A game has two sets of eight icons. One icon from each set is randomly chosen to be displayed on each turn, starting with set 1. The icons for each set are as follows. For example, if a three-interlocking-circles icon is chosen from set 1 and a sailing-boat icon is chosen from set 2, the display would be (three-interlocking-circles)(sailing-boat).

1. Find the probability of getting the required icon from each set and multiply them.

Set 1 has $4$ three-interlocking-circles icons out of $8$ icons.

Set 2 has $2$ three-interlocking-circles icons out of $8$ icons.

$P(\text{three-interlocking-circles from Set 1 and Set 2})=\frac{4}{8}\times\frac{2}{8}=\frac{8}{64}=\frac{1}{8}$

1. Find the probability that no three-interlocking-circles icon is chosen from either set.

Set 1 has $4$ non-three-interlocking-circles icons out of $8$, and Set 2 has $6$ non-three-interlocking-circles icons out of $8$.

$P(\text{no three-interlocking-circles})=\frac{4}{8}\times\frac{6}{8}=\frac{3}{8}$

2. Use the complement to find the probability of at least one three-interlocking-circles icon.

$P(\text{at least one three-interlocking-circles})=1-P(\text{no three-interlocking-circles})$

$=1-\frac{3}{8}=\frac{5}{8}$

1. Write the first-stage probabilities for choosing each icon from Set 1.

Let
$C$ = three-interlocking-circles, $S$ = sailing-boat, $R$ = rocket, $G$ = graduation-cap.

Set 1 has $4C, 1S, 2R, 1G$, out of $8$ icons.

So the first branches are
$P(C)=\frac{4}{8},\quad P(S)=\frac{1}{8},\quad P(R)=\frac{2}{8},\quad P(G)=\frac{1}{8}.$

2. Find the second-stage probabilities after each possible Set 1 choice, removing the same icon type from Set 2.

Set 2 originally has $2C, 2S, 3R, 1G$.

If $C$ is chosen from Set 1, remove $2C$ from Set 2. Remaining: $2S,3R,1G$, total $6$.
$P(S\mid C)=\frac{2}{6},\quad P(R\mid C)=\frac{3}{6},\quad P(G\mid C)=\frac{1}{6}.$

If $S$ is chosen from Set 1, remove $2S$ from Set 2. Remaining: $2C,3R,1G$, total $6$.
$P(C\mid S)=\frac{2}{6},\quad P(R\mid S)=\frac{3}{6},\quad P(G\mid S)=\frac{1}{6}.$

If $R$ is chosen from Set 1, remove $3R$ from Set 2. Remaining: $2C,2S,1G$, total $5$.
$P(C\mid R)=\frac{2}{5},\quad P(S\mid R)=\frac{2}{5},\quad P(G\mid R)=\frac{1}{5}.$

If $G$ is chosen from Set 1, remove $1G$ from Set 2. Remaining: $2C,2S,3R$, total $7$.
$P(C\mid G)=\frac{2}{7},\quad P(S\mid G)=\frac{2}{7},\quad P(R\mid G)=\frac{3}{7}.$

3. Present the complete tree diagram structure with all possible non-matching displayed pairs.

The tree diagram should show:

$$\begin{array}{c|c} \text{Set 1 branch} & \text{Set 2 branches after removal} \\ \hline C\;\frac{4}{8} & S\;\frac{2}{6},\; R\;\frac{3}{6},\; G\;\frac{1}{6} \\ S\;\frac{1}{8} & C\;\frac{2}{6},\; R\;\frac{3}{6},\; G\;\frac{1}{6} \\ R\;\frac{2}{8} & C\;\frac{2}{5},\; S\;\frac{2}{5},\; G\;\frac{1}{5} \\ G\;\frac{1}{8} & C\;\frac{2}{7},\; S\;\frac{2}{7},\; R\;\frac{3}{7} \end{array}$$

No branch has the same icon in both positions.