Think of a math problem about probability for yourself

The exact answer to this question is: 1 2!-1/3!+1/4!-1/5!+…1/60!(Note the relationship between the plus and minus signs and n);

Let ai mean that the ith person gets his own test paper, i=1,2,......60。At least one person who has their own test paper can be expressed as:

a=a1ua2u……ua60。By the addition formula: p(a)=p(a1ua2u......ua60)=1-1/2!+1/3!-1/4!+…1/60!。

Thus, the probability that no one will get their own test paper is: p=1-p(a)=1 2!-1/3!+1/4!-1/5!+…1/60!

Suppose the total number of misaligned arrangements of n elements is f(n).

When one of the elements A is arranged to the position where B is located, B has two cases:

1. B is arranged to A, then the remaining n-2 elements have F(N-2) arrangement.

2. B is arranged to the position of other elements other than A and B, so that is to say, after removing A, there are F(N-1) kinds of arrangement of N-1 elements left.

Therefore, there are f(n-1) + f(n-2) ways to arrange a to b.

And because the position of a has b, c, d...And so on, a total of n-1 choices.

So f(n)=(n-1)[f(n-1)+f(n-2)] This is a recursive formula, and the initial value condition is easy to obtain:

i.e. f(1)=0 and f(2)=1

The general formula for f(n) can be expressed in the form of a series:

f(n)=n![∑1)^n)/n!n>2) The above is copied, take f(60) and it is OK

This probability is roughly E -1 equals.

This is a matter of misalignment.

Exact probability = 1-1 2!+1/3!-1/4!+…1/60!e-1 (approximate Taylor).

Go to Euler to get the wrong envelope.

One factorial fraction of sixty.

Friend, you are so sorry! The detailed process is like the defeat of Mintu RT, Xiyanzhi hopes to help you solve the problem.

1 to 20 prime-hole silver take 1,2,3,5, 7,11,13,17,19, a total of 9, the probability of 9 20, draw 8 times, to three times, then c 3 8 9 20 3 11 20 5.

In the range of 1 to 20, the prime number has , a total of 8 celery limb mu, and the composite number is 20-8 = 12.

There are 8 times of extraction, the probability of drawing 3 prime numbers p1=(8 20) 3, and the other 5 times of drawing the probability of the composite number p2=(12 20) 5.

Therefore, the probability of drawing exactly 3 primes:

p = p1*p2

You can catch a hundred fish first, mark them, and put them back in the water. In a few days, catch another hundred fish. The number of marked fish caught is x. Then there is a fish about [100 divided by (x 100)].

As for fishing for a few, it can be analogous and casual.

Catch a number of fish (set to m) from the pond and mark them with a mark that will not fall off easily. After a certain period of time, n fish are caught again, and the number of marked fish n is recorded, then the total number of fish in the fish pond is estimated x = (m n) * n

Scoop up a fish first, mark it, and put it back in the pond. After a while, pick up 50 (any number) to see how many of the 50 have marks, as b bars, use a b 50), okay.

If you think it's right, please adopt it as soon as possible, I'm playing with someone else today, thank you.

A is an event and B is an event, which may or may not contain each other, and if A contains B, then A is the right, but not necessarily. Similarly, options B and D are not necessarily correct. In C, ab represents an event that both A and B satisfy, which is either equal to a or less than A, so adding A is equal to A.

There are also two matches against the first, wins, wins, losses, losses, so it's 75%.

The disagreement lies in the format, and under normal circumstances, if one side wins, there is no need to compare, and you are 50% right.

There are also games that must be played, even if you are ahead of many and win early, you have to finish the game before it ends, which is probably what is said in the question.

According to you, let's assume that if A wins the first game, he can win without playing.

Then A's second game is premised on a defeat in the first game, in other words, "the probability that A will win in this game is 50%" is just a conditional probability, and the condition is "A's first defeat in the first game".

Therefore, the probability of A winning the championship is still 50%x50%+50%=75%.

50% is obviously wrong, 50% is only the probability of A winning the next game and not the probability of winning the championship, even if A loses the next game, he still has a chance to win the championship, so he wins the championship (1) he wins the next game directly, that is, 50% (2) he loses the next game, and wins the next game, which is 50%*50%=25%. (1) + (2) is his probability of winning the game.

I think so, A's win rate is 50%, which means that the first straight win is 50%, and if you say that his probability is 50%, you are actually ignoring a problem. Even if A loses the first game, it doesn't mean that he loses, because B has to win two games. Your thinking seems to be distributed according to two points, either winning or losing.

But it's not a two-point distribution, because he still has one chance to win if he loses the first game. If you stop at losing the first game and directly judge it as a defeat, then the thing is not done, because there is a second game. My own opinion.

Please forgive me for being a little verbose.

With the continuous draw, the total number of lottery tickets is constantly decreasing, so.

First. Zhang Zhong.

The chance is 1 20 = 5%.

The odds in the second card are 1 19

The odds in the nineteenth card are 1 2 = 50%.

The odds in the last one are 100%.

Originally, yes.,It's very simple.,Each one that doesn't hit is 95%,So 20 don't hit is 20 95% multiplied is.,It's very simple.,Common sense.。

Hello! (1) The probability of winning for the player who touches the ball is 1*(2 5)*(1 4)=1 20

2) The probability of the stall owner winning is 1-19 20=19 20

3) Therefore, within a month, the stall owner makes 30 * 100 * (19 20) * 1-30 * 100 * (1 20) * 5 = 2100 yuan.

1.The probability of finding three white balls is 3 6*2 5*1 4=1 202The probability of finding 2 yellow and 1 white is 3 6 * 2 5 * 3 4 = 3 203

As long as the color of the ball is the same, you can get money, the three white balls are 1 20, and the three yellow balls are also 1 20, which means that the probability of losing money is 1 10, and the probability of winning is 9 10.

The expected value of a single win is (-5 * 1 10) + (1 * 9 10) = 100 people a day, 30 days a month, and the expected value of winning is yuan.

It's hard to make money.

Of course, this is not necessarily.

95% similarity refers to the similarity of a pair of **, which does not mean that 95% of this ** may be the same person.

The scope of the statistical study is: if 95% of this ** is likely to be the same person, then 1 million pairs of ** with a score of 95% have been found, and about 950,000 pairs of ** should be the same person.

That is to say: to see whether 95% similarity can represent 95% possibility, this is basically not mathematically possible, this has to be related to the structure of the human face, etc., which can only be measured by experimental methods, and cannot be determined mathematically.

95% similarity means that the similarity between these 1 million** and the portrait you want to test is more than 95%.

Or that the person in the 1 million is more than 95% similar to the person you're looking for; As for whether the people represented in these 1 million ** are from the same person, and how many are from the same person, there is no basis. For example, twins A and B. 1,000,000 sheets of A's ** are more than 95% similar to B, but it cannot be proved that these ** are B's.

No. Probability is not defined that way. Probability is only a divisor, not an exact number.

This kind of question is to consider the worst-case scenario.

The worst-case scenario is: after taking the blue and red socks, start taking the black socks There are a total of 31 + 81 = 112 blue and red socks.

After taking it, there are only black socks left, and then take two to complete the task, 112 + 2 = 114 can't have a worse situation, and taking 114 can only guarantee that there are 2 black socks to choose e

This is a nine-fold Benuni experiment, drawing 1 is event a, then p(a)=, drawing 0 is event b, then p(b)=, so the probability of success is 1 minus the probability of unsuccess, that is, taking 1 minus the probability of drawing 9 times to 0.

i.e. p= to the power of 9 =