Answer to Question #261071 in Statistics and Probability for sandhya

Question

The time taken X by a garage to repair a car is a continuous rv with pdf f(x) = { 3x/4 (2 − x); 0 ≤ x ≤ 2 0; elsewhere If, on leaving his car, a motorist goes to keep on an engagement lasting for a time Y, where Y is a continuous rv independent of X, with pdf f(y) = { 1/2 y; 0 ≤ y ≤ 2 0; elsewhere . Determine the probability that the car will not be ready on his return.

Answer 1

Since $X$ and $Y$ are independent, their joint probability density function is given by

$$f_{X,Y}(x,y)=\begin{cases} \dfrac{3}{8}xy(2-x) & 0\leq x,y\leq2 \\ 0 & elsewhere \end{cases}$$

The probability that $X$ exceeds $Y$

$$P(X>Y)=\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{x}\dfrac{3}{8}xy(2-x)dydx$$

$$=\dfrac{3}{16}\displaystyle\int_{0}^{2}\bigg[x(2-x)y^2\bigg]\begin{matrix} x \\ 0 \end{matrix}dx$$

$$=\dfrac{3}{16}\displaystyle\int_{0}^{2}(2x^3-x^4)dx$$

$$=\dfrac{3}{16}[\dfrac{2x^4}{4}-\dfrac{x^5}{5}]\begin{matrix} 2 \\ 0 \end{matrix}$$

$$=\dfrac{3}{16}(8-\dfrac{32}{5})=\dfrac{3}{10}=0.3$$