4 ms·
I don't think you are calculating the same thing. Your calculations are that an assult will happen, whereas belorn was calculating for a specific woman, whether
by Flenser 13y ago
I don't think you are calculating the same thing. Your calculations are that an assult will happen, whereas belorn was calculating for a specific woman, whether she would be the victim.
- yummyfajitas 13y agoI calculated the probability of an individual human being assaulted.
- antimagic 13y agoLet's use your equation with k = 20, alpha = 0.5 and beta = 0.1. This gives the probability of a person being assaulted as => 1 - pow(1 - alpha, kbeta(1 - alpha)) => 1 - pow(0.5, 20 * 0.1 * 0.5) => 1 - pow(0.5, 1) => 0.5 There's a 50% chance that a person will be assaulted. Divide by the number of women (10) gives a probability of 0.05 that a particular woman will be assaulted. Now let's do the same thing, but changing alpha to 0.05 1 - pow(0.95, 20 * 0.1 * 0.95) => 1 - pow(0.95, 1.9) => 0.09 However as there is only one woman , that 0.09 probability is all her. In other words her odds of being assaulted have just about doubled.
- datr 13y agoYou've calculated the probability that _some_ individual will be assaulted. The probability that a given female will be assaulted though would be (I think): P = 1 - \sum_{m=0}^K \choose{m}{K} (1-beta)^m (1-alpha)^m alpha^{K-m}