Question 1: As shown in the figure, the rectangle S represents the whole event, the red ellipse A represents a special event in S, the blue ellipse B represents another event, and AB is an independent event. The purple part is the intersection of event a and event B. Then the blank part P(x)=P(s)-P(a)-P(b)+P(ab).

First of all, the landlord has a habit of asking for this S first, whether it is useful or not. Then analyze whether it conforms to this picture. Obviously, the first question needs.

In ...

Event s is "two rows of six people standing in front and back" P(s)=A(3/6)*A(3/3).

Event A is "A stands in the front row" P (a) = C (1/3) * A (2/5) * A (3/3).

Event b is "b standing in the back row" p (b) = p (a)

Then event AB is "A is standing in the front row and bilibili is in the back row" P (AB) = C (1/3) * A (2/4) * C (1/3) * A (2/2).

Substitute it into the previous formula and you will get it. As a result, everything upstairs has been solved, so I won't go into details. Although the method is more troublesome, it is more common. The analysis of total events and independent sub-events is the key to permutation, combination and probability.

The second question: This question is also to analyze the total event S first, where S is "12 cars can line up in the parking lot. After stopping eight different cars, there are four vacant seats. The special activity is "Four vacant seats together".

Let's take a look at the practice upstairs after this analysis.

Also, multiple-choice questions should have "blinding" skills, and don't waste too much time, but it doesn't mean "blinding". Each problem should have its own unique methods and tricks. If it's really not enough, do it with the general plan, and don't waste time looking for smart ones. This is not worth the loss. Big problems should be organized. . The method I introduced is not suitable for multiple-choice questions in the exam, but just tells you how to do problem analysis and so on. . I hope it helps you.