Probability of drawing two colors after four draws You have a bin with 5 balls of different colors: Red, Orange, Yellow, Green, and Blue, all of which are equally likely to be drawn. After a ball is drawn, it is put back afterwards. What is the probability of drawing both the yellow and red balls at least once after drawing 4 times from the bin?
I attempted this problem by finding the probability that a yellow and red ball is not drawn as 3/5 for a single draw. Then, I did  $1-(3/5)^4$ to find the probability that a yellow and red ball is drawn at least once in the 4 draws. Is this correct?
 A: As Flip Tack pointed out in the comments, you have calculated the probability that at least one yellow ball or at least one red ball is selected.  We wish to find the probability that ball of both colors are selected.
Method 1:  We use the Inclusion-Exclusion Principle.
Let $R$ be the event that a red ball is chosen; let $Y$ be the event that a yellow ball is chosen.  We wish to find
$$\Pr(R \cap Y) = 1 - \Pr(R' \cup Y')$$
By the Inclusion-Exclusion Principle,
$$\Pr(R' \cup Y') = \Pr(R') + \Pr(Y') - \Pr(R' \cap Y')$$
Hence,
$$\Pr(R \cap Y) = 1 - \Pr(R') - \Pr(Y') + \Pr(R' \cap Y')$$
$\Pr(R')$:  Since four of the five balls are not red, the probability that a red ball is not selected in four draws is
$$\Pr(R') = \left(\frac{4}{5}\right)^4$$
$\Pr(Y')$:  Since four of the five balls are not yellow, the probability that a yellow ball is not selected in four draws is
$$\Pr(Y') = \left(\frac{4}{5}\right)^4$$
$\Pr(R' \cap Y')$:  Since three of the five balls are neither red nor yellow, the probability that neither a red nor yellow is selected in four draws is
$$\Pr(R' \cap Y') = \left(\frac{3}{5}\right)^4$$
Therefore, the probability that at least one red ball and at least one yellow ball are selected after drawing four times from the bin with replacement is
$$\Pr(R \cap Y) = 1 - \left(\frac{4}{5}\right)^4 - \left(\frac{4}{5}\right)^4 + \left(\frac{3}{5}\right)^4$$
Method 2: In the comments, Vadim Chernetsov asked how to perform a direct count.  We count directly.
One red, one yellow, and two from the other three colors:  Choose one of the four positions for the red ball and one of the remaining three positions for the yellow ball.  There are then three choices for each of the remaining two positions.  Hence, there are
$$\binom{4}{1}\binom{3}{1} \cdot 3^2$$
such cases.
Two red, one yellow, and one from the other three colors:  Choose two of the four positions for the red balls and one of the remaining two positions for the yellow ball.  There are three choices for the remaining position.  Hence, there are
$$\binom{4}{2}\binom{2}{1} \cdot 3$$
such cases.
One red, two yellow, and one from the other three colors:  By symmetry, there are
$$\binom{4}{2}\binom{2}{1} \cdot 3$$
such cases.
Two red and two yellow:  Choose two of the four positions for the red balls.  The other positions must be filled by yellow balls.  Hence, there are
$$\binom{4}{2}$$
such cases.
Three red and one yellow:  Choose three of the four positions for the red balls.  The other position must be filled by a yellow ball.  Hence, there are
$$\binom{4}{3}$$
such cases.
One red and three yellow:  By symmetry, there are
$$\binom{4}{3}$$
such cases.
Since there are five choices for each of the four positions in the sequence, there are $5^4$ possible outcomes in the sample space.
Hence, the probability that at least one red ball and at least one yellow ball are selected in four draws with replacement from the bin with one ball each from the five colors red, orange, yellow, green, and blue is
$$\Pr(R \cap Y) = \frac{\dbinom{4}{1}\dbinom{3}{1}\cdot 3^2 + 2\dbinom{4}{2}\dbinom{2}{1} \cdot 3 + \dbinom{4}{2} + 2\dbinom{4}{3}}{5^4}$$
