Q.4.17 - A First Course in Probability

Question

Suppose that the distribution function of X given by

$F\left(b\right)=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b<0\\ \frac{b}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le b<1\\ \frac{1}{2}+\frac{b-1}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le b<2\\ \frac{11}{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le b<3\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\le b\end{array}\right.$

(a) Find $P\{X=i\},i=1,2,3$.

(b) Find $P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}$.

Step-by-Step Solution

Answer

(a) The value for the P{x=i} if i=1,2,3 are

$P\left\{1\right\}=\frac{1}{4}$

$P\left\{2\right\}=\frac{1}{6}$

$P\left\{3\right\}=\frac{1}{12}$

(b) The value for the $P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}$ is $\frac{1}{2}$.

Step 1: Given information (Part a)

The distribution function of X given as

$F\left(b\right)=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b<0\\ \frac{b}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le b<1\\ \frac{1}{2}+\frac{b-1}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le b<2\\ \frac{11}{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le b<3\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\le b\end{array}\right.$

Step 2: Solution (Part a)

The calculation is given below,

$\mathrm{P}(\mathrm{X}=1)=\mathrm{F}(1+)-\mathrm{F}(1-)$

$=\frac{1}{2}+\frac{1-1}{4}-\frac{1}{4}$

$=\frac{1}{4}$

$\mathrm{P}(\mathrm{X}=2)=\mathrm{F}(2+)-\mathrm{F}(2-)$

$=\frac{11}{12}-\frac{1}{2}-\frac{2-1}{4}$

$=\frac{11}{12}-\frac{1}{2}-\frac{1}{4}$

$=\frac{11-6-3}{12}$

$=\frac{1}{6}$

Similarly,

$P(X=3)=1-\frac{11}{12}$

$=\frac{1}{12}$

Step 3: Final answer (Part a)

The value for the P{x=i} if i=1,2,3 are

$P\left\{1\right\}=\frac{1}{4}$

$P\left\{2\right\}=\frac{1}{6}$

$P\left\{3\right\}=\frac{1}{12}$

Step 4: Given information (Part b)

Given in the question that 

$F\left(b\right)=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b<0\\ \frac{b}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le b<1\\ \frac{1}{2}+\frac{b-1}{4}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le b<2\\ \frac{11}{12}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le b<3\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\le b\end{array}\right.$

Step 5: Solution (Part b)

The calculation is given below,

$\mathrm{P}\left(\frac{1}{2}<X<\frac{3}{2}\right)=\mathrm{F}(3/2)-\mathrm{F}(1/2)$

$=\left[\frac{1}{2}+\frac{\frac{3}{2}-1}{4}\right]-\left[\frac{\frac{1}{2}}{4}\right]$

$=\frac{1}{2}+\left(\frac{1}{2}\times \frac{1}{4}\right)-\left(\frac{1}{2}\times \frac{1}{4}\right)$

$=\frac{1}{2}$

Step 6: Final answer (Part b)

The value for the $P\left\{\frac{1}{2}<X<\frac{3}{2}\right\}$ is $\frac{1}{2}$.