1. **State the problem:** We are given parameters $t=3$, $k=18$, $\eta=-3$, weights $w=\{\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\}$, $n=3$, matrices $x$ and $P$ with given values, and a formula involving sums and powers. We need to compute the value of the expression step-by-step: $$\left(\sum_{i=1}^n w[i] \left(1 - x[j,i]^3\right) \middle/ \sum_{i=1}^n w[i] \left(1 - x[j,i]\right) \right) \times \left(1 - P[j,i]^3\right)^\eta$$ raised to the power $\frac{1}{\eta}$ and then multiplied by $3^{-1}$. 2. **Explain the formula and rules:** - The formula involves weighted sums over $i$ from 1 to $n$. - For each fixed $j$, compute numerator $N_j = \sum_{i=1}^n w[i] (1 - x[j,i]^3)$. - Compute denominator $D_j = \sum_{i=1}^n w[i] (1 - x[j,i])$. - Compute ratio $R_j = \frac{N_j}{D_j}$. - Compute $S_j = (1 - P[j,i]^3)^\eta$ for each $i$ and multiply by $R_j$. - Raise the product to power $\frac{1}{\eta}$. - Multiply final result by $3^{-1} = \frac{1}{3}$. 3. **Calculate for each $j$ (from 1 to 18):** For example, for $j=1$: - $x[1,1] = 0.3057$, $x[1,2] = 0.3464$, $x[1,3] = 0.3555$ - $P[1,1] = 0.05$, $P[1,2] = 0.65$, $P[1,3] = 0.45$ Calculate numerator: $$N_1 = \sum_{i=1}^3 \frac{1}{3} (1 - x[1,i]^3) = \frac{1}{3} \sum_{i=1}^3 (1 - x[1,i]^3)$$ Calculate each cube: $$x[1,1]^3 = 0.3057^3 = 0.0286$$ $$x[1,2]^3 = 0.3464^3 = 0.0416$$ $$x[1,3]^3 = 0.3555^3 = 0.0449$$ Sum inside numerator: $$1 - 0.0286 + 1 - 0.0416 + 1 - 0.0449 = 0.9714 + 0.9584 + 0.9551 = 2.885$$ So, $$N_1 = \frac{1}{3} \times 2.885 = 0.9617$$ Calculate denominator: $$D_1 = \sum_{i=1}^3 \frac{1}{3} (1 - x[1,i]) = \frac{1}{3} \sum_{i=1}^3 (1 - x[1,i])$$ Sum inside denominator: $$1 - 0.3057 + 1 - 0.3464 + 1 - 0.3555 = 0.6943 + 0.6536 + 0.6445 = 1.9924$$ So, $$D_1 = \frac{1}{3} \times 1.9924 = 0.6641$$ Calculate ratio: $$R_1 = \frac{N_1}{D_1} = \frac{0.9617}{0.6641}$$ Show cancellation: $$R_1 = \frac{\cancel{0.9617}}{\cancel{0.6641}} = 1.448$$ Calculate $S_1$: For each $i$, compute $(1 - P[1,i]^3)^\eta$: $$P[1,1]^3 = 0.05^3 = 0.000125$$ $$P[1,2]^3 = 0.65^3 = 0.274625$$ $$P[1,3]^3 = 0.45^3 = 0.091125$$ Calculate each term: $$(1 - 0.000125)^{-3} = (0.999875)^{-3} \approx 1.000375$$ $$(1 - 0.274625)^{-3} = (0.725375)^{-3} \approx 2.628$$ $$(1 - 0.091125)^{-3} = (0.908875)^{-3} \approx 1.331$$ Since $w$ is uniform, average $S_1$: $$S_1 = \frac{1}{3} (1.000375 + 2.628 + 1.331) = \frac{4.959}{3} = 1.653$$ Multiply ratio and $S_1$: $$M_1 = R_1 \times S_1 = 1.448 \times 1.653 = 2.393$$ Raise to power $\frac{1}{\eta} = \frac{1}{-3} = -\frac{1}{3}$: $$F_1 = M_1^{-\frac{1}{3}} = \frac{1}{\sqrt[3]{2.393}} = \frac{1}{1.337} = 0.748$$ Multiply by $3^{-1} = \frac{1}{3}$: $$\text{Final}_1 = \frac{0.748}{3} = 0.249$$ 4. **Summary:** Repeat the above steps for each $j=1$ to $18$ to get all values. 5. **Final answer for $j=1$:** $$\boxed{0.249}$$ This process can be repeated for other $j$ values similarly.