Questions dealing with cdf (cumulative distribution function) $F(x)=(k-\frac{1}{\lfloor x\rfloor})[x \ge 1],$

Question

Recall that the "floor" of a real number $x$ , denoted $\lfloor x \rfloor$ , is the largest integer $\leq x$ .

$$F(x)= \begin{cases} k-\frac{1}{\lfloor x\rfloor}, & x \ge 1,\\ 0, & x \lt 1,\end{cases}$$ is a cummulative distribution function (cdf) for some fixed number k . Find: k. and P(X=4)

i am able to find k as it is cdf so as x->infinity, F(x)=1 but i am not able to find P(X=4) why P(X=4) will not be equal to 3/4?

Answer 1

Here is my method to solve this problem:

$P(X <= 4) = P(X = 4) + P(X < 4)$ .... eq(1)

Therefore, $P(X = 4) = P(X <= 4) - P(X < 4)$

i.e.,$P(X = 4) = P(X <= 4) - P(X <= 3)$

$P(X = 4) = F(4) - F(3)$

$P(X = 4) = 1/4 - 1/3 = (1/12)$