David J. Weiss
Solutions for Probability Problem Set 2
1. 3/6 = ½. Using the formula for conditional probability, we find P(odd sum½”four” on first) = P(odd sum Ç “four” on first)/ P(“four” on first) = (3/36)/(1/6) = 18/36, which is also equal to ½.
2. 3/51, 4/51.
3. The second and third bills must both be fives. Treat this as an urn problem. On the second draw, there are six bills in your pocket, four of which are fives. Then there are five bills, three of which are fives. (4·3)/(6·5) = 12/30.
4. We must get the intersection of two fours and two non-fours, in any order. The probability of FFxx is 1/6 x 1/6 x 5/6 x 5/6 = 25/1296. The possible orders are FFxx, FxFx, FxxF, xFFx, xFxF, xxFF. All orders have the same probability, so the answer is given by 6 (for the six possible orders) x (25/ 1296) = 150/1296 = .116.
5. The target is hit if either shooter hits it. This is a union, so we want P(father hits target) + P (son hits target) – P(both hit target). We can reasonably assume the shots to be independent, so the probability of the intersection is the product of the probabilities. The solution is ¾ + 4/7 – (3/4 x 4/7) = .893.
6. Picking the correct key is independent of whether the door is locked. I’ll get in if the door is unlocked (p = ½) no matter which key I pick. If it is locked (p = ½), then I’ll have to pick the correct key (p = 1/3). So the answer is ½ + (½ x 1/3) = 2/3.