M15-18

Question

A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?

A.  \frac{2}{21}  
B.  \frac{3}{25}  
C.  \frac{1}{6}  
D.  \frac{9}{28}  
E.  \frac{11}{24}

Bunuel · Accepted Answer

Official Solution:

A basket contains 3 blue marbles, 3 red marbles, and 3 yellow marbles. If 3 marbles are drawn at random from the basket, what is the probability that one marble of each color is among those drawn?

A.  \frac{2}{21}  
B.  \frac{3}{25}  
C.  \frac{1}{6}  
D.  \frac{9}{28}  
E.  \frac{11}{24}

The probability  P  can be calculated as  P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=\frac{9}{28} . The numerator represents the number of ways to select one blue marble out of three, one red marble out of three, and one yellow marble out of three. The denominator represents the total number of ways to select three marbles out of nine.

Answer: D

WillGetIt · Answer

Hello,

Old thread but might be useful to someone else.

filipTGIM your approach is partial.

You are assuming a defined sequence only i.e. Blue first, red second and yellow last. But don't you think there could be other possibilities also i.e. Red-Blue-Yellow...so on...

Basically three item can be arranged in 3*2*1= 6 ways and hence in the final step you need to multiply with 06.

Probability of extracting 1 blue marble : 3/9
Probability of extracting the next one as red : 3/8
And the next one as yellow : 3/7

Which equals 3/9 * 3/8 * 3/7 = 3/56 (not 9/56)

Then we need to multiply with 06:

3/56*6 = 18/56 or 09/28

Hope it helps!

xnthic · Answer

combinatorics is my weakness but correct my if im wrong, the numerator of the answer should be addition not multiplication that equals 9?

you have 1C3∗1C3∗1C3 / 3C9 = 9/28

Bunuel · Answer

No, it's multiplication, because of Principle of Multiplication. The result is 27/84, which when reduced gives 9/28.

Principle of Multiplication 
If an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘m*n’ ways.

Principle of Addition 
If an operation can be performed in ‘m’ different ways and another operation in ‘n’ different ways then  either  of these two operations can be performed in ‘m+n’ ways ( provided only one has to be done ).

Check below post for more:
 Probability Made Easy!

Theory on probability problems

Data Sufficiency Questions on Probability 
 Problem Solving Questions on Probability

Tough Probability Questions

Prateek101 · Answer

I think this is a  high-quality  question and I agree with explanation. The formula expression is incorrect

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Vikramaditya00 · Answer

I think this is a  high-quality  question and the explanation isn't clear enough, please elaborate. I understood 3c1, but 3c9 comes out to 84 so shoudnt the ans be 9/84 ( this ans not given in options)? I am nopt able to understand where is 328 coming from

Bunuel · Answer

P=\frac{C^1_3*C^1_3*C^1_3}{C^3_9}=

=\frac{3*3*3}{(\frac{9!}{6!3!})}=

=\frac{3*3*3}{7*4*3}=

=\frac{9}{28}

M15-18

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