A probabilistic problem, a probabilistic problem solved

The first office has a 9% chance of having a pen, the second office has a 9% chance of not having a pen (i.e., 3 people don't have a pen with a pen) is (1-3%) (1-3%), and the chance of having a pen is 1-(1-3%) (1-3%) (1-3%)

So there is a good chance that the first office will have a fountain pen.

There is a 9% chance that the first office will have a fountain pen

The chance that someone in the second office will not have a pen is 1-3%=97%, so the chance that someone in the second office will not have a pen is 97%*97%*97%=, that is, the chance that someone in the second office will have a pen to use.

There is a good chance that the first office will have a fountain pen.

The chance of not having a pen in the first office is 1-9%=91%.

The chance of not having a pen in the second office (i.e., if there is no pen, three people at the same time) is 97%x97%x97%=

The first one has a high chance of having a pen.

A person has a good chance of that.

In another office, the probability that everyone in the office will not have a pen is <

Office 1: p=1*

Office 2: p=>

So there is a high probability that someone from another office is there. Do you know how to calculate it yourself?

The leakage rate of the locust on the edge of the plastic plate pressed by the small round plate is general.

The probability that the small round plate is pressed against the apex of the plastic plate is.

4*( 2+1*2 Burn 2) 9 2=(2 +4) 81

First of all, each ball has 4 ways to put it, so the total way to put it has 4 * 4 * 4 = 64 kinds of we consider it from the negative side: there is a box with 2 balls, and the reverse side is that there is no box with 2 balls

1: All three balls are placed in the same box, and there are 4 ways to put them.

2: If there are no two of the three balls in the same box, then there are 4a3 ways to put it, so the probability of tails is (4+4a3) 64=7 16, then heads: the probability of having 2 balls in a box is 1-7 16=9 16

9/16

The first one is put casually.

The second one has the same probability as the first one 1 4, and the third one is placed in the other three 1 4 3 4 = 3 16

The second box is different from the first, the probability is 3 4, and the third is the same box as the first two with the probability of 2 43 4 2 4 = 6 16

The total probability is 3 16 + 6 16 = 9 16

How can the probability of 3 balls, 4 boxes, and 2 balls being together be more than 1 in 2.

Are the 3 balls in the question the same?

This belongs to the geometric generalization.

Establish a Cartesian coordinate system. The x-axis represents the moment when A arrives, and the y-axis represents the moment when B arrives. With 10 points as the origin, the value of any point in a square with a side length of 30 can represent the moment when A and B arrive (in this case, a square with a side length of 3).

The two met in 15 minutes at the following shade:

Then the probability of meeting is divided by the shadow area by the entire area, which is <>

To put it simply, if the probability of each event occurring is only proportional to the length (area or volume) of the area that makes up the event, then such a probabilistic model is called a geometric probabilistic model, or geometric generalization for short. For example, for a randomized trial, we understand each basic event as a random point taken from a specific geometric region, and each point in that region has the same chance of being taken; The occurrence of a random event is understood as taking exactly a point in a specified area within the region described above.

The area here can be line segments, flat figures, three-dimensional figures, etc. Randomized trials are treated in this way, known as geometric generalizations. Geometric generalizations, as opposed to classical generalizations, extend the concept of equal possible events from the finite to the infinite.

This concept has been introduced in junior high school mathematics in China. The main difference between classical generalizations and geometric generalizations is that geometric generalizations are another type of equal possible generalizations, and the difference between them and classical generalizations is that the results of the experiment are not finite, and it is easy to give examples of events with a probability of 0 that are not impossible events, and events with a probability of 1 are not inevitable events.

You can use a two-dimensional graph to understand this problem deeply, draw the conditional equation into the graph, and see what the difference between x and y is in each region.

Gain a deeper understanding of geometric generalizations.

Is the answer authoritative? I think the probability of two people meeting is 25%, and this value includes the fact that A and B can count the last time of waiting between 10 and 10:30, or the time they just arrived. In other words, the actual stay time of A and B at the meeting place is 9:

This is a geometric probability problem, and the coordinates of each point (x,y) that satisfy the condition are qualified. So divide the area of the eligible part by the entire area (i.e., the area of a square with a length and width of 30). It's the probability of seeing each other.

Paint takes the shaded part of the crossing.

Let the probability of the second term being passed be x.

The probability of passing the first and third terms is (x-1 8) and (x+1 8), according to the title, (1-x+1 8)*x

9 32 solution x1 = 3 8 (rounded), x2 = 3 4, x=3 4, so the probability of each item is 5 8, 6 8, 7 8 and the probability of passing all is 105 256

As shown in Fig. I drew it myself. I hope you don't laugh at me, the title says that a coin with a diameter equal to 2cm is thrown into this triangle, which means that the range of activity at the center of the circle is an equilateral triangle, and the area with at most one intersection point with the triangle is the blue part, for the probability problem of area type, the boundary of the area is drawn as a dotted line is the same, it is the same area.

So you divide the blue part by the total. Using the condition, we can find that the side length of the equilateral triangle in blue is 2 3

Divide the area by the large equilateral triangle and get 1 4.

It is assumed that the center of the coin circle has an equal probability of falling into a region of the same area anywhere in the plane.

Make a small equilateral triangle in each triangle grid, so that the distance between the side of the small equilateral triangle and the grid line is 1cm (as shown in the red triangle), when the center of the coin falls into the red triangle, the coin has no common point with the grid line. It is easy to find the side length of the red triangle is half of the side length of the grid triangle, so the area of the red triangle is 1 4 of the area of the grid triangle, so the probability is found.

p = (sum of red triangle areas) (sum of grid triangle areas) = 1 4

Solution: Basis: The center of the circle is determined by the position of the circle, and the radius is determined by the size of the circle, as shown in the figure: only when the center of the circle falls within the equilateral def, there is no common point between the side of the coin with a diameter equal to 2cm and the side of the equilateral abc.

From the figure, it is easy to obtain the side length of the equilateral def equal to 2 times the root number 3cm.

So, s abc=[(root number 3) 4]*(4 times root number 3) = 12 times root number 3s def=[(root number 3) 4]*(2 times root number 3) = 3 times root number 3 So, the probability that the coin will not have a common point with the side of the equilateral triangle after it falls:

p=s def s abc = 3 times root number 3 12 times root number 3 = 1 4

Solution: x=3, it means that you didn't get the correct key the first two times. So you can only take out any two of the four that are left with the correct key removed.

c(4,2) means 4 handfuls that cannot be opened, and 2 handfuls are taken out.