Probability — Question 1

Question 1 – Solution

Step 1: Use the normalization condition. A probability density must have total area $1$: $$1=c\int_{-1}^{1}(1-x^2)\,dx.$$

Step 2: Evaluate the normalization integral. $$\begin{align*} \int_{-1}^{1}(1-x^2)\,dx &=\left[x-\frac{x^3}{3}\right]_{-1}^{1}\\ &=\frac23-\left(-\frac23\right)=\frac43. \end{align*}$$ Thus $$1=c\left(\frac43\right), \qquad \boxed{c=\frac34}.$$

Step 3: Translate the event. $$|X|<\frac12 \quad\Longleftrightarrow\quad -\frac12<X<\frac12.$$

Step 4: Integrate the density over this interval. $$\begin{align*} P\left(|X|<\frac12\right) &=\frac34\int_{-1/2}^{1/2}(1-x^2)\,dx\\ &=\frac34\left[x-\frac{x^3}{3}\right]_{-1/2}^{1/2}\\ &=\frac34\left(\frac{11}{12}\right) =\frac{11}{16}. \end{align*}$$ $$\boxed{P\left(|X|<\frac12\right)=\frac{11}{16}}$$