Is the probability of the number 1 appearing greater than that of other numbers 5

Because it starts with 1, and then it is the other numbers from 2 to 9. In a single digit, 0 appears once less than any other number.

Continue in tens of digits, first in the dozens, starting with 10, counting to 19, then 20, 21, 22, ,......Until 99. Overall, the number of occurrences of the ten-digit number from 1 to 9 is the same. 0 appears 9 times less than the other numbers.

And then 100. In this way, when counting from 1 to 100, 1 is one more time than other non-zero numbers.

Therefore, if it must be 1, the probability of occurrence is the greatest.

Or, to put it another way, you can use a formula to calculate the probability of a number appearing. In the single digit, 0 is only valid when the tens and hundreds are non-zero, and the probability of 1-9 is the same; In the tens digits, 0 is only valid when the hundred digits are 1 (100), and the probability of occurrence from 1 to 9 is the same; In the hundreds, only 1 can appear. Therefore, the number from 1 to 100 is counted coherently, and only 1 has the greatest probability.

The number itself is only 10 numbers, 0-9

In fact, the probability of each number appearing should be the same, and the reason why you are talking about the situation is because of the range of options you offer.

If there are 100 numbers from 0 to 99, and each number is written according to the 2-digit rule, 0 is written as 00, 1 is written as 01, and 2 is written as 02, then each number appears the same number of times. 20 times.

But according to what you said, first of all, the 0 in front is not written, so there are fewer 0s than the others.

Depending on your range from 1-x, 1 is the first to appear in any position, everyone, ten, hundred, etc., then the probability of 1 appearing is more than the others, at least definitely not less than the others.

It's a bit weird, why don't you start from scratch? What's the point of your sequence being sequential? Start with 0 with zero and start with 2 with 2 with the highest, and start with whoever is the highest. Or do I not understand what you mean?

The probability of each number appearing from 1 to 10 is 1 10.

1. There will be 10 outcomes when all possible outcomes appear, namely: 1 10;

2. The number of times each digital result appears is 1 time;

3. The probability of each number appearing is: 1 10 = 1 10.

The probability of each number appearing from 1 to 10 is 1 10.

1. There will be 10 outcomes when all possible outcomes appear, namely: 1 10;

2. The number of times each digital result appears is 1 time;

3. The probability of each number appearing is: 1 10 = 1 10.

The concept of natural numbers refers to non-negative integers used to measure the number of pieces of a thing or to express the number of pieces of a thing. Natural numbers start from 0 and follow each other to form an infinite collective. From the concept of natural numbers and the meaning of this question, it can be obtained:

From 1 to 100 in natural numbers, the number "1" appears 21 times. [Analysis] (1) 1 in 1 appears 1 time; (2) 1 of 10-19 appeared 11 times, and 1 of 1 appeared a total of 8 times; (4) 1 in 100 occurrences 1 time; So (1) + (2) + (3) + (4) = 1 + 11 + 8 + 1 = 21 (times).

Number One appears a total of 20 times.

The number two appears 220 times.

The number three appears 21 times.

Method: You can start by calculating the frequency of the number 1. But in fact, all three numbers are the same algorithm. Starting from 200, it can be calculated from the change of ten digits, and when the ten digit is zero, 1 appears once. 2 appears ten times. Three appear once.

When ten is one, one appears eleven times. Two appear eleven times. The late arrivals of the multi-lease items are 211 and 212 respectively. Three appear once.

And so on, it is very simple to get, the number one, which appears a total of 20 times) The number two appears a total of 220 times.

The number three appears 21 times.

The numbers are 1-6, which means that the minimum number added is 3 and the maximum is 18

3 has only 1 case (1+1+1), and 18 is also 1 (6+6+6) and so on, 3 and 18

The probability is the same, 4 and 17, 5 and 16....9 and 12, 10 and 11 are also the same.

And the closer the numbers, the greater the probability of occurrence.

Answer: 10 and 11 appear the largest, followed by 9 and 12, then 8 and 13, and so on.

So. The total may be 6*6*6=216 species;

s(k=3)=1;

s(k=4)=3

s(k=5)=3+3=6

s(k=6)=3+6+1=10

s(k=7)=3+6+3+3=15

s(k=8)=3+6+6+3+3=21

s(k=9)=6+6+3+3+6+1=25s(k=10)=6+6+3+6+3+3=27s(k=11)=6+3+6+6+3+3=27s(k=12)=6+6+3+3+6+1=25s(k=13)=3+6+6+3+3=21

s(k=14)=3+6+3+3=15

s(k=15)=3+6+1=10

s(k=16)=3+3=6

s(k=17)=3

s(k=18)=1;

It can be seen that the sum is 10 and 11

is the most probable; This is followed by 9 and 12

8 and 13 can actually be seen.

The probability of an sum of x is as great as that of a sum of 21-x; This is the symmetry of this kind of problem;

In general, problems related to binomial or binomial distribution tend to increase first and then decrease;

That is to say, the maximum point of this kind of problem can generally be passed.

a[n]>=a[n+1]

a[n]>=a[n-1] to calculate n; Thus the position of the peak is obtained;

The problem is simple, with symmetry in terms of the number of occurrences from 3 to 18.

There are fewer on both sides and more in the middle, and the symmetrical position occurs the same number of times.

So 10 and 11 occur the most.

9 and 12 second.

8 and 13 again

It's to roll three dice, if it's two, the probability of 7 is the largest, and I'll calculate three.

If 000-999 is considered, then the odds are the same for all numbers.

Because there can be no 1,2 digits of Bo Jingyin starting with 0, and there are no three digits starting with 9, and there is only 1 3 digits starting with 8, so the least probability may be that the manuscript is 0 and 9,8, considering that there are 100 digits in 1,2 digits, but there are 2 more base halls 0 and 1 8 in 800, and the three digits starting with 9 have 900-999 for a total of 100, so 9 loses the most numbers, so the probability of the number 9 appearing is the smallest.

Hello: In ordinary calculations, the probability of a number appearing in the back is as follows:

The bridge deck is 0, accounting for 10%.

The 1 followed by 10%.

The next 2 accounts for 10%.

The next is 3 of 10%.

The next 4 accounts for 10%.

The next 5 accounts for 10%.

The next 6 accounts for 10%.

The next is 7 for 10%.

The back is 8 talking about the sail accounting for 10%.

The next is 9 of 10%.

The total is 100%.