1/2. This problem is actually a bit tricky. The marbles are in the bags. When he picked one white marble, the bag is certain and fixed because the question asks the remaining marble in the same bag. You cannot mix the marbles in two bags for the probability. There are only two bags that are relevant, the remaining marble is either white (the two-white-marble bag) or black (the one-white one-black bag). So the answer is 1/2.
from chatgpt: To solve this problem, we can use conditional probability. Let''s calculate the probability step by step. There are three possible bags that Matt could have chosen: A, B, or C. We''ll denote the event of choosing a particular bag as A, B, or C. P(A) = Probability of choosing Bag A = 1/3 (since there are three bags and they are equally likely to be chosen) P(B) = Probability of choosing Bag B = 1/3 P(C) = Probability of choosing Bag C = 1/3 Now, we want to find the probability that the remaining marble from the same bag is also white, given that he picked a white marble. Let''s denote this event as W. We can calculate P(W|A), P(W|B), and P(W|C) as follows: P(W|A) = Probability of drawing a white marble from Bag A and the remaining marble being white = 1 (since Bag A contains only white marbles) P(W|B) = Probability of drawing a white marble from Bag B and the remaining marble being white = 0 (since Bag B contains only black marbles) P(W|C) = Probability of drawing a white marble from Bag C and the remaining marble being white = 1/2 (since Bag C contains one white and one black marble) Now, we can use Bayes'' theorem to calculate the probability of the remaining marble being white, given that he picked a white marble: P(W) = Probability of drawing a white marble = P(A) * P(W|A) + P(B) * P(W|B) + P(C) * P(W|C) = (1/3) * 1 + (1/3) * 0 + (1/3) * (1/2) = 1/3 + 0 + 1/6 = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2
To find the probability that the remaining marble from the same bag is also white, we can use conditional probability. Let''s define the events: A: Bag A is chosen. B: Bag B is chosen. C: Bag C is chosen. W: A white marble is drawn. We want to find P(A|W), the probability that Bag A is chosen given that a white marble is drawn. Using Bayes'' theorem, we have: P(A|W) = (P(A) * P(W|A)) / P(W) We need to calculate three probabilities: P(A), P(W|A), and P(W). P(A) is the probability of choosing Bag A initially. Since there are three bags and we pick one randomly, P(A) = 1/3. P(W|A) is the probability of drawing a white marble given that Bag A is chosen. Bag A contains two white marbles, so the probability of drawing a white marble from Bag A is 2/2 = 1. P(W) is the probability of drawing a white marble regardless of the bag. To calculate this, we need to consider all possible scenarios in which a white marble is drawn. There are two scenarios: drawing a white marble from Bag A and drawing a white marble from Bag C. P(W) = P(A) * P(W|A) + P(C) * P(W|C) P(C) is the probability of choosing Bag C initially. Similarly, since there are three bags and we pick one randomly, P(C) = 1/3. P(W|C) is the probability of drawing a white marble given that Bag C is chosen. Bag C contains one white marble and one black marble, so the probability of drawing a white marble from Bag C is 1/2. P(W) = (1/3) * 1 + (1/3) * (1/2) = 1/3 + 1/6 = 1/2 Now, we can calculate P(A|W) using Bayes'' theorem: P(A|W) = (P(A) * P(W|A)) / P(W) = (1/3 * 1) / (1/2) = 2/3 Therefore, the probability that the remaining marble from the same bag is also white, given that a white marble is drawn, is 2/3 or approximately 0.67.
P(remaining marble is also white | picked a bag and picked a white ball) = P (second ball is white, picked a bag and picked a white ball) / P(picked a bag and picked a white ball) = (P(picked the two-whites bag) * 1 + P(picked the 1-black-1-white bag) * 0) / (P(picked the two-whites bag) * 1 + P(picked the 1-black-1-white bag) * 1/2 ) = 1/3 / (1/3+1/3*1/2) = 2/3
Correct answer - 50% - is commonly considered wrong, but really - past events have no influence on future probabilities at all (You cannot count "You have picked bag #1" twice. You just either did it or not). Right now You have one white marble in your hand and knowledge, that another one is either black or white depending on which bag (out of two) you selected. Challenge: You essentially just start by picking a marble--and there are three white marbles. Two of those white marbles are in the same bag as another white marble; so, given that you''''''''ve picked one of the three white marbles, 2/3 of the time its pair matches--right? Rebuttal: You''''''''re thinking about this wrong - since you picked a white marble, you know you have chosen from either bag #1 or bag #3. Now, there are only two possible outcomes - either you have bag #1, in which case the other marble is white, or bag #3 in which case it''''''''s black. And the probability is 50% that the other marble in your chosen bag is white. https://en.m.wikibooks.org/wiki/Puzzles/Statistical_puzzles/3_Bags_of_Marbles/Solution TWBear 发表于 2023-05-17 16:16
回复 64楼TWBear的帖子 "Now, there are only two possible outcomes - either you have bag #1, in which case the other marble is white, or bag #3 in which case it''''s black. And the probability is 50% that the other marble in your chosen bag is white." 这段是混淆概念，" only two possible outcomes" doesn't mean the probabilities of these two possible outcomes are equal.
