Math problems problems of possibility .

The black and white area is 1 3, and the probability of winning is 1 3.

The white area accounts for 1 2 of the whole, and the black area accounts for 1 4

So the probability of Xiaofang winning is 1 4, and the probability of Xiaoli winning is 1 2;

Fairness means that two people have the same probability of winning, so they should use 1 hypothesis to turn to n and get the prize of M square.

If 1<=n<=5, then count n cells to get m=2n, that is, to get even cells.

If 6<=n<=10, then 2n>10

So it's actually up to the cell m=2n-10.

2n-10 = 2 (n-5), which is still an even number.

So you can make a table.

n=1,m=2

n=2,m=4

n=3,m=6

n=4,m=8

n=5,m=10

n=6,m=2

n=7,m=4

n=8,m=6

n=9,m=8

n=10,m=10

So 2, 4, 6, 8, 10 appear 2 times each.

1, 3, 5, 7, 9 no.

So 2,.4, 6, 8, 10 probability are all 1 5, 1, 3, 5, 7, 9 is 0

The sum of 2 numbers must be even, and the even numbers are worthless things.

So the chances of getting 2468 10 are all 1 5 and the chance of getting 13579 is 0

Answer: (1) Use the number wheel to play the game, and the probability of winning Xiaofang and Xiaoli is 1 3;

2) The probability of winning Xiao Fang with the number wheel is 1 4, and the probability of Xiao Li winning is 1 2;

3) The game should be played with a number wheel, and everyone has the same probability of winning, so that it is fair.

The principle is that odd + odd = even, even + even = even, and even 10 more than even, so the number where you finally hear it must be even.

Question 1: Answer:

1) Play the game with the number wheel, and the probability of winning for Xiaofang and Xiaoli is 1 3;

2) The probability of winning Xiao Fang with the number wheel is 1 4, and the probability of Xiao Li winning is 1 2;

3) If you are the organizer of the game, from the point of view of fairness, you should use the number wheel to make the game, so that everyone has the same probability of winning.

If you let Xiaofang choose, you should choose No. 1; If you let Xiaoli choose, of course choose No. 2. Question 2: The purpose of the person who sets the prize is to make money, and the jackpot is just a bait for people to be fooled, and no one can get the grand prize.

Those who spend 5 yuan will always only get prizes corresponding to the numbers 2, 4, 6, 8, and 10, and although those who spend money will definitely win the prize (with a 100% overall chance, which is also a bait), the actual value of the prizes corresponding to the numbers 2, 4, 6, 8, and 10 is very small (far less than 5 yuan). The reason for this phenomenon is easy to get by mathematical knowledge: any positive integer number must be added to itself to get an even number, so no matter where the pointer stops, it must be an even number after adding the corresponding number, and it is impossible to get the prize corresponding to the number.

The probability of winning the corresponding prize is equal, both are 20% (or 1 5) for reference! Jiangsu Wu Yunchao wishes you progress in your studies.

1) It's all 1/3

2) It's all 1/4

3) Use the number 1 because all 3 parts are the same.

2.Can only go to, the odds are 20% because odd odd = even even + even = even so only that small prize knows odd odd = even + even = odd 0

In fact, this is a deceptive game, and you won't get any prizes. You can only get prizes corresponding to 2, 4, 6, 8, and 10 numbers, and the odds are 20%. The rule is that the prizes are all even-numbered prizes.

I don't understand! The first one, it should be the choice of the second!

1. (1) Put 30*2 5=12 red balls. Similarly (2), put in 6 yellow balls. (3) 12-year green ball.

2. (1) 1 12 1 3 is different, A is 1 3, and A of hearts is 1 12.

The appearance of such a topic in elementary school is indeed beyond the curriculum. The concept of probability, which was learned in high school, is too early to appear. Think of it this way, divide 30 balls into 5 parts, each part is 6, and take out two red balls from it, that is, 12 red balls and 18 yellow balls.

In the same way, 25% is 1 4, divided into 4 parts, 6 per portion, yellow balls account for 6 parts, and the others are red balls.

The green ball accounts for 1 3, that is, the red ball and the yellow ball account for a total of 2 3, that is, there are 24 parts of 2, 12 pieces each. That is, you need to put 12 green balls.

The rest is a matter of probability. Try to find out for yourself.

1. (1) 12 red balls and 18 yellow balls;

2) 18 red balls and 6 yellow balls;

3) Put 12 green balls;

1. The probability of touching hearts is: 1 in 4;

The probability of touching other suits is: 3 in 4;

2. The probability of touching "A" is: 1 in 3;

The probability of touching a "2" is: 1 in 3;

The probability of touching a "3" is: 1 in 3;

3. The possibilities are different

Because there is only one "ace of hearts" and four "aces";

1. Put 9 spades and 3 other suits;

2. 2 of the four clubs and 2 squares;

Question 1:

1) 12 red balls, 18 yellow balls;

2) 18 red balls and 6 yellow balls;

3) Put 12 green balls; Question 2:

1. The probability of touching hearts is: 1 in 4;

The probability of touching other suits is: 3 in 4;

2. The probability of touching "A" is: 1 in 3;

The probability of touching a "2" is: 1 in 3;

The probability of touching a "3" is: 1 in 3;

3. The possibilities are different

Because there is only one "ace of hearts" and four "aces";

1. Put 9 spades and 3 other suits;

2. 2 of the four clubs and 2 squares;

In the third condition of the first question, 24 balls refer to 24 balls each or 24 in total?

Question 2: (1).

12 = 1 4, the probability of touching hearts is 1 4, and the probability of other suits is 3 4, respectively, 1 4;

2, because the number of a three cards are the same, so 4 12 = 1 3, all account for 1 3;

3. Not the same, because there is only one ace of hearts, and there are 4 cards for a total of 4 different numbers, and the possibilities are different.

2) 1. As long as the number of spades accounts for 3 4, there are many methods;

2. You can make 2 clubs and 2 squares, or 1 each, and the other two at will.

The first question is ambiguous. I don't know if it can help you.

1 (1) Red Ball: 30x2 5=12 (pcs) Yellow Ball: 30x(1-2 5) = 18 (pcs).

2) Yellow ball: 24x25%=6 (pieces) Red ball: 24x(1-25%)=18 (pieces).

3) 24x1 2=12 (pcs).

2 (1) hearts: 3 12=1 4 Other suits: 9 12=3 4(2)A: 1 3=1 3 2:1 3=1 3 3:1 3=1 3

3) Not the same, because the "Ace of Hearts" has 1 card"a"There are 3 sheets.

1) Spades: 12x3 4=9 (cards) 3 cards any (2) Clubs: 4x1 2=2 (cards) Diamonds: 4x1 2=2 (cards).

The probability that the product of numbers on two cards is singular is.

The probability that the product of numbers on two cards is an even number is.

All answers: 5*5=25(

There are 25 types in total.

20 doubles, 5 singles.

Double: 20 25 = 4 5 = 80&

Single: 5 25 = 1 5 = 20%.