1. **Problem statement:** (b) Given the probability of scoring a penalty is $0.7$, find: (i) Probability first three penalties are scored. (ii) Probability the first missed penalty is the third one taken. (iii) Probability only one of the first three penalties is scored. (c) Given a spinner with sectors €0 (210°), €10 (90°), and €12 (60°), complete the table and find: (ii) Expected win/loss for one play (cost €5). (iii) Probability Laura neither wins nor loses money after two plays. --- 2. **Formulas and rules:** - For independent events, probability of all occurring is product of individual probabilities. - Probability of first failure at $n^{th}$ trial: $P(\text{fail at } n) = (p)^{n-1}(1-p)$ where $p$ is success probability. - Binomial probability: $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. - Expected value: $E = \sum (\text{value} \times \text{probability})$. - Probability from angle: $P = \frac{\text{angle}}{360^\circ}$. --- 3. **(b)(i) Probability first three penalties scored:** $$P = 0.7 \times 0.7 \times 0.7 = 0.7^3 = 0.343$$ --- 4. **(b)(ii) Probability first missed penalty is third:** First two scored, third missed: $$P = 0.7 \times 0.7 \times (1-0.7) = 0.7^2 \times 0.3 = 0.147$$ --- 5. **(b)(iii) Probability only one of first three penalties scored:** Number of ways to score exactly one penalty out of three is $\binom{3}{1} = 3$. Probability for one specific sequence (one scored, two missed): $$0.7 \times 0.3 \times 0.3 = 0.063$$ So total probability: $$3 \times 0.063 = 0.189$$ --- 6. **(c) Complete the table:** - Angle for €0 is given as 210°. - Probability for €0: $$P = \frac{210}{360} = \frac{7}{12}$$ - Probability for €12: $$P = \frac{60}{360} = \frac{1}{6}$$ - Probability for €10 is given as $\frac{1}{4}$ (which matches $\frac{90}{360}$). Table: | Win | €0 | €10 | €12 | |---|---|---|---| | Angle | 210° | 90° | 60° | | Probability | $\frac{7}{12}$ | $\frac{1}{4}$ | $\frac{1}{6}$ | --- 7. **(c)(ii) Expected win/loss for one play:** Expected win: $$E = 0 \times \frac{7}{12} + 10 \times \frac{1}{4} + 12 \times \frac{1}{6} = 0 + 2.5 + 2 = 4.5$$ Cost to play is 5, so expected net: $$4.5 - 5 = -0.5$$ Player expects to lose 0.5 per game on average. --- 8. **(c)(iii) Probability Laura neither wins nor loses money after two plays:** Neither winning nor losing means net zero, so total winnings equal total cost. Cost for two plays: $2 \times 5 = 10$. We want probability total winnings = 10. Possible winnings per play: 0, 10, 12. To get total 10 in two plays, possible pairs: - (0,10) or (10,0) - (12, -2) not possible since -2 not a winning amount So only (0,10) or (10,0) sum to 10. Calculate: $$P = 2 \times P(0) \times P(10) = 2 \times \frac{7}{12} \times \frac{1}{4} = \frac{7}{24} \approx 0.2917$$ --- **Final answers:** (i) $0.343$ (ii) $0.147$ (iii) $0.189$ (c)(ii) Expected net loss per game: $-0.5$ (c)(iii) Probability neither win nor lose after two plays: $\frac{7}{24}$