# Complement Rule for Probability

Four marbles are randomly selected from a bag that contains 6 red marbles, 5 blue marbles, and 4 green marbles. What is the probability of selecting at least one red marble and at least one blue marble?

Let $$E$$ = "at least one red marble and at least one blue marble". Then $$E'$$ = "no red marble and no blue marbles". Therefore, the Complement Rule for Probability yields: $$\begin{eqnarray*} P(E) & = & 1 - P(E') \\ & = & 1 - \frac{C(4,4)}{C(15,4)} \\ & = & 1 - \frac{1}{1365} \\ & = & \frac{1365}{1365} - \frac{1}{1365} \\ & = & \frac{1364}{1365}. \end{eqnarray*}$$

Is this correct or should my complement be $$E'$$ = "no red marble or no blue marbles" in which case I need to use the Additive Rule?

The complement of "is this and is that" is "is not this or is not that".

\begin{align}&~~~~~~\mathsf P(\text{at least one red }\textit{and}\text{ at least one blue})\\&=1-\mathsf P(\text{no red }\textit{or}\text{ no blue})\\&=1-\mathsf P(\text{no red})-\mathsf P(\text{no blue})+\mathsf P(\text{no red }\textit{and}\text{ no blue})\end{align}

"no red marble or no blue marbles" is the correct description of the complement (De Morgan's Law).

However, computing $$P(E')$$ is not as simple as using the additive rule because the intersection "no red marble and no blue marbles" is nonempty. You will need to use inclusion-exclusion to compute $$P(E')$$.

If two events $$A$$ and $$B$$ are complementary, then two things will be true:

1. There is no overlap, i.e. there are no events that count as satisfying both $$A$$ and $$B$$;

2. There is no gap, i.e. any event that can occur is either part of $$A$$, or part of $$B$$.

If $$A$$ is "at least one red and at least one blue" then the event "no red and no blue" doesn't cover the possibility of 1 red and 0 blue, nor does it cover 0 red and 1 blue, so it cannot be its complement.

More generally, when we use set or logic notation we use the rule that "NOT (A AND B)" is the same as "(NOT A) OR (NOT B)", so the complement of "at least one red and at least one blue", will be "(NOT (at least one red)) OR (NOT (at least one blue))", i.e. "either no red or no blue (or none of either)".