A certain college gives aptitude tests in the sciences and the humanities to all entering freshmen. If X and Y are, respectively, the proportions of correct answers that a student gets on the tests in the two subjects, the joint probability distribution of these random variables can be approximated with the joint probability density

` $$ \begin{aligned} f(x,y)=\left\{ \begin{array}{ll} \frac{2}{5}(2x+3y) & \hbox{$0< x< 1, 0< y<1$;}\\ 0, & \hbox{Otherwise} \end{array} \right. \end{aligned} $$ `

What are the probabilities that a student will get

(a) less than 0.40 on both tests;

(b) more than 0.80 on the science test and less than 0.50 on the humanities test?

#### Solution

(a) The probability that a student will get less than 0.40 on both tests is

` $$ \begin{aligned} P(X < 0.40, Y < 0.40) &= \int_0^{0.40}\int_0^{0.40} f(x,y) \; dx dy\\ &= \int_0^{0.40}\int_0^{0.40} \frac{2}{5} (2x+3y) \; dx dy\\ &= \frac{2}{5} \int_0^{0.40}\int_0^{0.40} (2x+3y) \; dx dy\\ &= \frac{2}{5} \int_0^{0.40}\bigg[\int_0^{0.40} (2x+3y) \; dx \bigg] \; dy\\ &= \frac{2}{5} \int_0^{0.40}\bigg[2\frac{x^2}{2} +3xy\bigg]_0^{0.40} \; dy\\ &= \frac{2}{5} \int_0^{0.40}\bigg[0.40^2 +3\times 0.40 y\bigg] \; dy\\ &= \frac{2}{5} \int_0^{0.40}\bigg[0.16 +1.2y\bigg] \; dy\\ &= \frac{2}{5} \bigg[0.16 y +1.2\frac{y^2}{2}\bigg]_{0}^{0.40}\\ &= \frac{2}{5} \bigg[0.16\times 0.40 +1.2\frac{0.40^2}{2}\bigg]\\ &= \frac{2}{5} \bigg[0.64 +0.096\bigg]\\ &= 0.2944 \end{aligned} $$ `

(b) The probability that a student will get more than 0.80 on the science test and less than 0.50 on the humanities test is

` $$ \begin{aligned} P(X > 0.80, Y < 0.50) &= \int_{0.80}^1\int_0^{0.50} f(x,y) \; dy dx\\ &= \int_{0.80}^1\int_0^{0.50} \frac{2}{5} (2x+3y) \; dy dx\\ &= \frac{2}{5} \int_{0.80}^1\int_0^{0.50} (2x+3y) \; dy dx\\ &= \frac{2}{5} \int_{0.80}^1\bigg[\int_0^{0.50} (2x+3y) \; dy \bigg] \; dx\\ &= \frac{2}{5} \int_{0.80}^1\bigg[2xy +3\frac{y^2}{2}\bigg]_0^{0.50} \; dx\\ &= \frac{2}{5} \int_{0.80}^1\bigg[2x*0.50 +3\frac{0.50^2}{2}\bigg] \; dx\\ &= \frac{2}{5} \int_{0.80}^1\bigg[x +0.375\bigg] \; dx\\ &= \frac{2}{5} \bigg[\frac{x^2}{2}+0.375 x\bigg]_{0.80}^1\\ &= \frac{2}{5} \bigg[\frac{1^2}{2}+0.375*1 -\frac{0.80^2}{2}-0.375*0.80\bigg]\\ &= \frac{2}{5} \bigg[0.64 +0.096\bigg]\\ &= 0.255 \end{aligned} $$ `