1. **Problem statement:** We have two independent random variables $Y_1$ and $Y_2$ with probability density functions (pdfs) $$f_1(y_1) = 6y_1(1 - y_1), \quad 0 \leq y_1 \leq 1$$ $$f_2(y_2) = 3y_2^2, \quad 0 \leq y_2 \leq 1$$ We want to find the probability $P(Y_1 Y_2 \leq 0.46)$. 2. **Key idea:** Since $Y_1$ and $Y_2$ are independent, the joint pdf is the product: $$f_{Y_1,Y_2}(y_1,y_2) = f_1(y_1) f_2(y_2)$$ We want to find $$P(Y_1 Y_2 \leq 0.46) = \iint_{y_1 y_2 \leq 0.46} f_1(y_1) f_2(y_2) \, dy_1 dy_2$$ 3. **Set up the integral:** For fixed $y_2$, the condition $y_1 y_2 \leq 0.46$ implies $$y_1 \leq \frac{0.46}{y_2}$$ Since $y_1$ is between 0 and 1, the upper limit is $$\min\left(1, \frac{0.46}{y_2}\right)$$ 4. **Write the probability as an integral over $y_2$:** $$P = \int_0^1 \left[ \int_0^{\min(1, \frac{0.46}{y_2})} 6 y_1 (1 - y_1) \, dy_1 \right] 3 y_2^2 \, dy_2$$ 5. **Split the integral over $y_2$ at $y_2 = 0.46$ because for $y_2 \leq 0.46$, $\frac{0.46}{y_2} \geq 1$:** - For $0 \leq y_2 \leq 0.46$, upper limit for $y_1$ is 1. - For $0.46 < y_2 \leq 1$, upper limit for $y_1$ is $\frac{0.46}{y_2}$. So, $$P = \int_0^{0.46} \left[ \int_0^1 6 y_1 (1 - y_1) \, dy_1 \right] 3 y_2^2 \, dy_2 + \int_{0.46}^1 \left[ \int_0^{\frac{0.46}{y_2}} 6 y_1 (1 - y_1) \, dy_1 \right] 3 y_2^2 \, dy_2$$ 6. **Calculate inner integral for $y_1$ from 0 to 1:** $$\int_0^1 6 y_1 (1 - y_1) \, dy_1 = 6 \int_0^1 (y_1 - y_1^2) \, dy_1 = 6 \left[ \frac{y_1^2}{2} - \frac{y_1^3}{3} \right]_0^1 = 6 \left( \frac{1}{2} - \frac{1}{3} \right) = 6 \times \frac{1}{6} = 1$$ 7. **Calculate inner integral for $y_1$ from 0 to $\frac{0.46}{y_2}$:** Let $a = \frac{0.46}{y_2}$, $$\int_0^a 6 y_1 (1 - y_1) \, dy_1 = 6 \int_0^a (y_1 - y_1^2) \, dy_1 = 6 \left[ \frac{a^2}{2} - \frac{a^3}{3} \right] = 3 a^2 - 2 a^3$$ 8. **Substitute back into the integral:** $$P = \int_0^{0.46} 1 \times 3 y_2^2 \, dy_2 + \int_{0.46}^1 (3 a^2 - 2 a^3) 3 y_2^2 \, dy_2$$ Recall $a = \frac{0.46}{y_2}$, so $$3 a^2 - 2 a^3 = 3 \left( \frac{0.46}{y_2} \right)^2 - 2 \left( \frac{0.46}{y_2} \right)^3 = \frac{3 \times 0.46^2}{y_2^2} - \frac{2 \times 0.46^3}{y_2^3}$$ Multiply by $3 y_2^2$: $$\left( \frac{3 \times 0.46^2}{y_2^2} - \frac{2 \times 0.46^3}{y_2^3} \right) 3 y_2^2 = 3 \times 3 \times 0.46^2 - 2 \times 3 \times 0.46^3 \frac{1}{y_2} = 9 \times 0.46^2 - 6 \times 0.46^3 \frac{1}{y_2}$$ 9. **Rewrite $P$ as:** $$P = \int_0^{0.46} 3 y_2^2 \, dy_2 + \int_{0.46}^1 \left(9 \times 0.46^2 - \frac{6 \times 0.46^3}{y_2} \right) dy_2$$ 10. **Calculate the first integral:** $$\int_0^{0.46} 3 y_2^2 \, dy_2 = 3 \left[ \frac{y_2^3}{3} \right]_0^{0.46} = (0.46)^3 = 0.097336$$ 11. **Calculate the second integral:** $$\int_{0.46}^1 9 \times 0.46^2 \, dy_2 = 9 \times 0.2116 \times (1 - 0.46) = 1.9044 \times 0.54 = 1.0284$$ $$\int_{0.46}^1 \frac{6 \times 0.46^3}{y_2} \, dy_2 = 6 \times 0.097336 \times \int_{0.46}^1 \frac{1}{y_2} \, dy_2 = 0.584 \times \ln\left(\frac{1}{0.46}\right) = 0.584 \times 0.7765 = 0.4539$$ 12. **Combine the second integral parts:** $$1.0284 - 0.4539 = 0.5745$$ 13. **Add both parts to get $P$:** $$P = 0.097336 + 0.5745 = 0.6718$$ **Final answer:** $$\boxed{P(Y_1 Y_2 \leq 0.46) \approx 0.672}$$