Question

Let (X,Y) have the following joint p.d.f.
f(x,y) = \left\{\begin{matrix} 6xy^2 ,0\leq x\leq 1,0\leq y\leq 1\\0,otherwise \end{matrix}\right.

Find P(2Y + X > 1).

14 Feb 2018
Answer :
Word Count : 564
To solve for \( P(2Y + X > 1) \), we need to evaluate the given joint probability density function (p.d.f.) and integrate over the region where \( 2Y + X > 1 \). ### Given Joint p.d.f: \[ f(x, y) = \begin{cases} 6xy^2, & \text{for } 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0, & \text{otherwise} \end{cases} \] We are tasked with finding the probability \( P(2Y + X > 1) \), which can be expressed as: \[ P(2Y + X > 1) = \int \int_{2y + x > 1} f(x, y) \, dx \, dy \] ### Step 1: Set the limits of integration for \( 2Y + X > 1 \) The inequality \( 2y + x > 1 \) defines a region in the \( (x, y)
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Related Questions

Let the p.d.f. of a continuous random variable X be given by
f(x) = \beta e^{^{-\beta x}},x\geq 0
Find the maximum likelihood estimator of \beta based on a sample X1 ,X2 ,....,Xn .

Let (X,Y) have the following joint p.d.f.
f(x,y) = \left\{\begin{matrix} 6xy^2 ,0\leq x\leq 1,0\leq y\leq 1\\0,otherwise \end{matrix}\right.

Find P(2Y + X > 1).

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