Since the word density provides notion of a bunch or interval, your message Possibility density function suggests for that probability of a consistent variable.  Employed in computing the possibilities of continuous hit-or-miss specifics.
Generally, the equation employed to go into detail a ongoing probability distribution is named a probability incident function. Sometimes, it is called a thickness purpose, a PDF. For a frequent probability distribution. The probability incident (occurence) function (PDF) P(x) of this continuous distribution is termed the derivative in the (cumulative) distribution intent D(x).

A function f(x) is said to be the probability density function (pdf) of a continuous random variable X if,
  1. f(x) ≥ 0
  2. $\int_{-\infty }^{\infty }f(x)dx$ = 1

In general the probability on the continuous random variable assuming values in the interval [a, b] is written by
P(a ≤ x ≤ b) = $\int_a ^b f(x)dx$.

Let X be a uniform random changing assuming all values from the finite interval (a, b).
Then this probability density purpose of X is due to
f(x) = $\frac{1}{b-a}$ for a < x < b
= 0 otherwise.
The probability of X assuming a value in the sub interval (α, β) is given by
P(α<X<β) = $\int_{\alpha }^{\beta }\frac{1}{b-a}dx$ = $\frac{\beta -\alpha }{b-a}$
The Probability distribution function of a Normal variable is described by the parameters μ the mean and σ2 the variance of
random variable.

f(x) = $\frac{e^{\frac{-(x-\mu )^{2}}{2\sigma ^{2}}}}{\sigma \sqrt{2\pi }}$

The pdf of the standard normal distribution with μ = 0 and σ =1 is

f(x) = $\frac{e^{\frac{-z^{2}}{2}}}{\sqrt{2\pi }}$
The transformation formula for the standard normal variable is z = $\frac{x-\mu }{\sigma }$
The actual pdf of of the exponential probability occurrence function is defined with the parameter λ.

f(x) = $\lambda e^-\lambda x $. regarding x > 0.

The mean and variance of the exponential distribution will be given by $\frac1 \lambda $ and $\frac1 \lambda ^{2} $.
A function f(x, y) would be the joint probability density function of the two dimensional constant random variable (X, Y) in the event,

f(x, y) ≥ 0 for many x, y
$\int_-\infty ^\infty \int_-\infty ^\infty f(x, y)dxdy$ = 1.
Suppose (X, Y) is really a two dimensional steady random variable having joint probability occurrence function f(x, y).
Next the marginal probability occurrence functions of X and Y usually are respectively,
g(x) = $\int_-\infty ^\infty f(x, y)dy$ and also h(y) = $\int_-\infty ^\infty f(x, y)dx$.
Suppose (X, Y) is a two dimensional continuous random variable with joint probability density function f(x, y).
Then the conditional probability density function of X = x, given that Y = y is given by,

g($\frac{x}{y}$) = $\frac{f(x,y))}{h(y)}$ where h(y) is the marginal probability density function of Y.

And the conditional probability density function of Y = y, given that X =x is given by,

h($\frac{y}{x}$) = $\frac{f(x,y))}{g(x)}$
Imagine (x, Y) is a two dimensional ongoing random variable utilizing joint probability denseness function f(x, y).
Then the conditional probability denseness function of X= x, since Y = y is published by,

g($\frac x y $) = $\fracf(x, y) h(y)$  where h(y) will be the marginal probability denseness function of B.

And the conditional possibility density function connected with Y = y simply, given that X =x is published by,

h($\fracy x $) = $\fracf(x, y) g(x)$.

Solved Example

Question: A continuous random variable has the probability density function
f(x) =  kx                    for 0 ≤ x < 10
      = k(30 - x)           for 10 ≤ x < 30
      = 0                      Otherwise.
(1) Find the value of k
(2) Find the probability that X lies between 20 and 30
(3) Find the probability that X lies between 100 and 300
Solution:
 
Since f(x) is a probability density function f(x) ≥ 0 for all x
$\int_{-\infty }^{\infty }f(x)$ = 1
$\int_{-\infty }^{\infty }f(x)$ = $\int_{0}^{10}kxdx + \int_{10}^{30}k(30-x)dx$ = 1

                                         = $\frac{kx^{2}}{2}|_{0}^{10}$ + k(30x-$$\frac{x^{2}}{2})_{10}^{30}$ = 1

                                         = 50 k + 200k = 1        ⇒      k = $\frac{1}{250}$

P(20 < x < 30) = $\int_{20}^{30}\frac{30-x}{250}$$dx$

                       = $[\frac{30x}{250}-\frac{x^{2}}{500}]_{20}^{30}$

                       = $\frac{100}{500}$ = $\frac{1}{5}$

P(100 < X < 300) = 0     as f(x) = 0   for x > 30.
 

The graphs of probability density functions of some continuous random variables are shown below.
 Uniform Distribution The graph of pdf f(x) = $\frac{1}{4}$ is shown here
which is defined for 1 < x < 5. The total area of the shaded region =1 which represents the probability P(1 < X < 5)
The actual graph of pdf of a normal random varying
is shown in this article. The graph will be symmetric about
this mean and G (X < μ) = P(X > μ) = 0. 5.
 Normal Curve
 Exponential Graph The graph in the pdf for this exponential
distribution f(x) = 0. 8e$^{-0. 8x}$ can be shown here.
The region under the curve for that interval x > 0
is comparable to 1.