2/3. Bayes’s therorem. 在抽到白球的前提下，袋子里另一个也是白球。 第一个球是白球的概率是1/3+ 1/3*1/2=1/2。 抽到 Bag A的概率是1/3， 这个前提下抽到白球概率 1。 抽到 bag C概率1/3， 这个前提下抽到白球概率 1/2。
一个袋子两个白球的概率是1/3。 所以 1/3 除以1/2。
这题我孩子五年级的时候，补习班里就开始有了。不知道啥level
三个白球编号1.2.3，1.2在一个袋子里，3和黑球在一起 抽到每个白球的概率均等都是1/3，只有抽到1.2时候另外一个球是白球。
所以概率是1/3+1/3是2/3
。。。
一共6个球，3黑3 白，第一个球是白的概率是1/2。
抽到第一个白球后，两黑球的袋自动被过滤掉。 剩下的球必然是两白一黑，只有3种可能的组合。
1号白 2号白 2号白 1号白 3号白 3号黑
因此抽到白球的概率是2/3。 选择是按照球，而不是按袋子。选1/2就是被袋子干扰了
正确答案是2/3. define two events, Event A= W, event B=bag A. 接下来贝叶斯。 P(B|A)=P(B^A)/P(A)=P(B)/P(A)=(1/3)/(1/2)=2/3
上面的想法是不对的。正确的小学生能理解的解释是
袋子A两个黑marble BB 袋子B一个白一个黑marble WB 袋子C两个白marble WW
随便选一个袋子先后取出两个marble的所有情况如下 B，B B，B ———— W，B B，W ———— W，W W，W
第一个是白色的情况下，第二个也是白色的概率是2/3
在第一球是白球的条件下，那个全黑球的袋子不在这个条件下，那么只有两个袋子才具备这个条件。在这个条件下，第二球是白球的只有一个袋子。所以1/2
。。。
我咋觉得是(1/3)/(2/3) = 1/2……
如果拿第一个球以后没看， 那拿第二个球是白球的概率就是1/2。如果看了并知道是白球， 这时候概率就上升到2/3了。多看一眼 概率就变了 有意思。
答案是1/2。 同样的例子，某人生男生女的概率各是二分之一，已知他家有2个孩子，其中一个是女孩，那另一个也是女孩的概率是多少？答案是三分之一。 已知他家老二是女孩，那老大是女孩的概率是多少。答案是二分之一。
我写了程序模拟了一下，原来是我错了，对不住给我点赞的各位亲了。正确答案应该是2/3。
我知道错在哪里了，把那个装两个白色石头的袋子称作A袋，一黑一白是B袋，然后应该把A袋中两块石头编号分别为1号2号，所以就可以分成三种情况： 第一种：第一次抽到A袋1号，第二次抽到A袋2号 第二种：第一次抽到A袋2号，第二次抽到A袋1号 第三种：第一次抽到B袋白石头，第二次是黑石头
所以三次中有两次都是在同一个袋子中抽到白色，所以应该2/3
不是。是在第一个球是白球的前提下拿到全白一袋的概率。是条件概率
如果你有天看到某人带个女孩看电影， 他告诉你那是他女儿。 在这个条件下， 他有两个女儿的概率是二分之一
贝叶斯，条件几率啥的就算了，真有能听懂的小学生，恐怕是高斯No. 2了吧。
这应该是conditional probability. 不用从头算。而且小学的题，不会这么复杂。
这是被球干扰了。不该想剩下的球是两白一黑，应该想剩下的就是两个袋子
“也就是从3个球里拿一个白球的几率是多少。” 题目不是这样问的。下一个球只有两个可能：一是白球（说明袋子里原来装的是两个白球），二是黑球（袋子里是一黑一白）。1/2。不可能有第三个颜色。跟三个球选没关系。
如果袋子不编码，那第二次还是随机从三个袋子选，那概率就是1/3。
概率问题 的确是检验智商的 最好工具
这个是正解。虽然bayes thereom 对小学生来说有点超纲了
可是那个装两个白球的袋子抽取到不同的白球的次序要考虑。我开始也把两个白球的次序忽略了
把题目改成 A 是 100个白球 B是 99个白球 1个黑球 C是100个黑球 拿了一个球是白球 问 A和 B 概率是多少
这个题目应该考虑的是球，而不是袋子，我开始也觉得是1/2，就是把同一个袋子两次抽取不同的白球看作一次。那两个球虽然同为白球，但是确是不同的两个球，所以两个白球要编号然后再抽取
To solve this problem, we can use conditional probability. Let''s calculate the probability step by step. There are three possible bags that Matt could have chosen: A, B, or C. We''ll denote the event of choosing a particular bag as A, B, or C. P(A) = Probability of choosing Bag A = 1/3 (since there are three bags and they are equally likely to be chosen) P(B) = Probability of choosing Bag B = 1/3 P(C) = Probability of choosing Bag C = 1/3 Now, we want to find the probability that the remaining marble from the same bag is also white, given that he picked a white marble. Let''s denote this event as W. We can calculate P(W|A), P(W|B), and P(W|C) as follows: P(W|A) = Probability of drawing a white marble from Bag A and the remaining marble being white = 1 (since Bag A contains only white marbles) P(W|B) = Probability of drawing a white marble from Bag B and the remaining marble being white = 0 (since Bag B contains only black marbles) P(W|C) = Probability of drawing a white marble from Bag C and the remaining marble being white = 1/2 (since Bag C contains one white and one black marble)
