# Let $X$ be a continuous variable with probability density function $kx(1-x)^2$ for $0<x<1$

Let $X$ be a continuous variable with probability density function $f(x)=kx(1−x)^2$ over $0< x <1$, zero otherwise.

• $(a)$ Find a value of $k$ so that $f(x)$ is a proper density.
• $(b)$ Find the cumulative distribution function of $X$.
• $(c)$ Find $\mathbb{P}\left[\frac{1}{4} < X < \frac{3}{4}\big|X > 0.1\right]$.

For $(a)$ I found that $k= 12$.

For $(b)$ the cdf is $F(x)=3x^4-8x^3+6x^2$.

For $(c)$ I calculated $\mathbb{P}[X > 0.1] = 0.0523$ and $\mathbb{P}\left[0.25 < X < 0.75\right] = F(.75)-F(.25) = 0.6875$.

Now how do I get $\mathbb{P}\left[0.25 < X < 0.75\big|X > 0.1\right]$? I'm stuck here.

• By definition (since $0.1<0.25$), $\mathbb{P}\left[0.25 < X < 0.75\mid X > 0.1\right]=\mathbb{P}\left[0.25 < X < 0.75\right]/\mathbb P\left[X > 0.1\right]$ – Did Jun 28 '15 at 13:21

## 2 Answers

$$\mathbb{P}\left[\frac{1}{4}<X<\frac{3}{4}\,\Big|\, X>\frac{1}{10}\right]=\frac{\int_{\frac{1}{4}}^{\frac{3}{4}}x(1-x)^2\,dx}{\int_{\frac{1}{10}}^{1}x(1-x)^2\,dx}=\frac{6875}{9477}\approx \color{red}{72.54 \%}.$$

P(a given b) = p(a and b)/ p (b)

p(x is between 0.25 and 0.75 and x is greater than 0.1) = p(x is between 0.25 and 0.75)

p(x is between 0.25 and 0.75 given x is greater than 0.1) = p(x is between 0.25 and 0.75)/p(x is greater than 0.1)

• I calculatedP(X>.01) incorrect. Does it equal P(.01<X<1) – Joz Jun 28 '15 at 14:02