probability - 随机选择大理石的可能性(Probability of randomly selecting marbles [duplicate])

Question

This question already has an answer here:

(这个问题已经在这里有了答案：)

• Probability: marbles in bag question 2 answers

  (概率：袋装弹珠问题 2的答案)

if you have 5 black marbles and 1 white marble in one bag, when randomly selecting 4 out of the 6 marbles at once without replacement, what is the probability of the white marble being included in the 4 marbles selected

(如果您在一个袋子中有5个黑色大理石和1个白色大理石，则一次从6个大理石中随机选择4个而不进行更换时，选择的4个大理石中包含白色大理石的概率是多少)

  ask by Jaysen Cong translate from so

Answer 1

Imagine the marbles were labeled with unique names: A, B, C, D, E and F, where F is the white marble.

(想象一下，大理石上都标有唯一的名称：A，B，C，D，E和F，其中F是白色大理石。)

There are 6 x 5 x 4 x 3 ways to choose 4 of these 6 marbles.

(有6 x 5 x 4 x 3种方法可以选择这6种大理石中的4种。)

Only some of these have the white marble;

(其中只有一些带有白色大理石。)

5 x 4 x 3 x 2 do not.

(5 x 4 x 3 x 2不。)

So the probability of getting the white marble is 1 - 5x4x3x2 / 6x5x4x3 = 1 - 2/6 = 2/3.

(因此，获得白色大理石的概率为1-5x4x3x2 / 6x5x4x3 = 1-2 / 6 = 2/3。)

Using the same argument we find the probability of getting a white when drawing 2 marbles is 1/3.

(使用相同的参数，我们发现绘制2个大理石时获得白色的概率为1/3。)

This makes sense since it's the complement of our experiment.

(这是有道理的，因为这是我们实验的补充。)

Also, drawing 6 gives us 3/3 = 1 and drawing 0 gives us 0/3 = 0. Because drawing the white in one draw is 1/6, drawing the white in five draws must be 5/6.

(同样，图6给我们3/3 = 1，图0给我们0/3 =0。由于在一次绘制中绘制白色为1/6，所以在五次绘制中绘制白色必须为5/6。)

By symmetry, drawing in three draws must be the same as its complement, 3/6 = 1/2.

(通过对称性，三个平局中的平局必须与其补数相同，即3/6 = 1/2。)

In summary, the probability of drawing the white marble when drawing D of the 6 marbles is D/6 for 0 <= D <= 6.

(总而言之，当0 <= D <= 6时，绘制6个大理石中的D时绘制白色大理石的概率为D / 6。)