3878 – jobyme88

We have a function $f(x) = 3x^2, x \in [0, 1]$ as the probability density function.

What is the probability of $P(X = 0.2)$ ?

What is the probability of $P( \frac{1}{2} \leq X \leq 1)$ ?

Solution:

(1) $\begin{align*}P( \frac{1}{2} \leq X \leq 1) = \int_\frac{1}{2}^1 3x^2 dx \\& = x^3_1 - x^3_\frac{1}{2} \\& = 1- \frac{1}{8} \\ & = \frac{7}{8}\end{align*}$

(2) $\begin{align*}P( X = 0.2) = \int_\frac{0.2}{0.2}^1 3x^2 dx \\& = x^3_0.2 - x^3_0.2 \\& = 0\end{align*}$

Conclusion:

$P(X=a) = 0$

$P(a \leq X \leq b) = P(a \leq X < b) = P(a < X \leq b) = P(a <X < b)$

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