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Anastaziya [24]
2 years ago
12

If 4 marbles are drawn at random all at once from a bag containing 8 white and 6 black marbles. in how many ways can we draw the

4 marbles so that 2 will be white and 2 will be black?
Mathematics
1 answer:
Andreas93 [3]2 years ago
3 0

You started out as if it was going to be a probability problem, but it turned out to be a simple binary counting problem.

It really doesn't matter how many of each color are in the bag, or how many  marbles there are in the bag all together.  I think you're just asking how many ways there are to pull out two white ones and two black ones.  That's exactly the same question as "How many 4-bit binary numbers can be written with two 1's and two 0's ?"

It's easy to draw up that list.  You start out like this:

0011,  0101,  1001,  0110,  1010,  1100 .

Gosh.  I just wanted to start out doing the list, but I think that's all of them.

There are six (6) ways.

WWBB,  WBWB,  BWWB,  WBBW,  BWBW, and  BBWW .

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I don’t understand… but thank you if u do answer my question :))
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3 years ago
There are 8 2/3 pounds of walnuts in a container which will be divided equally into containers that hold 1 1/5 pounds this would
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we'll do the same as before, turning the mixed fractions to improper and do the division, keeping in mind that is simply asking how many times 1⅕ goes into 8⅔.


\bf \stackrel{mixed}{8\frac{2}{3}}\implies \cfrac{8\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{26}{3}}\\\\\\\stackrel{mixed}{1\frac{1}{5}}\implies \cfrac{1\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{6}{5}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\\cfrac{26}{3}\div \cfrac{6}{5}\implies \cfrac{26}{3}\cdot \cfrac{5}{6}\implies \cfrac{130}{18}\implies \cfrac{126+4}{18}\implies \cfrac{126}{18}+\cfrac{4}{18}\\\\\\\boxed{7+\cfrac{4}{18}}\implies 7\frac{4}{18}

7 0
3 years ago
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