1. **State the problem:** We have a Venn diagram with sets R, T, and V, and integers a and b representing unknown quantities in the intersections. 2. **Given probabilities:** - $P(T \mid R) = \frac{3}{5}$ means the probability of being in T given the item is in R. - $P(V \mid R) = \frac{2}{3}$ means the probability of being in V given the item is in R. 3. **Identify the elements in R:** - Elements in R are $b$ (only R), $a$ (R and T only), 4 (R and V only), and 6 (R, T, and V). - Total in R: $b + a + 4 + 6 = b + a + 10$. 4. **Calculate $P(T \mid R)$:** - Elements in both T and R are $a$ and 6. - So, $P(T \mid R) = \frac{a + 6}{b + a + 10} = \frac{3}{5}$. 5. **Calculate $P(V \mid R)$:** - Elements in both V and R are 4 and 6. - So, $P(V \mid R) = \frac{4 + 6}{b + a + 10} = \frac{10}{b + a + 10} = \frac{2}{3}$. 6. **Set up equations:** - From $P(T \mid R)$: $\frac{a + 6}{b + a + 10} = \frac{3}{5}$ - From $P(V \mid R)$: $\frac{10}{b + a + 10} = \frac{2}{3}$ 7. **Solve for $b + a + 10$ from the second equation:** $$\frac{10}{b + a + 10} = \frac{2}{3} \implies 10 \times 3 = 2 \times (b + a + 10)$$ $$30 = 2b + 2a + 20 \implies 2b + 2a = 10 \implies b + a = 5$$ 8. **Substitute $b + a = 5$ into the first equation:** $$\frac{a + 6}{5 + 10} = \frac{3}{5} \implies \frac{a + 6}{15} = \frac{3}{5}$$ $$5(a + 6) = 45 \implies 5a + 30 = 45 \implies 5a = 15 \implies a = 3$$ 9. **Find $b$:** $$b + a = 5 \implies b + 3 = 5 \implies b = 2$$ **Final answer:** $$a = 3, \quad b = 2$$