An interesting probability question 40

Because the probability behind the three doors is the same, each is 1 3, even if you tell me that there is nothing in the two doors, the door I chose is also 1 3, which shows why ** is the principle of fairness. "Suppose you choose a door A, and now I'm telling you that B has no prize"This sentence illustrates a problem, it is that A is chosen, and B has no prize, so the probability of A is 1 3

Everyone agrees that wow, the answer is 1 3

This is a conditional probability problem (the university mathematics department has studied probability theory and mathematical statistics, and after studying this book, with the idea of analysis, this topic is very simple. ）

I'm going to do it in elementary ways. First of all, the probability of classifying 1 and choosing a is 1 3, at this time you tell me that b does not, then the prize is in a, c, and the probability of being in c is 1 3 * 1 2 = 1 If I don't choose A, then the probability is 2 3, there is no condition, the prize is in a, b, c, then the probability of being in c at this time is 2 3 * 1 3 = 2 9.

Therefore, the probability of the prize in c is 1 6 + 2 9 = 7 18 positive solution, and this answer is also obtained by the advanced method.

2 3, the probability of each door at the beginning is 1 3, you choose a then the probability of a is 1 3, the probability in bc is 2 3, now exclude b, the probability of a is still 1 3, the probability of c will be 2 3

I think it's 1 2. Excluding b, only A and C are possible, and they are equal to each other, so each is 1 2.

This topic is inherently ambiguous, if the premise is: "If you choose a door A, and now I tell you that B has no prize", then the prize is in the middle of A and C, and the probability of choosing A and C is 1 2!

I despise LS pretending to be profound, and I won't be able to read the title, hum!

1 6, the probability in front is 1 3, when you say b is not, it is 1 2, and multiplying it is 1 6

2 3 can be listed b is 1 3 This is a question studied by American scientists.

2 3 It seems that I have seen it in my teachings.

The probability problem is this: Suppose there is a disease with an incidence of 1 1000 and the accuracy of the test results is quite high. If you get the disease and the test result is positive, the accuracy rate is ; If you don't get sick, the test result is negative, and the accuracy rate is also .

Q: If someone tests positive, what are the chances that they will get sick?

The answer given by most people is; Some people will be alert that the problem may not be so simple, and after thinking about it, they don't seem to understand, so they just guess a number.

From the beginning of the question, the earliest information we receive about the incidence of the disease, followed by the accuracy of the test, and finally the positive result of a person's test, and then the return to determine the probability that the person will get the disease. We get the information gradually, and when we get the new information, our judgment of the whole thing changes.

Most people believe that the accuracy of the test result is the probability that the person will get the disease (the answer given by most people is . It can be seen that the information received later accounts for a very large proportion of people's judgment, and even makes the mind ignore the original known factual basis. As we progress through the development of events, new information comes in, and in the end, our judgments are so dependent on recent information and experience that they are biased.

This problem is indeed quite complicated from the perspective of probability, and not many people can calculate it by relying on the brain alone, so you need to borrow a calculator to calculate it. If we think in terms of probability to frequency, the problem is simplified.

Regardless of whether the test result is negative or positive, its accuracy is only , so there is a misjudgment. 1 out of 1000 people is sick, so this 1 person goes to test for infection and may be positive. The remaining 999 people who were not sick were likely to test positive (999x) due to the presence of test misjudgments.

For every 1000 people, there are positive test results ( of which only are infected and the rest are misjudged, so the true probability of getting sick is: 1, which is much less than .

In real life, our situation is similar, constantly receiving new information and knowledge, updating our old concepts, judgments, etc., is there often a large deviation, thinking about this probability problem helps us to correct the past decision-making bias, think about where it can be used.

Although mathematics is a lagging subject in the student days, the effective and widespread application of mathematical knowledge in life and work is undeniable and certainly interesting, but we don't always generalize and understand it this way.

The law of large numbers, which we often talk about, is one of the interesting facts about probability. Using a coin toss as an example, generally speaking, the probability of heads should be 50%. However, the second time you toss, does it have to be the opposite?

Not necessarily. Maybe you even tossed a dozen times in a row, all on the front of the file. However, if you keep throwing, 100 throws, 1000 times, and finally the number of heads and tails, it should be the same.

That is, as long as you toss enough times, the result should be stable.

Isn't it possible to think of the right thing this way: the right thing to do over and over again? If you fail to succeed in 1 or 2 times, do it a few more times, and the probability of success will be greatly increased. That is, to turn the uncertainty of the individual into the certainty of the group. This is an interesting application of "probabilistic" thinking.

When we go with He Hong, we must also face up to and be in awe of "extreme probability". Because no matter how much you increase the probability, you can't reach 100%. In other words, although the probability is small, there is still an extreme probability. There will still be surprises.

Life and work are infinite games, and you must always promise not to let yourself out. Winning or losing is temporary, and you can continue to play. But once you're out, it's game over.

Extreme events are extreme because they are very unlikely to occur, but they are extremely destructive. Last time, you'll lose all your luck. So we have to try to isolate that risk as much as possible.

Keep your own choice, know where the road is, know why you chose this path, know how to go, and why be afraid?

