1. **State the problem:** We have a Venn diagram with sets $\mathbb{R}$, $\mathbb{T}$, and $\mathbb{V}$, and integers $a$ and $b$ representing unknown quantities in the diagram. 2. **Given probabilities:** - $P(\mathbb{T} \mid \mathbb{R}) = \frac{3}{5}$ means the probability of being in $\mathbb{T}$ given the item is in $\mathbb{R}$. - $P(\mathbb{V} \mid \mathbb{R}) = \frac{2}{3}$ means the probability of being in $\mathbb{V}$ given the item is in $\mathbb{R}$. 3. **Identify the parts of $\mathbb{R}$:** - Only in $\mathbb{R}$: $b$ - In $\mathbb{R} \cap \mathbb{T}$ but not $\mathbb{V}$: $a$ - In $\mathbb{R} \cap \mathbb{V}$ but not $\mathbb{T}$: 4 - In $\mathbb{R} \cap \mathbb{T} \cap \mathbb{V}$: 6 4. **Calculate total in $\mathbb{R}$:** $$ |\mathbb{R}| = b + a + 4 + 6 = b + a + 10 $$ 5. **Calculate $P(\mathbb{T} \mid \mathbb{R})$:** Items in $\mathbb{T}$ and $\mathbb{R}$ are $a$ and 6. $$ P(\mathbb{T} \mid \mathbb{R}) = \frac{a + 6}{b + a + 10} = \frac{3}{5} $$ 6. **Calculate $P(\mathbb{V} \mid \mathbb{R})$:** Items in $\mathbb{V}$ and $\mathbb{R}$ are 4 and 6. $$ P(\mathbb{V} \mid \mathbb{R}) = \frac{4 + 6}{b + a + 10} = \frac{10}{b + a + 10} = \frac{2}{3} $$ 7. **Solve for $b + a + 10$ from the second equation:** $$ \frac{10}{b + a + 10} = \frac{2}{3} \implies 10 \times 3 = 2 (b + a + 10) \implies 30 = 2b + 2a + 20 $$ $$ 2b + 2a = 10 \implies b + a = 5 $$ 8. **Use $b + a = 5$ in the first equation:** $$ \frac{a + 6}{b + a + 10} = \frac{3}{5} \implies \frac{a + 6}{5 + 10} = \frac{3}{5} \implies \frac{a + 6}{15} = \frac{3}{5} $$ $$ 5(a + 6) = 45 \implies 5a + 30 = 45 \implies 5a = 15 \implies a = 3 $$ 9. **Find $b$ using $b + a = 5$:** $$ b + 3 = 5 \implies b = 2 $$ **Final answer:** $$ a = 3, \quad b = 2 $$