Penalty Shootout

Solution

Answer: 0.332

Let \(X\) be the number of goals scored in the remaining \(16\) penalties. Each penalty has probability \(p = 0.3\) of being scored, so \[ X \sim B(16, 0.3). \] You already have \(4\) goals and you need at least \(10\) goals in total to win. Therefore the number of additional goals required is \[ 10 - 4 = 6. \] Thus we need the probability of scoring at least \(6\) goals from the remaining \(16\) penalties: \[ P(\text{win}) = P(X \ge 6). \] For a binomial distribution, \[ P(X=k) = \binom{16}{k}(0.3)^k(0.7)^{16-k}. \] So \[ P(X \ge 6) = \sum_{k=6}^{16} \binom{16}{k}(0.3)^k(0.7)^{16-k}. \] It is easier to compute this using the complement: \[ P(X \ge 6) = 1 - P(X \le 5). \] Hence \[ P(X \ge 6) = 1 - \sum_{k=0}^{5} \binom{16}{k}(0.3)^k(0.7)^{16-k}. \] Evaluating the terms, \[ P(X=0) = (0.7)^{16} \approx 0.003323 \] \[ P(X=1) = 16(0.3)(0.7)^{15} \approx 0.022790 \] \[ P(X=2) = \binom{16}{2}(0.3)^2(0.7)^{14} \approx 0.073251 \] \[ P(X=3) \approx 0.146502 \] \[ P(X=4) \approx 0.205941 \] \[ P(X=5) \approx 0.216238 \] Summing, \[ P(X \le 5) \approx 0.668045. \] Therefore \[ P(X \ge 6) = 1 - 0.668045 \approx 0.331955. \] \[ P(\text{win}) \approx 0.332. \]