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NDA 1 2024 Maths, question 105

Topic: Statistics

Q105

If two random variables XX and YY are connected by relation 2X−3Y5X+4Y=4\frac{2X-3Y}{5X+4Y}=4 and XX follows Binomial distribution with parameters n=10n=10 and p=12p=\frac{1}{2}, then what is the variance of YY ?

  1. (a) 810361\frac{810}{361}Correct answer
  2. (b) 919\frac{9}{19}
  3. (c) 21361\frac{21}{361}
  4. (d) 121361\frac{121}{361}

Solution.

Answer: (a) 810361\frac{810}{361}

  1. Cross-multiply: 2X−3Y=20X+16Y2X - 3Y = 20X + 16Y, so Y=−1819XY = -\frac{18}{19}X.
  2. Var(X)=np(1−p)=10×12×12=52Var(X) = np(1 - p) = 10 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{2}.
  3. Var(Y)=(1819)2×52=324361×52=810361Var(Y) = \left(\frac{18}{19}\right)^{2} \times \frac{5}{2} = \frac{324}{361} \times \frac{5}{2} = \frac{810}{361}.

Trap: The ratio is undefined when X=0X = 0; the relation is used in its cross-multiplied form, Y=−1819XY = -\frac{18}{19}X.