Problem
Imagine I play the lottery and have a $2\%$ rate of winning. My friend also plays the lottery and has a $1\%$ rate of winning. Whoever wins, we will share the prize.
We could think our chances are $3\%$ then.
But let's think on the other side. My $2\%$ chance of winning also are $98\%$ chance of losing, which is $0.98$. For my friend, that losing rate is $99\%$ which is $0.99$. If I multiply our chance of NOT winning, that's $0.9702$. This can be converted to $2.98\%$ chance of winning.
So, are our chances of winning are $3\%$ or $2.98\%?$
If $A$ is the event that you win, and $B$ is the event that your friend wins, then $A\cup B$ is the event that you share at least one prize, and $A\cap B$ is the event that you both win the lottery (so you share two prizes).
Using $P(A)$ for the probability that the event $A$ happens, $P(B)$ for the probability that $B$ happens, etc,
$$ P(A\cup B) = P(A) + P(B) - P(A\cap B). $$
If you and your friend are able to compare plans before you play and make sure you never play the same numbers, then you will never win two prizes simultaneously. That is, in that case $P(A\cap B) = 0$ and the probability that at least one of you wins a prize is actually $3\%.$
However, if you and your friend play completely independently with no mutual planning, it is reasonable to suppose that your chances of winning are completely independent. That is, it is reasonable to suppose that $P(A\cap B) = P(A)\cdot P(B).$ (This is the definition of what it means for events to be independent.) In that case $$ P(A\cap B) = 0.02 \cdot 0.01 = 0.0002 $$ (writing all the percentage probabilities in non-percentage decimal format, such as $0.02$ instead of $2\%$, so that we don't get confused about how to multiply $2\% \times 1\%$), and therefore $$ P(A\cup B) = 0.02 + 0.01 - 0.0002 = 0.0298, $$ which is $2.98\%.$