a. 2 of the marbles are white and the other 2 are blue?

b. They are not all of the same color?

a. 2 of the marbles are white and the other 2 are blue?

b. They are not all of the same color?

Tutors, please sign in to answer this question.

a. You can choose 4 marbles out of 9+7+5=21 by C_{4}^{21}=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 C_{2}^{7}=21 ways, 2 blue marbles can be chosen by C_{2}^{5}=10 ways. Overall number of ways to choose 2 white and 2 blue marbles is 21*10=210.

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)=C_{4}^{5}/5985=5/5985=1/1197;

P(marbles are all white)=C_{4}^{7}/5985=35/5985=1/171;

P(marbles are all red)=C_{4}^{9}/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:

The number of ways of choosing 4 balls out of 21 is _{21}C_{4}=5985

The number of ways of choosing 2 white out of 7 white is _{7}C_{2}=21

The number of ways of choosing 2 blue out of 5 blue is _{5}C_{2}=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 _{9}C_{4}/_{21}C_{4}=126/5985

The probability that they are all white is _{7}C_{4/21}C_{4}=35/5985

The probability that they are all blue is _{5}C_{4}/_{21}C_{4}=5/5985

The probability that they are not all of the same color =1-(126+35+5)/5985=(5985-166)/5985=.972264

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)

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

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**

Sergey O.

Computer Science/Programming Tutor, Math Ph.D.

New York, NY

4.9
(51 ratings)

Edward B.

STAT/MATH/Actuarial Science/MBA/Econ/Fin. - Ivy League Exp & Prof

New York, NY

4.9
(294 ratings)

John K.

Ardent Mathematics, Science, History, SAT and German Polymath

College Point, NY

5.0
(49 ratings)

## Comments

_{9}C_{4}=0 and so 1-0=1