Let X be a random variable with cumulative distribution function given by $\mathrm{F}(\mathrm{x})=\{\begin{array}{cc}0,& ifx<0\\ \frac{x+1}{3},& if0\le x<1\\ 1,& ifx\ge 1\end{array}$ Then the value of $\mathrm{P}(\frac{1}{3}<\mathrm{X}<\frac{3}{4})+\mathrm{P}(\mathrm{X}=0)$ is equal to

Concepts Used:

1. Cumulative Distribution Function (CDF):

The CDF F(x) gives the probability that the random variable X takes a value less than or equal to x . That is, F(x) = P(X ≤ x) .

2. Finding Probability Using the CDF:

The probability that the random variable X lies within a certain interval (a, b] is given by:
     
 P(a < X ≤ b) = F(b) - F(a)

3. Probability at a Specific Point (Jump Discontinuity):

The probability at a specific point x = c is the difference in the CDF just to the right and just to the left of c :
     
P(X = c) = F(c⁺) - F(c^-)

Explanation -

We are given a cumulative distribution function (CDF) F(x) of a random variable X as:

F(x) = $\{\begin{array}{ll}0,& \text{if }x<0\\ \frac{x+1}{3},& \text{if }0\le x<1\\ 1,& \text{if }x\ge 1\end{array}$

From the definition of the probability from the CDF:  P(a < X $\le$ b) = F(b) - F(a)

In this case, we need to calculate $F\left(\frac{3}{4}\right)andF\left(\frac{1}{3}\right)$.

Since $0\le \frac{1}{3}<1$  and $0\le \frac{3}{4}<1$ , we use the formula $F(x)=\frac{x+1}{3}$ for both 1/3  and 3/4 :

$F\left(\frac{1}{3}\right)=\frac{\frac{1}{3}+1}{3}=\frac{\frac{4}{3}}{3}=\frac{4}{9}$

$\Rightarrow F\left(\frac{3}{4}\right)=\frac{\frac{3}{4}+1}{3}=\frac{\frac{7}{4}}{3}=\frac{7}{12}$

Thus, the probability $P(\frac{1}{3}<X\le \frac{3}{4})$ is:

$P(\frac{1}{3}<X\le \frac{3}{4})=F\left(\frac{3}{4}\right)-F\left(\frac{1}{3}\right)=\frac{7}{12}-\frac{4}{9}$

= $\frac{21}{36}-\frac{16}{36}=\frac{5}{36}$

The probability at a point is the jump in the CDF at that point. We need to calculate F(0⁺) - F(0^-) .

From the CDF definition: F(0⁺) = F(0) =$\frac{0+1}{3}=\frac{1}{3}$

⇒ F(0^-) = 0

Thus, $P(X=0)=F(0^{+})-F(0^{-})=\frac{1}{3}-0=\frac{1}{3}=\frac{12}{36}$

Now, we add the two results:

$P(\frac{1}{3}<X\le \frac{3}{4})+P(X=0)=\frac{5}{36}+\frac{12}{36}=\frac{17}{36}$

Thus, the final answer is 17/36.