You fire 3 guns at a target. the first has a 20% probability of hitting, the second a 15% probability of hitting and the 3rd an 10% chance of hitting. what is sample space for the number of hits? what is the expected number of hits? what is the variance?

Sample space for hits: {0, 1, 2, 3} Expected hits: 0.45 Variance: 0.3775 We have 3 guns, each of which can either hit or miss. The sample space for the number of hits is {0, 1, 2, 3} The expected number of hits is 0.20 + 0.15 + 0.10 = 0.45 hits For the variance, let's calculate the probability of 0, 1, 2, or 3 hits. 0 hits: 0.80 * 0.85 * 0.90 = 0.612 1 hit: 0.8 * 0.85 * 0.1 + 0.8 * 0.15 * 0.9 + 0.2 * 0.85 * 0.9 = 0.068 + 0.108 + 0.153 = 0.329 2 hits: 0.8 * 0.15 * 0.1 + 0.2 * 0.85 * 0.1 + 0.2 * 0.15 * 0.9 = 0.012 + 0.017 + 0.027 = 0.056 3 hits: 0.2 * 0.15 * 0.1 = 0.003 Since we have 4 discrete outcomes, the variance is simply the sum of the probability of the event multiplied by the square of the difference between the value of the event minus the mean. So 0.612 * (0 - 0.45)^2 + 0.329 * (1 - 0.45)^2 + 0.056 * (2 - 0.45)^2 + 0.003 * (3 - 0.45)^2 = 0.612 * (-0.45)^2 + 0.329 * (0.55)^2 + 0.056 * (1.55)^2 + 0.003 * (2.55)^2 = 0.612 * 0.2025 + 0.329 * 0.3025 + 0.056 * 2.4025 + 0.003 * 6.5025 = 0.12393 + 0.0995225 + 0.13454 + 0.0195075 = 0.3775 So the variance is 0.3775 Note: I could have eliminated the step consolidating the probabilities of getting 0, 1, 2, or 3 hits and instead went directly into the calculation of the variance using the probability of each of the 8 possible events. But in doing so that would have made the calculations twice as large and still would have gotten the same effect. Effectively for the cases of 1 and 2 hits, I calculated (p1 + p2 + p3)*(1 - 0.45)^2 instead of p1*(1 - 0.45)^2 + p2*(1 - 0.45)^2 + p3*(1 - 0.45)^2 which the distributive property allows us and will give the same result.