For any random variable, the total probability of all its possible values is 1.

Core idea (discrete case)

If a random variable XXX can take values x1,x2,x3,…x_1,x_2,x_3,\dots x1​,x2​,x3​,… with probabilities P(X=xi)P(X=x_i)P(X=xi​), then

∑iP(X=xi)=1\sum_i P(X=x_i)=1i∑​P(X=xi​)=1

This is because those values cover every possible outcome, and something must happen in each trial.

Think of rolling a fair die:

  • Possible values: 1, 2, 3, 4, 5, 6
  • Each has probability 1/61/61/6
  • Total probability:

P(1)+P(2)+⋯+P(6)=6×16=1P(1)+P(2)+\dots+P(6)=6\times \frac{1}{6}=1P(1)+P(2)+⋯+P(6)=6×61​=1

So the sum of probabilities over all values of the random variable “die outcome” is 1.

Continuous case

For a continuous random variable with probability density function fX(x)f_X(x)fX​(x), we do not sum probabilities of individual points, but we integrate over all real values:

∫−∞∞fX(x) dx=1\int_{-\infty}^{\infty}f_X(x),dx=1∫−∞∞​fX​(x)dx=1

This integral plays the same role as the sum in the discrete case: it says the total probability over all possible values is 1.

Why it must be 1

By definition, probability is assigned on a sample space that contains every possible outcome of the experiment. The random variable just re-expresses those outcomes as numbers, but it cannot invent or lose outcomes. So the probabilities across all its possible values always add (or integrate) up to 1.

Answer in one line:
The total probability of all possible values of a random variable is always 1 (sum of all P(X=x)P(X=x)P(X=x) in the discrete case, or integral of the density over all xxx in the continuous case).

Information gathered from public forums or data available on the internet and portrayed here.