑x=0∞​(x+1)rx. Using 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. Thus,

=\frac{1}{\left(1-\frac13\right)^2} =\frac{1}{\left(\frac23\right)^2} =\frac{9}{4}.$$ Therefore, $$k\cdot \frac94=1 \implies k=\frac49.$$ 3. **Find $P(X\ge 2)$** We can compute $$P(X\ge 2)=1-P(X=0)-P(X=1).$$ Now, $$P(X=0)=k(0+1)3^0=k=\frac49,$$ $$P(X=1)=k(1+1)3^{-1}=\frac49\cdot 2\cdot \frac13=\frac{8}{27}.$$ So, $$P(X\ge 2)=1-\frac49-\frac{8}{27}.$$ Convert to denominator $27$: $$1=\frac{27}{27}, \quad \frac49=\frac{12}{27}.$$ Hence, $$P(X\ge 2)=\frac{27}{27}-\frac{12}{27}-\frac{8}{27}=\frac{7}{27}.$$ 4. **Check options** $$\frac{7}{27}$$ matches **Option C**. 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So the answer agrees with the stored correct answer.
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