1. **Problem statement:** We have a penalty shoot-out where the probability of scoring a penalty is $0.7$. We want to find: (i) The probability that the first three penalties are scored. (ii) The probability that the first penalty missed is the third one taken. (iii) The probability that only one of the first three penalties is scored. 2. **Formulas and rules:** - The probability of independent events all happening is the product of their individual probabilities. - The probability of the first failure occurring at a specific trial means all previous trials succeeded and the current one failed. - The probability of exactly $k$ successes in $n$ trials with success probability $p$ is given by the binomial formula: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ 3. **Calculations:** (i) Probability first three penalties are scored: $$P = 0.7 \times 0.7 \times 0.7 = 0.7^3 = 0.343$$ (ii) Probability first penalty missed is the third one: This means first two scored, third missed: $$P = 0.7 \times 0.7 \times (1 - 0.7) = 0.7^2 \times 0.3 = 0.147$$ (iii) Probability only one of the first three penalties is scored: Using binomial formula with $n=3$, $k=1$, $p=0.7$: $$P = \binom{3}{1} (0.7)^1 (0.3)^2 = 3 \times 0.7 \times 0.09 = 0.189$$ **Final answers:** (i) $0.343$ (ii) $0.147$ (iii) $0.189$