The probability is zero.

pdf = slope of the cdf. The value of the pdf at a given point is not a probability, it's the instantaneous rate of change of a probability. You need to integrate the pdf over a range to get a probability.

You could take the area under the pdf (i.e., integrate) for a window around a given x or use the area under the tail of the pdf past x (i.e., p-value).

You have an interval of real numbers [0, 1]. What's the probability of picking 0.3639401?

If we're using a uniform random number generator in [0, 1] with 32-bit IEEE754 numbers, the closest to 0.3639401 is

    2^-2 * (1 + 3823195*2^-23) = 0.3639400899410247802734375

The numbers directly before and after it are:

    2^-2 * (1 + 3823194*2^-23) = 0.363940060138702392578125
    2^-2 * (1 + 3823196*2^-23) = 0.36394011974334716796875

Giving a range of size 2.980232238769531250 * 10^-8 in which all numbers compare equal to 0.3639401 in IEEE754 32-bit floating point. And since we're looking at a domain of [0, 1], that's also immediately our probability.

(I'm fully aware this isn't what you were asking, but I found it fun either way)