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?
Best Answer
The complement of "is this and is that" is "is not this or is not that".
$$\begin{align}&~~~~~~\mathsf P(\text{at least one red }\textit{and}\text{ at least one blue})\\&=1-\mathsf P(\text{no red }\textit{or}\text{ no blue})\\&=1-\mathsf P(\text{no red})-\mathsf P(\text{no blue})+\mathsf P(\text{no red }\textit{and}\text{ no blue})\end{align}$$