It is obvious from the meaning of the title that the probability of winning is 3 4, and the probability of losing is 1 4 because if I win, I will get 1 yuan, and if I lose, I will get 1 yuan. So every time I earn: (3 4 times 1) (1 4 times the dollar, so I make a steady profit.

The key is that the students you bet on know what to eat for breakfast today, and if they do, then they will only guess one of the other 3, that is, the probability of winning is 1 3, the probability of failure is 2 3, and the probability of failure is 2 times that of China.

So don't lose or earn.

If you don't know what to eat for breakfast today, then the probability of winning is 1 4, the probability of failure is 3 4, and the probability of failure is 3 times that of medium.

Then you've earned.

Porridge buns are not available in the sky, so I don't think about it.

The key is fried noodles, fried rice, noodles and soup powder.

And according to what you said, you won't have the same two days, so your classmate is not a fool, and he will definitely not guess what he ate today, but only the remaining three, and the chance of getting it right is one-in-three, and the chance of getting it wrong is two-thirds, and the chances of making it are greater, but there's no guarantee that you'll make money every time.

It shows that you don't understand the probability well.

The probability of each positive is 1 2, and the probability of at least one positive shot ten times is 1-(1-1 2) 10 = 1023 1024, this probability means: do 1024 times and throw ten times in a row, about 1023 times there will be heads, and there may be one time (ten throws) are tails.

This does not contradict 1 2 at a time.

The concept is not clear, what you said about the probability of ten times is large, which means that the probability of at least one out of ten times is large, and the reverse of this is ten times are all negatives, this probability is very small, it and the probability of at least one time out of ten times is 1, so in fact, the probability of at least one occurrence of ten times is very large, and the probability of a single positive appearance is always 1 in 2

Probability, refers to the possibility, throw a coin, before throwing, you can know that after throwing, it is either heads or tails (ignoring the situation that the edge is just standing), the probability of the positive and negative sides is 1 2, but a certain number of votes, the number of heads and tails is not necessarily exactly half, only when the number of votes is infinite, the number of heads and tails will be almost equal, and the frequency of occurrence will be close to 1 2, which is the law of large numbers.

Looking back, look at the high probability, the high probability refers to the high probability of something happening, such as ten votes, the positive may not necessarily be 1 2, but at least one positive probability is very high, it can be said to: ten times in a row, the positive situation is a high probability event. (The probability is 1023 1024.)

The problem of probability is logical inductive reasoning. For example. There are 100 times.

It was cloudy, and then it rained. Then people will say that a cloudy day produces rain. But inductive reasoning is not entirely inevitable.

Completely closed reasoning. Because if you stretch it out to 100 million times. There may be tens of millions of cloudy days when it doesn't rain.

Probability is such an uncertain subject.

Cast 10 times, not necessarily the chance of the front appearing, or the chance of the back side being more (because the number of tests is too small)...

But if you cast a very large number of times, then the result is an infinite number of heads and tails of nearly 1 in 2.

solution, every appearance of the front is 1 2. No matter how many times you vote.

But the probability of at least one occurrence is related to the number of votes n times.

1-(1 2) n, the larger n is, the larger n is.

When n=10, then =1-(1 2) 10.

solution, the probability of each occurrence is constant.

But the probability of a 10 head is.

n=1-(1 2) 10, this probability is.

Very big.

Agree with the algorithm on the first floor, which is a problem of probability calculation in high school math. Just calculate the probability that each of the ten times is not positive, and subtract it by 1.

It seems to be 1 2 to the 10th power. You can try it yourself, for example, two coins, there are four cases, heads and tails, heads and tails, heads and tails, then heads are 1 4. What about three. You can list them all down to find out the pattern.

First of all, 1 2 heads are thrown, and heads have appeared 10 times, these are two propositions.

The probability is that he will appear once in many times, just like the insurance company buys insurance, invest 5 yuan, and insure 200,000 They also use the probability, if it is a human black box operation, it is difficult to say.

To solve this problem, you first need to know what the corresponding probability is for the occurrence of different points, and this problem is suitable for using the partition method to find the probability. There are a total of 6 4 cases for rolling four dice, and there is only one case when the number of points is 4 (all 1); The number of points is 5 (there is a 2) with C4, 3 cases; There are 6 points (two are 2 or one is 3), there are c5, 3 cases....The number of points is 24 and there is c23, and in 3 cases, divide the number of cases by the total number of cases 6 4 is the probability of this happening. To see who is cost-effective, you need to look at the average benefit of each, that is, the expected size of e, which should consider both the amount of winning x and the probability of x taking each value.

The sum of the number of money won x multiplied by the probability of this situation is expected e.

e=50*1/6^4+20*c4,3/6^4+10*c5,3/6^4+..1)*c9,3/6^4+(-1)*c10,3/6^4+..1)*c17,3/6^4+1*c18,3/6^4+2*c19,3/6^4+..

50*c23,3 6 4 When the number is positive, you win, when the result is negative, the dealer wins, and the number (expected value) is your average benefit, that is, how much you win. Welcome, remember to add points and reviews, hehe!