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This is the question I wish to solve:

A random variable X has PDF given by $$f(x) = \left\{ \matrix{ 6x(1 - x),0 \le x \le 1 \hfill \cr 0,otherwise \hfill \cr} \right\}$$

(a)Determine and sketch the CDF

(b) Find Pr(0.25 < X < 0.75)

(c) Find Pr(X > 0.75|X > 0.25)

(d) Find Var(X)

My Attempt-:

(a)This is ${x^2}(3 - 2x)$ between 0 and 1 anyway.

(b)This is obviously ${{11} \over {16}}$

(c) I thought I could calculate this by $$1 - \Pr (X \le 0.75) + \Pr (X \le 0.25)$$

My reasoning being that it is the complement of the probability being less than or equal to 0.75 but then you have to add the correction since it isn't less than or equal to 0.25. I don't get the right answer though so it seems I am wrong.

(d) I tried to use that $Var(X) = E[{X^2}] - {(E[X])^2}$ but $E[X] = \int_{ - \infty }^\infty {6x(1 - x)dx} $, right? That doesn't converge.

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For a), what about $x<0$ and $x>1$?.
For c), use the rule \begin{align*} \mathbb{P}(A | B ) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}. \end{align*} And for d), you have for all $n \in \mathbb{N}$, so in particular for $n=1,2$ \begin{align*} \mathbb{E}(X^n) = \int_{-\infty}^\infty x^n f(x) \text{d}x = \int_0^1 x^n f(x) \text{d}x. \end{align*}

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