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An urn contains 9 red marbles, 7 white marbles, and 5 blue marbles. You grab 4 of the marbles. What is the probability that:

a. 2 of the marbles are white and the other 2 are blue?
b. They are not all of the same color?

5 Answers by Expert Tutors

Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
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a. You can choose 4 marbles out of 9+7+5=21 by C421=21!/(4!*(21-4)!)=21!/(4!*17!)=18*19*20*21/24=
=3*19*5*21=5985 ways. 2 white marbles can be chosen by C27=21 ways, 2 blue marbles can be chosen by C25=10 ways. Overall number of ways to choose 2 white and 2 blue marbles is 21*10=210.

Probability to choose 2 white and 2 blue marbles is:
P=210/5985=2/57≈0.0035=0.35%

b. The probability that marbles are not of the same color is 1-P(marbles are of the same color). Let us figure the probability that marbles are of the same color. It is the sum of probabilities that marbles are all blue, or white, or red.

P(marbles are all blue)=C45/5985=5/5985=1/1197;
P(marbles are all white)=C47/5985=35/5985=1/171;
P(marbles are all red)=C49/5985=126/5985=2/95;
P(marbles are of the same color)=P(marbles are all blue)+P(marbles are all red)+P(marbles are all white)=(5+35+126)/5985=166/5985≈0.0277=2.77%
Probability that they are not of the same color is:
P=1-166/5985=5819/5985≈0.9723=97.23%
Michael F. | Mathematics TutorMathematics Tutor
4.7 4.7 (6 lesson ratings) (6)
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The number of ways of choosing 4 balls out of 21 is 21C4=5985
The number of ways of choosing 2 white out of 7 white is 7C2=21
The number of ways of choosing 2 blue out of 5 blue is 5C2=10
The number of favorable draws  21×10=210  Thank you Richard P
I now realize that for each of the 21 ways of drawing 2 white out of 7 white I have 10 ways of drawing 2 blue out of 5 blue.
The number of possible draws 5985
The probability =210/5985 = .035088

The probability that they are not of the same color is 1- the probability that they are all of the same color
The probability that they are all red is 9C4/21C4=126/5985
The probability that they are all white is 7C4/21C4=35/5985
The probability that they are all blue is 5C4/21C4=5/5985
The probability that they are not all of the same color =1-(126+35+5)/5985=(5985-166)/5985=.972264
Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (629 lesson ratings) (629)
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The total number of marbles is 21.  So the number of ways to choose four marbles is  21 choose 4.  This is the combinatorial coefficient   21!/( 4! 17!) =  5985.

{Note N! = N (N-1)(N-2) ... 3 2 1    }

Of these 5985 the number of ways to choose 2 white and 2 blue is
[7 choose 2 ] [ 5 choose 2]  =  [ 7! /(2! 5!) ] [ 5! /(2! 3!) ]  = 210

So the answer to part A  is   210 / 5985 = 2 /57 =  about 0.035

For part B, the number of ways to choose all four white is   7 choose 4  = 7!/(4! 3!) = 35
so the probability of all white is   35/5985 =  1/ 171
The number of ways to choose all four blue is   5 choose 4  = 5!/(4! 1!) = 5
so the probability of all blue is   5/5985 = 1/1197The number of ways to choose all four red is   9 choose 4 = 9!/(4! 5!) =  126
so the probability of all red is  126/5985  =  2/95

The answer to part B is  1 - 1/171 - 1/1197 - 2/95)

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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There are 21 marbles.

There are C(21,4)=21!/4!/16!=5985 ways of choosing 4 of the 21 marbles.

a)  There are C(7,2)=21 ways of choosing 2 of the 7 white marbles. There are C(5,2)=10 ways of choosing 2 of the 5 blue marbles. Therefore there are 21*10=210 ways of choosing 2 white and 2 blue marbles. Therefore the probability is 210/5985=42/1197=14/399≈0.035.

b) The probability they are not of the same color is 1 minus the probability they are of the same color, i.e., all white, all red, or all blue. That's

1- ( C(7,4)+C(5,4)+C(9,4) )/5985=1-166/5985≈0.97

Muhammad C. | Muhammad the grand math tutorMuhammad the grand math tutor
4.9 4.9 (216 lesson ratings) (216)
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A) (7/25)(6/24)(5/23)(4/22) = 7/2530

B) (9/25)(7/24)(5/23)(4/22) = 252/60,720 = 21/5060

I do not think you are correct. Try to solve it your way for 3 white, 3 red, and 3 blue marbles. If I got your idea correctly, you would get (3/9)*(3/8)*(3/7)*(4/6)=1/28 for probability that not all of them are of the same color. At the same time it is obvious that such probability is 1.
The probability that for 3 red 3 white 3 blue there is any favorable case for drawing 4 of any color is 0
and so all my terms would be 0/9C4=0 and so 1-0=1
Muhammad, could you please outline your analysis.  Sorry I did not ask this with my previous comment
Michael F
Yep....I did it wrong but not for that reason. I read it wrong & there are 21 total marbles. Therefore:

A) (7/21)(6/20)(5/19)((4/18) = 1/171
B) all red = (9/21)(8/20)(7/19)(6/18) = 2/95
all blue = (5/21)(4/20)(3/19)(2/18) = 1/1197
all white= (7/21)(6/20)(5/19)(4/18) = 1/171

Probability of the marbles not all being the same color is 1 - (2/95 + 1/1197 + 1/171) = .973