⧼exchistory⧽
63 exercise(s) shown, 0 hidden
BBy Bot
Jun 09'24
Prove that if [math]B_1,B_2, \ldots,B_n[/math] are mutually disjoint and collectively exhaustive, and if [math]A[/math] attracts some [math]B_i[/math], then [math]A[/math] must repel some [math]B_j[/math].
BBy Bot
Jun 09'24
- Suppose that you are looking in your desk for a letter from some time ago. Your desk has eight drawers, and you assess the probability that it is in any particular drawer is 10% (so there is a 20% chance that it is not in the desk at all). Suppose now that you start searching systematically through your desk, one drawer at a time. In addition, suppose that you have not found the letter in the first [math]i[/math] drawers, where [math]0 \le i \le 7[/math]. Let [math]p_i[/math] denote the probability that the letter will be found in the next drawer, and let [math]q_i[/math] denote the probability that the letter will be found in some subsequent drawer (both [math]p_i[/math] and [math]q_i[/math] are conditional probabilities, since they are based upon the assumption that the letter is not in the first [math]i[/math] drawers). Show that the [math]p_i[/math]'s increase and the [math]q_i[/math]'s decrease. (This problem is from Falk et al.[Notes 1])
- The following data appeared in an article in the Wall Street Journal.[Notes 2] For the ages 20, 30, 40, 50, and 60, the probability of a woman in the U.S.\ developing cancer in the next ten years is 0.5%, 1.2%, 3.2%, 6.4%, and 10.8%, respectively. At the same set of ages, the probability of a woman in the U.S. eventually developing cancer is 39.6%, 39.5%, 39.1%, 37.5%, and 34.2%, respectively. Do you think that the problem in part (a) gives an explanation for these data?
Notes
- R. Falk, A. Lipson, and C. Konold, “The ups and downs of the hope function in a fruitless search,” in Subjective Probability, G. Wright and P. Ayton, (eds.) (Chichester: Wiley, 1994), pgs. 353-377.
- C. Crossen, “Fright by the numbers: Alarming disease data are frequently flawed,” Wall Street Journal, 11 April 1996, p. B1.
BBy Bot
Jun 09'24
Here are two variations of the Monty Hall problem that are discussed by Granberg.[Notes 1]
- Suppose that everything is the same except that Monty forgot to find out in advance which door has the car behind it. In the spirit of “the show must go on,” he makes a guess at which of the two doors to open and gets lucky, opening a door behind which stands a goat. Now should the contestant switch?
- You have observed the show for a long time and found that the car is put behind door A 45% of the time, behind door B 40% of the time and behind door C 15% of the time. Assume that everything else about the show is the same. Again you pick door A. Monty opens a door with a goat and offers to let you switch. Should you? Suppose you knew in advance that Monty was going to give you a chance to switch. Should you have initially chosen door A?
Notes