If two random variable X and Y are connected by relation 2 X-3 Y 5 X+4 Y =4 and X follows Binomial distribution with parameters n=10 and p=(1)/(2) , then what is the variance of Y ?

Correct answer: A. (810)/(361) (2 x-3 y)/(5 x+4 y)=4 aligned & 2 x-3 y=20 x+16 y & 19 y=18 x & y=(18 x)/(19) & var .(x)=n p q=n p(1-p) & =10 × (1)/(2) × (1)/(2)=(5)/(2) & Now, var. (y)=var .(-(18)/(19) x) & var. ·(y)=((18)/(19))^2 × (5)/(2)=(810)/(361) aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Probability

NDA 2024 · Mathematics · Probability

If two random variable X and Y are connected by relation 2 X-3 Y 5...

If two random variable X and Y are connected by relation

\frac{2 \mathrm{X}-3 \mathrm{Y}}{5 \mathrm{X}+4 \mathrm{Y}}=4

and X follows Binomial distribution with parameters

n=10

and

p=\frac{1}{2}

, then what is the variance of Y ?

Options

A $\frac{810}{361}$
B $\frac{9}{19}$
C $\frac{21}{361}$
D $\frac{121}{361}$

Correct answer

A. $\frac{810}{361}$

Step-by-step solution

\because \quad \frac{2 x-3 y}{5 x+4 y}=4

\begin{aligned} & \Rightarrow \quad 2 x-3 y=20 x+16 y \\ & \Rightarrow \quad 19 y=18 x \\ & \Rightarrow \quad y=\frac{18 x}{19} \\ & \operatorname{var} .(x)=n p q=n p(1-p) \\ & =10 \times \frac{1}{2} \times \frac{1}{2}=\frac{5}{2} \\ & \text{ Now, } \quad \text{ var. }(y)=\operatorname{var} .\left(-\frac{18}{19} x\right) \\ & \Rightarrow \quad \text{ var. } \cdot(y)=\left(\frac{18}{19}\right)^2 \times \frac{5}{2}=\frac{810}{361} \end{aligned}

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