Now, we can use Bayes'' theorem to calculate the probability of the remaining marble being white, given that he picked a white marble:
P(W) = Probability of drawing a white marble = P(A) * P(W|A) + P(B) * P(W|B) + P(C) * P(W|C) = (1/3) * 1 + (1/3) * 0 + (1/3) * (1/2) = 1/3 + 0 + 1/6 = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2
我也问了chatgpt. 答案很长， 但跟你的不一样是2/3：
To find the probability that the remaining marble from the same bag is also white, we can use conditional probability. Let''s define the events: A: Bag A is chosen. B: Bag B is chosen. C: Bag C is chosen. W: A white marble is drawn. We want to find P(A|W), the probability that Bag A is chosen given that a white marble is drawn. Using Bayes'' theorem, we have: P(A|W) = (P(A) * P(W|A)) / P(W) We need to calculate three probabilities: P(A), P(W|A), and P(W). P(A) is the probability of choosing Bag A initially. Since there are three bags and we pick one randomly, P(A) = 1/3. P(W|A) is the probability of drawing a white marble given that Bag A is chosen. Bag A contains two white marbles, so the probability of drawing a white marble from Bag A is 2/2 = 1. P(W) is the probability of drawing a white marble regardless of the bag. To calculate this, we need to consider all possible scenarios in which a white marble is drawn. There are two scenarios: drawing a white marble from Bag A and drawing a white marble from Bag C. P(W) = P(A) * P(W|A) + P(C) * P(W|C) P(C) is the probability of choosing Bag C initially. Similarly, since there are three bags and we pick one randomly, P(C) = 1/3. P(W|C) is the probability of drawing a white marble given that Bag C is chosen. Bag C contains one white marble and one black marble, so the probability of drawing a white marble from Bag C is 1/2. P(W) = (1/3) * 1 + (1/3) * (1/2) = 1/3 + 1/6 = 1/2 Now, we can calculate P(A|W) using Bayes'' theorem: P(A|W) = (P(A) * P(W|A)) / P(W) = (1/3 * 1) / (1/2) = 2/3 Therefore, the probability that the remaining marble from the same bag is also white, given that a white marble is drawn, is 2/3 or approximately 0.67.
你是输入原题吗？我是输入exact question，然后加了一个“5年级数学”。
不能不考虑袋子吧，第二次抽是在同一个袋子中抽
在这个袋子中，你先抽中1号球再抽中2号球，和你先抽中2号球再抽中1号球，这就算两次在同一个袋子里抽中白球的机会，虽然两次机会都是针对同一个袋子。你不能因为这两次机会都是同一个袋子你就只算成一次机会。
