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There are six red, blue and yellow glass balls of the same size in the box,
There are 25 glass balls, which are divided into red, yellow and blue. The number of red glass balls is 1 / 6 of that of yellow. How many blue glass balls are there at least?
There should be a formula
Because the number of red glass balls is 1 / 6 of that of yellow,
The number of balls must be an integer
therefore
Yellow is a multiple of six, if you want to
Blue is the least, so
Yellow and red as much as possible, but
Yellow = 24, red = 4, more than 25,
therefore
Yellow at most = 18, at this time
Red = 18 × 1 / 6 = 3
Blue at least = 25-18-18 × 1 / 6 = 4
There are 80 red, yellow and blue glass balls of the same size. They are always arranged in the order of 3 red balls, 2 yellow balls and 4 basketball balls. How many red balls account for the total number
3x+2x+4x=80
x=80/9
The red ball is 3x
So red ball accounts for (3 * 80 / 9) / 80 = 640 / 3 of the total
Li Hong and Wang Fei are good friends. She often goes to her home to do her homework
The two references to "she" are not clear
Li Hong and Faye Wong are good friends. Li Hong often goes to Faye Wong's house to do her homework
Li Hong and Wang Fei are good friends. Wang Fei often goes to Li Hong's home to do her homework
There are 10 red balls, 9 white balls, 8 yellow balls and 2 basketball balls. Only when you touch at least a few balls at a time can you ensure that there are 5 balls of the same color
There are 10 red balls, 9 white balls, 2 blue balls and 8 yellow balls in one bag. How many at least at a time can guarantee that five of them are of the same color?
Suppose you take four red balls for the first time
Four white balls for the second time
Four yellow balls for the third time
Two basketball for the fourth time
That is to say, none of the 14 balls are of the same color
So take at least 15
There are 10 red balls, 9 white balls, 2 blue balls and 8 yellow balls in a cloth bag with the same size but different colors. How many balls can be taken at least at a time to ensure that four of them are of the same color ()
A. 4B. 10C. 13D. 12
The four colors of red, white, yellow and blue are regarded as four drawers. Consider the worst case: touch out 11 balls: touch out all 2 blue balls, and touch out 3 red balls, 3 white balls and 3 yellow balls respectively. At this time, if you touch out another ball at will, you can ensure that there are 4 balls in one drawer, so 11 + 1 = 12. A: at least 12 balls at a time, you can ensure that there are 4 ball colors It is the same as D
There are four red balls, four blue balls, five white balls and seven yellow balls in the pocket. How many balls should be taken out of the pocket to ensure that there are white balls?
16
Four red balls first, four basketball and seven yellow balls second, and the last white ball
4 + 4 + 7 + 1 = 15
Please accept!
There are 5 red balls, 10 white balls and 15 yellow balls in the bag. How many red balls and how many white balls should be added to make the possibility of touching red balls one third?
Ten red balls. Five white balls
There are 7 white balls and 24 yellow balls in the bag. How many more white balls need to be added to make the number ratio of white balls and yellow balls in the bag 5:3?
24x5/3-7
=40-7
=33 (pieces)
You need to add 33 white balls
1. There are four white balls and two yellow balls in the pocket. If you touch two balls at a time, all the white balls you touch will be returned
There are four white balls and two yellow balls in the pocket. If you touch two balls at a time, the white balls you touch will be returned to the pocket and the yellow balls will be kept. When you touch two yellow balls at the nth time, that is to say, only the white balls you touch at the nth + 1st time, then the probability of this kind of situation is PN. Find PN
My practice is not to consider the situation of touching together twice
Let the probability of 4 white and 1 yellow in n-2 times bag be Q
Then n-1 time is still 4 white and 1 yellow, that is, the probability of touching two white balls in n-1 time is t = Q * 4C2 / 5c2 (formula 1) (4C2 means combination, 4 is the number in the lower right corner of C, which can't be typed)
The probability of touching a yellow ball and a white ball is PN-1 = Q * 4C1 * 1C1 / 5c2 (formula 2)
So t = PN-1 * 4C2 / 4 (1,2 combined)
So PN = t * 4 / 5c2 = PN-1 * 4C2 / 5c2
So without considering the two touching together, PN is an equal ratio sequence, and the common ratio is 3 / 5
The re examination rate is two, PN = x * (3 / 5) ^ n + (4C2 / 6c2) ^ (n-1) * 1 / 6c2
Help: what's wrong with the above solution
X is a certain number. I'm too lazy to calculate it. It's easy to get X by substituting n = 1 or 2
PN is the probability of getting all the yellow balls out for the nth time
Let the probability of remaining 4 white after the nth operation be a [n], the probability of remaining 4 white and 1 yellow be B [n], and the probability of remaining 4 white and 2 yellow be C [n]
A[n]=A[n-1]+B[n-1]×(C(4,1)/C(5,2))+C[n-1]×(1/C(6,2))=A[n-1]+(2/5)B[n-1]+(1/15)C[n-1] (1)
B[n]=B[n-1]×(C(4,2)/C(5,2))+C[n-1]×(C(2,1)C(4,1)/C(6,2))=(3/5)B[n-1]+(8/15)C[n-1] (2)
C[n]=C[n-1]×(C(4,2)/C(6,2))=(2/5)C[n-1] (3)
And a [0] = B [0] = 0, C [0] = 1
From (3), we know that C [n] = C [0] × (2 / 5) ^ n = (2 / 5) ^ n
Substitute (2) to get B [n] = (3 / 5) B [n-1] + (8 / 15) (2 / 5) ^ (n-1), namely B [n] + (8 / 3) (2 / 5) ^ n = (3 / 5) (B [n-1] + (8 / 3) (2 / 5) ^ (n-1)), get B [n] + (8 / 3) (2 / 5) ^ n = (B [0] + 8 / 3) (3 / 5) ^ n, get B [n] = (8 / 3) ((3 / 5) ^ n - (2 / 5) ^ n)
Obviously, P [n] = a [n] - a [n-1] = (2 / 5) B [n-1] + (1 / 15) C [n-1] = (16 / 15) ((3 / 5) ^ (n-1) - (2 / 5) ^ (n-1)) + (1 / 15) (2 / 5) ^ (n-1) = (16 / 15) (3 / 5) ^ (n-1) - (2 / 5) ^ (n-1)
There must be a problem in your solution. You only consider the case of 4 white and 1 yellow in the n-2nd bag. When 4 white and 2 yellow and 4 white appear, you completely ignore it
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