1. **Problem 15:** A biased pyramid-shaped die has faces 1 to 5 with probabilities given by $P(x) = \frac{k - x}{25}$ where $k$ is a constant. 2. **Find the probability of scoring 5:** $$P(5) = \frac{k - 5}{25}$$ 3. **Find the probability of scoring less than 3 (i.e., 1 or 2):** $$P(1) + P(2) = \frac{k - 1}{25} + \frac{k - 2}{25} = \frac{(k - 1) + (k - 2)}{25} = \frac{2k - 3}{25}$$ 4. **Use the fact that probabilities sum to 1:** $$\sum_{x=1}^5 P(x) = 1 \implies \sum_{x=1}^5 \frac{k - x}{25} = 1$$ Calculate the sum inside: $$\sum_{x=1}^5 (k - x) = 5k - (1 + 2 + 3 + 4 + 5) = 5k - 15$$ So: $$\frac{5k - 15}{25} = 1 \implies 5k - 15 = 25 \implies 5k = 40 \implies k = 8$$ 5. **Substitute $k=8$ back into probabilities:** $$P(5) = \frac{8 - 5}{25} = \frac{3}{25}$$ $$P(1) + P(2) = \frac{2(8) - 3}{25} = \frac{16 - 3}{25} = \frac{13}{25}$$ 6. **The die is rolled three times, sum of scores less than 5:** Possible sums less than 5 are 3 and 4 (since minimum sum is 3). Possible triples $(x_1,x_2,x_3)$ with $x_i \in \{1,2,3,4,5\}$ and sum less than 5: - Sum = 3: only $(1,1,1)$ - Sum = 4: permutations of $(1,1,2)$ 7. **Calculate probability of sum = 3:** $$P(1)^3 = \left(\frac{8 - 1}{25}\right)^3 = \left(\frac{7}{25}\right)^3 = \frac{343}{15625}$$ 8. **Calculate probability of sum = 4:** There are 3 permutations of $(1,1,2)$: $(1,1,2), (1,2,1), (2,1,1)$. Probability for one permutation: $$P(1) \times P(1) \times P(2) = \left(\frac{7}{25}\right)^2 \times \frac{8 - 2}{25} = \frac{49}{625} \times \frac{6}{25} = \frac{294}{15625}$$ Total for sum=4: $$3 \times \frac{294}{15625} = \frac{882}{15625}$$ 9. **Total probability sum less than 5:** $$\frac{343}{15625} + \frac{882}{15625} = \frac{1225}{15625} = \frac{49}{625}$$ --- 10. **Problem 16:** A game board with colored squares that send the player back to start if landed on. The player rolls a fair 6-sided die and moves forward accordingly. Colored squares are at positions 3, 7, 11, and 15. 11. **a.i Probability the player is on 'start' after rolling once:** Player returns to start if they land on a colored square. Possible rolls to land on colored squares from start (position 0): 3 or 7 only (since 11 and 15 are further). Probability: $$P(\text{start after 1 roll}) = P(3) + P(7) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$ 12. **a.ii Probability the player is on 'start' after rolling twice:** Two ways to be on start after two rolls: - Land on colored square on first roll (probability $\frac{1}{3}$), then any roll on second roll (probability 1), total $\frac{1}{3}$. - Not land on colored square on first roll (probability $\frac{2}{3}$), then land on colored square on second roll from new position. Positions after first roll if not on start: 1,2,4,5,6 (since 3 and 7 send back to start). From each position, check if a roll leads to a colored square: - From 1: roll 2 to get to 3 (colored) - From 2: roll 1 to get to 3 (colored) - From 4: roll 3 to get to 7 (colored) - From 5: roll 2 to get to 7 (colored) - From 6: roll 1 to get to 7 (colored) Calculate probability of landing on colored square on second roll given position: - From 1: $\frac{1}{6}$ - From 2: $\frac{1}{6}$ - From 4: $\frac{1}{6}$ - From 5: $\frac{1}{6}$ - From 6: $\frac{1}{6}$ Probability of being at each position after first roll (not on start): each is $\frac{1}{6}$ Sum: $$\sum P(\text{pos}) \times P(\text{land colored from pos}) = 5 \times \frac{1}{6} \times \frac{1}{6} = \frac{5}{36}$$ Total probability on start after two rolls: $$\frac{1}{3} + \frac{2}{3} \times \frac{5}{36} = \frac{1}{3} + \frac{10}{108} = \frac{1}{3} + \frac{5}{54} = \frac{18}{54} + \frac{5}{54} = \frac{23}{54}$$ 13. **b.i Probability on square 18 after three rolls:** We analyze possible paths to 18 without landing on colored squares. From start (0), possible moves: - After first roll, positions: 1,2,4,5,6 (not start) - After second roll, positions reachable without landing on colored squares and not returning to start. We find that 18 is reachable only from 10 (second row) by rolling 8 steps, impossible in one roll. Check if 18 can be reached in three rolls without landing on colored squares: - One path: 0 -> 9 (roll 9 impossible), so no direct path. - Another path: 0 -> 9 (roll 9 impossible), then down to 10, then 18. Since 9 and 10 are not reachable in one roll, probability is 0. Hence, probability on 18 after three rolls is 0. 14. **b.ii Probability on square 17 after three rolls:** Similarly, 17 is on second row, reachable from 10 by rolling 7. Since 10 is not reachable in one roll, probability is 0. --- **Final answers:** - 15a.i $P(5) = \frac{k - 5}{25}$, with $k=8$, so $P(5) = \frac{3}{25}$. - 15a.ii $P(x<3) = \frac{2k - 3}{25}$, with $k=8$, so $P(x<3) = \frac{13}{25}$. - 15b $k=8$, probability sum of three rolls less than 5 is $\frac{49}{625}$. - 16a.i Probability on start after one roll is $\frac{1}{3}$. - 16a.ii Probability on start after two rolls is $\frac{23}{54}$. - 16b.i Probability on 18 after three rolls is 0. - 16b.ii Probability on 17 after three rolls is 0.