Choose 3 out of 9 signs, and * * has c (9,3) = 84 methods.

All three flags are red, only 1

So the probability is 1/84.

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My solution to the landlord:

"* * * * There are 3 3 = 27 species, and the probability that three faces are red = 1/27"

The landlord probably wants to solve the problem with a repeatable arrangement.

There are three ways to red, yellow and blue each time, ***3 times.

The difference between repeatable arrangement and permutation is that repeatable arrangement is repeatable, that is, I take a flag as the first side, and I can take it as the second side next time. This means "repeatable", but repeated selection is not allowed in ordinary arrangements.

In the arrangement and combination, we often encounter the situation of touching the ball, which is often divided into "repeatable" and "irreplaceable", which is the difference between repeatable and unrepeatable.

In addition, the difference between permutation and combination is that permutation emphasizes order, while combination does not need order. This problem needs three sides, not to mention the order, and can be solved in combination.

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To supplement your statement:

What you said is a bit complicated. Another condition of the repeatable permutation problem is that each selection is an independent sample point. For a topic like this, the number of flags to be selected is limited, so only one fixed flag can be selected as the sample point, not red, blue and yellow (in fact, the probability of blue, yellow and red is not necessarily equal every time, but the probability of each sample point must be equal). There is no way to solve the problem without telling the number of flags of each color.

In addition, there is another kind of topic in combinatorial mathematics, that is, independent repeated experiments. For example, the probability of getting the red flag every time is 1/3, so the probability of getting the red flag three times in a row is (1/3)? .

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"Color three rectangles randomly with three different colors, and each rectangle is painted with only one color. Find the probability that all three rectangles have the same color? "

This problem is a typical repeatable permutation.

Because it is randomly colored, the probability that a rectangle is colored by any of the three colors is the same, and a coloring method can be regarded as an independent sample point.

At the same time, the first rectangle is painted with a certain color, and the second rectangle can also be painted with the same color, and so on.

So, one * * * has three? A coloring method.

All three rectangles have the same color, and * * * has three coloring methods.

The probability is 3/3? = 1/9

The fundamental difference between this question and the first two questions lies in whether it can be repeated.

For example, the red, yellow and blue flags are X, Y, z Y and Z respectively, so the probability of taking the red flag is x/(x+y+z). If there are only x- 1 red flag for the first time, the probability of the second red flag becomes (X- 1)/(X+Y+Z-66.

Painting is different. No matter what color you painted the previous times, the probability of painting red is always 1/3.

You got it?

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What is the "experiment" of this problem and what is the basic event?

Then, for different solutions, the selected sample space is not necessarily the same.

According to my solution:

An experiment is to choose three flags from nine flags, and this process is an experiment.

Basic events represent a fixed selection method, such as 1, 2, 3, 4, 6, 9 and so on. However, it should be noted that my solution is not sequential. For example, selecting 1, 2,3 root and selecting 3,2, 1 root are the same basic event.

According to an experiment and basic event I defined, we can find the sample space. Because it can't be repeated and doesn't pay attention to order, combination can be used as the calculation method. So the size of the sample space is c (9,3)

The required number of sample points is 1 (because there is no order, it is the same event to take three colors of red).