对，这题最传统的是用bayes''''''''s therorem去解。为什么在bayes''''''''s therorem给出解答是2/3的基础上，还有人坚持1/2？类似于小学题目用方程求出解了，可以反过来推一个小学理解的过程但不能改变答案。
对于小学生，这题其实等同于问 三袋子，一袋两黑，一袋一白一黑，一袋两白 先取一个袋子拿手上 然后在袋子里摸到一个白球（这里跟原题一样） 问题，手上那个袋子是两白的概率是多少（转换一下问题，因为摸到一个白的剩下一个白的=手上袋子是两百）
不换袋子，不摸两次，拿出白球之后就直接猜手上的那个袋子是两白的概率。 这样简化了题目，只模一次了，但概率还是等于要求的答案
在摸到一个白球的概率下 手上袋子是两白的概率是2/3。
（摸出来的球可能是白球中的任何一个，每个白球有1/3的概率，所以手上袋子是两白的概率是2/3，手上袋子是一白一黑的概率是1/3，手上袋子是两黑的概率是0）
======= 至于chatGPT为什么会有时给1/2，有时给2/3 给2/3的那个是对的。 给1/2的那个，它的w只是求除了摸到一个白球的概率，不表示条件下事件，否则w|c毫无意义。表示条件概率a|w下来表示。
=== 这题在math is fun里面就有 https://www.mathsisfun.com/puzzles/bags-of-marbles-solution.html
它跟山羊换门monty hall problem其实是一样的 认为不用换门的都认为每个剩下门有羊的概率都是1/2（跟认为一黑一白袋子，和两白袋子被抽中白球概率是均等的一样）
如果不能理解为什么不是1/2，而是2/3，可以YouTube monty hall problem 想清楚了门为什么不是1/2，也就懂了袋子了。
可以这么想，6个球，摸到哪个的可能性都一样。 那么given摸到白球，摸到每个白球的可能性都是1/3。 那么，在A袋（两白球）的可能性是2/3。
有没有一种可能，那就是正确答案，楼主都说了这是小学题
抄的这段逻辑非常hand-waving，无法用数学逻辑写清楚的是很可能有谬误的。如果真的对，肯定可以有清晰的数学表达式的。
我觉得这里wikibooks上的解释是错误的。这两个袋子的概率并不是相等的，所以不是50%
对小孩来说，这题可能比monty hall problem更容易理解2/3从哪里来的。
"Now, there are only two possible outcomes - either you have bag #1, in which case the other marble is white, or bag #3 in which case it''''s black. And the probability is 50% that the other marble in your chosen bag is white." 这段是混淆概念，" only two possible outcomes" doesn't mean the probabilities of these two possible outcomes are equal.
Simulation 的design如果就是错的，结果没有说服力
过来人都知道美国的数学并不是传说中的那么简单
美国小学数学的范围还挺广的，概率统计还有finance方面的知识都会涉及到，只不过不是很深罢了
初中学三角函数，高中学微积分也是很正常的
就是跟着学校学而已，跟卷不卷还真没关系
2/3是对的，我学文科的，都看懂了。
前面那个朋友举得例子挺好的，我就是通过他的那个例子想明白的。这道题可以改一下。
假设有三个袋子，每个袋子里都是1万个球，一个袋子A里有1万个黑球，一个袋子B里有1个白球和9999个黑球，另一个袋子C里有10000个白球。
现在小朋友随机的拿起一个袋子，从袋子里面随机的拿出了一个球，发现是个白球，问这个袋子剩下的所有的球都是白球的概率是多少？
我保证你不会回答1/2。尽管只有两个袋子B和C里有白球。
这么一说确实不是1/2哈，还是要取极端例子才意识到问题，概率真是一门深奥的学问
不可教也
我尝试再解释和挽救一下。
哎，首先你能够理解原题不需要取两次，只需要取一次袋子，然后从中取一个球，取出了之后是白球就，问手上这个袋子是全白的概率是多少？
如果你能够理解原题可以简化成这种取一次的情况，那么就可以理解楼上有人的举例了
有三个外表看不出差异的袋子，一袋1万个黑球，一袋1万个白球，
（a）剩下的一袋装1个白球9999个黑球，三个袋子中随机取一个袋子，然后随机掏一个球，看到是白球， 问手中袋子是全白的概率是多少？
（b）剩下的一袋装2个白球9998个黑球，三个袋子中随机取一个袋子，然后随机掏一个球，看到是白球， 问手中袋子是全白的概率是多少？
（c）剩下的一袋装3个白球9997个黑球，三个袋子中随机取一个袋子，然后随机掏一个球，看到是白球， 问手中袋子是全白的概率是多少？ 。。。。 （...）剩下的一袋装n个白球1000-n个黑球，三个袋子中随机取一个袋子，然后随机掏一个球，看到是白球， 问手中袋子是全白的概率是多少？ （1<n<9999）
针对以上，你可以写出一个所求概率关于n的函数。
一半一半只是以上的特例。楼上设计万中取1，实际上是为了别人能够理解以上更generalize的情况。
================
我同意楼上有人提到的，intuition不行或者觉得有争议的，应该用公式取求P（A|W）（就是chatGPT第二个的的结果），如果学过条件概率的话，这题不难。一步步算出结果再来理解2/3的由来。 或者simulation1000次看看，也能得出差不多0.66的结果。
经典概率题在现有公式的情况下，基本都可以回归公式解。 先整明白正确结果是什么，然后再来理解。
数学的公式可以简洁表达所有的意思，各种intuition回归到数学才行。
第一次拿到白球的概率不同，我理解。 可是题目不是问，拿到第一个白球之后的概率么？之前的事件应该结束了。 这跟问拿到一个白球后再拿一个白球的概率，应该有区别吧