If $$A_{1}, A_{2}, A_{3}$$ are three mutually exclusive events, the probability of their union is equal to:

A

$$P\left(A_{1}\right) + P\left(A_{2}\right) + P\left(A_{3}\right) - P\left(A_{1}PA_{2}PA_{3}\right) $$

B

$$P\left(A_{1}\right)P\left(A_{2}\right) + P\left(A_{2}\right)P\left(A_{3}\right) + P\left(A_{3}\right)P\left(A_{1}\right)$$

C

$$P\left(A_{1}\right) + P\left(A_{2}\right) + P\left(A_{3}\right)$$

D

$$P\left(A_{1}\right)P\left(A_{2}\right)P\left(A_{3}\right)$$