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# Probability question about drawing marbles?

a jar containing 4 marbles: 2 red and 2 white. 2 marbles are chosen at random. What is the probability that the marbles chosen are different colors?

Im confused why my way doesnt work:

I did it sort of the permutation combination way:

I said the numerator will be 2X2 because how many ways we can pick each seperate group is 2 thus 4 total. Then in the denominator I performed (4X3)/2! The reason I did this is because it seems order shouldnt matter! Pulling two marbles out how is pulling Red1white1 different from pulling white1red1. But the answer seems to not agree with me and they have 4/(4X3) without the 2 factorial. Now if you are going to say that order matters in the denominator, by not putting the 2! then in the numerator you should be able to say that there are not 4 possibilities but now 8 ie R1W1 would be diff from W1R1. I am really sorry if i confused you but I dont know how else to explain it. It seems like in the denominator they make order matter then not in numerator

### 5 Answers

- Anonymous1 decade agoFavorite Answer
To be honest, you are making this a lot harder than it is. :)

You have the same number of both colors, so it doesn't matter which one you pick first. After the pick, there are 3 marbles left, and only one is the same color as the one you just picked. So, the odds of picking two different colors are 2 out of 3, or 2/3.

- Anonymous4 years ago
a million) there are 2 crimson marbles rationalization: if a 0.33 of the marbles are blue then which potential there are 4 blue marbles...and if the prospect of drawing a crimson/blue marble is a half then which potential there are 6 crimson and blue.. consequently, there are 2 crimson marbles 2)odds of finding out on a blue chip = 4/12 = a million/3 odds of finding out on a crimson or white = 8/12 = 2/3 (or u can say a million- a million/3 = 2/3) rationalization: there are a finished of 12 marbles (6 crimson, 4 blue, 2 white) so finding out on a blue could be 4 over the finished (12) that's a 0.33 and finding out on a crimson/white would be something (a million- a million/3) cuz there are not the different colours desire it facilitates ;)

- holdmLv 71 decade ago
there are C(4,2) = 6 ways to choose 2 marbles.

1 choice is RR

1 is WW

that leaves 4 for RW

so prob = 2/3

in this method, the order of marbles drawn is irrelevant in both numerator and denominator

think instead of 4 aces

BB = SC 1 way

RR = HD 1 way

RB = SH, SD, CH, CD 4 ways

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