Friday, July 31, 2009

I need help solving a finite mathematics problem.?

This is not my homework. This is from a sample test. I got 3/35 when i figured it out so i just selected 4/35. The answer is D. can anyone tell me how to do this?





A box contains 4 red balls and 3 blue balls. Four balls are selected simultaneously at random and their


colors are noted. A random variable X is defined to be the number of red balls minus the number of


blue balls. Find Pr[X = 2].





(a) 4/35 (b) 1/7 (c) 2/5 (d) 12/35 (e) 4/7 (f) none of the others

I need help solving a finite mathematics problem.?
To have 2 more red balls than blue balls, when only 4 balls are selected, you must choose 3 red balls and one blue ball.





Let C(n,k) = n!/[k!(n-k)!] be the binomial coefficient "n-choose-k", the number of ways to choose k objects from a set of n.





There are C(7,4) = 7!/[4!3!] = 35 ways to choose 4 balls. Of these ways, how many include 3 red balls and one blue ball? There are C(4,3) = 4 ways to choose the 3 red balls, and C(3,1) = 3 ways to choose the 1 blue ball, so there are 4*3 = 12 ways to choose a combination including 3 red balls and one blue ball. Therefore, 12 of the 35 possible combinations will give X=2.
Reply:The answer is simpler than it appears.





Take four red balls named a, b, c, d and three blue ones named e, f, g.





If you select four total balls and the number of red must be exactly 2 more than the number of blue, then you must select three red balls and one blue ball. It's the only combination that totals four with a difference of two.





There are only four ways to select three red balls from four (abc, abd, acd, bcd) and only three ways to select one blue ball from three (e, f, or g).





Four ways times three ways is twelve ways!





It looks like you already figured out where the 35 came from, but for those who didn't:





1. You have 7 choices to select the first ball, 6 choices for the second, then 5, and 4. That's 840 possibilities.





2. Then you divide this number by 24 (4 x 3 x 2 x 1) which eliminates the duplicate selections. If you select balls abcd this is exactly the same as selecting bacd.





3. 840/24 = 35.
Reply:You have to multiply that by 4 because the order doesn't matter.


4(4/7 * 3/6 * 2/5 * 3/4) = 4(3/35) = 12/35


You could have drawn the blue one first or second or third or fourth.


You could have


BRRR, RBRR, RRBR, or RRRB
Reply:Let us know what you found
Reply:You have to know which balls you are talking about, the balls picked or the balls left.

orange

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