Find $P(X \leq 1.25 | X \leq 1.5)$ C.D.F

Question

The p.d.f of a random variable X is given by

$$f_X(x) = \begin{cases} k(x-1)(2-x) & 1 < x < 2 \\ 0 & otherwise \\ \end{cases} $$

(a) Find the value of the constant $k$

$$1 = \int_{1}^{2}dx = k(-\frac{1}{3}x^3 + 3/2x^2 -2x)\bigg|_{1}^{2} = 1/6k$$

Therefore $k = 6$

(b) Find the c.d.f (comulative distribution function) of $X$

$$\int_{-\infty}^{x} 6(x-1)(2-x)dx = \int_{-\infty}^{1} 6(x-1)(2-x)dx + \int_{1}^{x} 6(x-1)(2-x)dx $$

$$= 0 + -2x^3 + 9x^2 -12x + 5 = -2x^3 + 9x^2 -12x + 5$$

$$F_X(x) = \begin{cases} 0 & x \leq 1 \\ -2x^3 + 9x^2 -12x + 5 & 1 \leq x < 2 \\ 1 &x \geq 2 \end{cases} $$

(c) Find $P(X \leq 1.25 | X \leq 1.5)$

$P(X \leq 1.25 | X \leq 1.5) = \frac{P(X \leq 1.25 \cap X \leq 1.5)}{P(X \leq 1.5)}$

$=\frac{\text{Confuse what to do here}}{F(1.5)}$

I'm not sure how to go about this. I know that

$P(X \leq 1.25 \cap X \leq 1.5) = \frac{P(X \leq 1.25)P(X \leq 1.5 | X \leq 1.25)}{P(X \leq 1.5)}$

Answer 1

You want to calculate the probability that $X\leq 1.25$ AND $X \leq 1.5$. But if $X\leq 1.25$, you know for sure that $X\leq 1.5$

Hence $P(X\leq1.5 \, \text{and} \, X\leq1.25) = P(X\leq 1.25)$

Answer 2

$X\leq 1.25$ and $X\leq 1.5$ if and only if $X\leq 1.25$. Because $[1;1.25]\subset[1;1.5]$.

ie: $\mathsf P(\{X\leq 1.25\}\cap\{X\leq 1.5\})=\mathsf P(\{X\leq 1.25\})$