In a game two dice are used.
Die A has 2 blue faces and 4 yellow faces.
Die B has 4 blue faces and 2 pink faces.
The two dice are thrown together.

1. Calculate the probability that both dice show blue

$P(\text{A blue})=\frac{2}{6}=\frac{1}{3},\qquad P(\text{B blue})=\frac{4}{6}=\frac{2}{3}$

Since both dice must show blue,
$P(\text{both blue})=\frac{1}{3}\times\frac{2}{3}=\frac{2}{9}.$

1. Identify the two ways that just one die can show blue

Just one blue means either:

$A$ is blue and $B$ is not blue, or $A$ is not blue and $B$ is blue.

$P(A\text{ blue})=\dfrac{2}{6}=\dfrac{1}{3}$, $P(A\text{ not blue})=\dfrac{4}{6}=\dfrac{2}{3}$

$P(B\text{ blue})=\dfrac{4}{6}=\dfrac{2}{3}$, $P(B\text{ not blue})=\dfrac{2}{6}=\dfrac{1}{3}$

2. Calculate and add the probabilities of the two mutually exclusive cases

$P(\text{just one blue})=\left(\dfrac{1}{3}\times\dfrac{1}{3}\right)+\left(\dfrac{2}{3}\times\dfrac{2}{3}\right)$

$=\dfrac{1}{9}+\dfrac{4}{9}=\dfrac{5}{9}$

1. Identify the probabilities needed from the first throw and second throw

From earlier parts,

$P(\text{both blue})=\dfrac{2}{9}$

$P(\text{just one blue})=\dfrac{4}{9}$

For a new throw, the probability of both blue is still $\dfrac{2}{9}$.

2. Calculate the probability of winning on the second throw after just one blue on the first throw

$P(\text{just one blue first, then both blue second})=\frac{4}{9}\times\frac{2}{9}=\frac{8}{81}$

3. Add the mutually exclusive winning probabilities

$P(\text{wins})=P(\text{both blue first})+P(\text{just one blue first, then both blue second})$

$=\frac{2}{9}+\frac{8}{81}$

$=\frac{18}{81}+\frac{8}{81}=\frac{26}{81}$