How can I prove this inequation $\Pr\{X+YCould you please help me to prove the inequality probability as follows:
$\Pr\{X+Y<t\} \le \Pr\{X<t\} \Pr\{Y<t\}$
where $X$ and $Y$ are non-negative independent random variables with common distribution.
Many thanks for your helps

If $X(\omega)+Y(\omega)<t$, since $X(\omega)\geq 0$ we have $Y(\omega)<t$ and since $Y(\omega)\geq 0$ we have $X(\omega)<t$ so $\{X+Y<t\}\subset \{X<t\}\cap \{Y\cap t\}$ and taking the probabilities, thanks to independence
$$P(\{X+Y<t\})\leq P(\{X<t\}\cap \{Y<t\})=P(\{X<t\})\cdot P(\{Y<t\}).$$
Note that the fact that $X$ and $Y$ have the same distribution wasn't used.
