A bag has several red and blue balls, if 2 red balls are removed from it

Suppose that the total number of balls is x

Then the number of red balls with the first method is 2+(x-2) 6, and the number of red balls with the second method is (x-3) 5

Because the number of red balls does not change twice, the equation is obtained

2+(x-2)/6=(x-3)/5

The total number of balls calculated is 68

So the number of red balls is (68-3) 5=13

The number of blue balls is 68-13=55

Let x be a red ball and y be a basketball.

1/6(x+y)=x

1/5(x+y)=y

Just ask for a solution.

Let the total number of balls be m, and after taking out the red balls, the number of red balls in the bag will be x

Then: (m-2) 6=x

After taking out the basketball: (m-3) 5=x+2

The solution of the two formulas is: x=15 m=88

Number of Basketballs = M-(X+2) = 71

That is, there were 17 red balls and 71 blue balls in the bag.

I agree with zwjtk - assistant level 2.

Let the total number of balls be m, and after taking out the red balls, the number of red balls in the bag will be x

Then: (m-2) 6=x

After taking out the basketball: (m-3) 5=x+2

The principle is correct, but the answer is not, so I did the math again based on your principle.

As Shi Munan - Assistant Level 2.

Red 13 Blue 55 The total is 68 :)

The topic is relatively simple.

Solution: Set x red balls and y basketballs.

x+y-2)/6+2=x （1）

x+y-3)/5=x （2）

So x+y-2=6x-12 i.e. 5x-y-10=0 (3)x+y-3=5x i.e. 4x-y+3=0 (4) so from (3)-(4) we get x=13 y=55 Answer: There are 13 red balls and 55 blue balls in the bag.

Summary. The other bag has a total of three balls in it, and then two of them are red balls, so the probability of you touching the red ball is 2 3

If one bag is full of red balls, and another bag has two red balls and one yellow ball, then it's likely.

This question is up to me, it takes a little time to type, so please be patient.

There is a 100% chance of touching a red ball inside the first bag

The probability of touching the red ball inside the second bag is 2 3

If all of them are red balls, they will be touched, and they can only be red balls.

The other bag has a total of three balls in it, and then two of them are red balls, so the probability of you touching the red ball is 2 3

Because there are only two colored balls in the bag, of which 6 3, that is, the number of basketballs in the bag is far greater than the number of red balls, so the possibility of touching the blue ball is greater;

So the answer is: blue

(1) In the case of taking any 3 of the eleven balls, there are c3, 11 = 165 kinds (2) there are c3 in the case of taking all blue balls, 5 = 10 kinds (3), then there are 165-10 = 155 kinds of at least one red ball (4), so the probability of taking at least one red ball is about 155 165x100% = 94%.

Suppose the original hail covers the x-ball.

The red bench oak has: 1+(x-1) 7

The red source coarse state ball has: (x-2) 5

So: 1 + (x-1) 7 = (x-2) 535 + 5x-5 = 7x-14

2x=44x=22

Only binary primary bond equations can be used.

There are x red balls and y yellow balls.

Then 7(x-1)=x+y-1

5(y-2)=x+y-2

The solution is x=32 jujube 23, y=54 23

There's a problem with the topic of your stool.

Set the one-year difference to destroy a total of x balls.

Take out 1 red. There are also (x-1) 7 reds.

Take out 2 blue. There are also (x-2) 5 Red Celebrity.

So the red ball 1 + (x-1) is 7 = (x-2) 5 and is multiplied by 3535 + 5x-5 = 7x-14

2x=44x=22

So there are 22 balls.

The topic is relatively simple.

Solve the Wang key to change: set x red balls and y basketballs.

x+y-2)/6+2=x

x+y-3)/5=x

2) Bright Enlightenment. So.

x+y-2=6x-12

i.e. 5x-y-10=0

x+y-3=5x

That is, 4x-y+3=0

So from (3)-(4) we get x=13

y=55A: There are 13 red balls and 55 blue balls in the bag.

b。According to the method of finding probability, find two points: the total number of smirks in all possible scenarios; the number of eligible cases; The ratio of the two is the probability of its occurrence. Because of this, the file is destroyed, first find the total number of balls, and then solve it according to the probability Sun Bei formula:

3 red balls, 2 blue balls, a total of 5 balls, take out a random ball from the bag, the probability of taking out the red ball is.

Therefore, choose B.

Suppose that the total number of balls is x

then use the first method.

Red balls. The number is 2+(x-2) 6

The number of red balls with the second method is (x-3) 5

Because the number of red balls does not change twice, the equation is obtained

2+(x-2) grinding excitation 6=(x-3) 5

The total number of balls calculated is 68

So the number of red balls is (68-3) 5=13

Blue balls. The number refers to the game game 68-13 = 55

Let the total number of balls be m, take out the red balls Qi Menghou Min ridge, and the number of red balls in the Gaonaqiao bag is x, then: (m-2) 6=x

After taking out the basketball: (m-3) 5=x+2

The two formulas are combined, and the solution is: x=15

m = 88 basketball count = m - (x + 2) = 71

That is, there were 17 red balls and 71 blue balls in the bag.

Every time you take it out, there are 6-4=2 pieces left in the white ball.

A total of 12 2 = 6 times.

It turns out that the red and white balls each have: 6 6 = 36.

It turns out that there are 36 2 = 72.