Question

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?

Answer 1

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 .

Answer 2

Using the combination formula, it is found that there are 420 ways to draw the marbles.

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• The order in which the marbles are chosen is not important, which means that the combination formula is used to solve this question.

Combination formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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We want:

• 2 white, from a set of 8.
• 2 black, from a set of 6.

Thus:

$C_{8,2} \times C_{6,2} = \frac{8!}{2!6!} \times \frac{6!}{2!4!} = 420$

There are 420 ways to draw the marbles.

A similar problem is given at https://brainly.com/question/16638977