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Let X denote the reaction time, in seconds, to a certain stimulus and } Y denote the temperature ( in degrees Fahrenheit at which a certain reaction starts to take place. Suppose that two random variables X and Y have the joint density f(x,y)=4xy, for 0<x<10 \lt x \lt 1, 0<y<10 \lt y \lt 1, f(x,y)=0, elsewhere. Find (a) P(0X12 and 14Y12);P(0 \leq X \leq \frac{1}{2} \text{ and } \frac{1}{4} \leq Y \leq \frac{1}{2}); (b) P(X<Y)P(X \lt Y).

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(a)\textbf{(a)}

P(0X12 and 14Y12)=0121412f(x,y)dydx=01214124xy dydx=0122x(14122y dy)dx=0122x(y2)1412dx=((12)2(14)2)0122x dx=(14116)(x2)012=316(12)2=364\begin{align*} P(0 \leq X \leq \frac{1}{2} \text{ and } \frac{1}{4} \leq Y \leq \frac{1}{2}) &= \int_0^{\frac{1}{2}} \int_{\frac{1}{4}}^{\frac{1}{2}} f(x,y) dydx \\ &= \int_0^{\frac{1}{2}} \int_{\frac{1}{4}}^{\frac{1}{2}} 4xy \text{ } dydx \\ &= \int_0^{\frac{1}{2}} 2x \biggr( \int_{\frac{1}{4}}^{\frac{1}{2}} 2y \text{ } dy \biggr) dx \\ &= \int_0^{\frac{1}{2}} 2x \biggr( y^2 \biggr) \biggr |_{\frac{1}{4}}^{\frac{1}{2}} dx \\ &= \biggr( \biggr(\frac{1}{2 }\biggr)^2 - \biggr(\frac{1}{4} \biggr)^2 \biggr) \int_0^{\frac{1}{2}} 2x \text{ } dx \\ &= \biggr( \frac{1}{4} - \frac{1}{16} \biggr) \biggr( x^2 \biggr) \biggr |_0^{\frac{1}{2}} \\ &= \frac{3}{16} \cdot \biggr( \frac{1}{2} \biggr)^2 \\ &= \boxed{ \frac{3}{64} } \end{align*}

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