Originally Posted by

**Robb** Suppose that $\displaystyle Y$ has density function

$\displaystyle f(y)=\left\{\begin{array}{cc}ky(1-y),&0\leq y\leq 1\\0, & \mbox{elsewhere}\end{array}\right. $

a. Find the value of $\displaystyle k$that makes $\displaystyle f(y)$ a probabilty density function.

No problem, found $\displaystyle k=6$

b. Find $\displaystyle P(.4 \leq Y \leq 1) = \int^{1}_{.4}6y(1-y)dy=.648$

the part I am having trouble with is;

d. Find $\displaystyle P(Y \leq .4 | Y \leq .8)$

not too sure if $\displaystyle P(Y \leq .4 | Y \leq .8)=\frac{P(.4 \leq Y \leq .8)}{P(Y \leq .8)}$ Mr F says: This should be $\displaystyle {\color{red}P(Y \leq .4 | Y \leq .8)=\frac{P(Y \leq .4 \, \text{ and } \, Y \leq .8)}{P(Y \leq .8)}}$.

or how to calculate this conditional probability...