Answer to Question #338483 in Statistics and Probability for ken

Question

The joint density function of the random variables X, Y is given as

f(x, y) = { 0.4 (2x + 3y), 0 < x < 1, 0 < y< 1

0, elsewhere

1)Determine P [ 0.33 < X < 2/3 |Y > 3 /4 ]

Answer 1

1). Check, whether $f(x,y)$ is a valid joint density function: $\int_0^1\int_0^10.4(2x+3y)dydx=0.4\int_0^1(2xy+\frac32y^2)|_0^1dx=0.4\int_0^1(2x+\frac32)dx=0.4(x^2+\frac32x)|_0^1=1$Thus, it is a valid joint density function.

Remind the formula: $P(A|B)=\frac{P(A\cap B)}{P(B)}$. From the latter we get: $P(0.33<X<\frac23|Y>\frac34)=\frac{P(0.33<X<\frac23,Y>\frac34)}{P(Y>\frac34)}$.

Consider separately each probability in the last formula: $P(0.33<X<\frac23,Y>\frac34)=0.4\int_{0.33}^{\frac23}\int_{\frac34}^{1}(2x+3y)dydx=0.4\int_{0.33}^{\frac23}(2xy+\frac32y^2)_{\frac34}^{1}dx=0.4\int_{0.33}^{\frac23}(2x+\frac32-\frac32x-\frac{27}{32})dx=0.4\int_{0.33}^{\frac23}(\frac12x+\frac{21}{32})dx=0.4(\frac14x^2+\frac{21}{32}x)_{0.33}^{\frac23}\approx0.1219$

The latter value is rounded to $4$ decimal places.

$P(Y>\frac34)=0.4\int_0^1\int_{\frac34}^1(2x+3y)dydx=0.4\int_0^1(2xy+\frac32y^2)|_{\frac34}^1dx=0.4\int_0^1(2x+\frac32-\frac32x-\frac{27}{32})dx=0.4\int_0^1(\frac12x+\frac{21}{32})dx=0.4\left(\frac{1}{4}x^2+\frac{21}{32}x\right)|_0^1=0.4\left(\frac14+\frac{21}{32}\right)=\frac{29}{80}$

Thus, the probability is the following: $P(0.33<X<\frac23|Y>\frac34)\approx0.1219\cdot\frac{80}{29}\approx0.336$

Answer: $P(\frac13<X<\frac23,Y>\frac34)\approx0.336$ (it is rounded to $3$ decimal places)