David J. Weiss
Solutions for Probability Problem Set 1
- Each die has to not have a "four". The probability that a single die does not have a "four" is 5/6, so that probability that all four do not (an intersection) is 5/6 x 5/6 x 5/6 x 5/6.
- Option (b) is better. The probability of getting a club on the first draw is ¼. You don’t even need to calculate the probability for part (b), since it calls for a union. The probability must be ¼ + ¼ - the probability that both will occur, which is obviously less than the probability that either will occur.
- The chance that you will have a wild ride is ¼, since you are equally likely to draw any of the four horses. The chance that one of your group will ride Hitler is ¼ + ¼ + ¼ = ¾ ; it’s just a union, since only one person gets assigned to each horse. If there were four of you, someone would be certain to ride Hitler, but your chance would still be ¼. There is not enough information to assess your chances of survival, since we don’t know how good a rider you are.
- To get a sum of three, each of the three dice must yield a "one". This is an intersection. The probability is 1/6 x 1/6 x 1/6 = 1/216. You can envision a three-dimensional die chart, which will contain 6 x 6 x 6 = 216 cells, of which only one will have three "ones." For all three dice to have three or more spots showing, calculate similarly. The probability for one die to have a three or more is 4/6, so the answer is 4/6 x 4/6 x 4/6 = 64/216 or 8/27.
- Consider all of the possible sets of four cards. There are 52 x 51 x 50 x 49 sets, all of which are equally likely. To get three kings, you must get KKKx, KKxK, KxKK, or xKKK, where x stands for any of the other twelve ranks. The number of sets of KKKx is 4 x 3 x 2 x 48 (48 = twelve ranks x four suits). The number of sets of KKxK is 4 x 3 x 48 x 2. Proceeding similarly for the other two cases, we find the number of sets of three kings to be 4(4 x 3 x 2 x 48). So the answer is 4608/6497400 = .0007.