This is a repaste of a problem posted by gmatacer: There

Question

This is a repaste of a problem posted by gmatacer: 

There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? 

Please suggest the easiest way to solve this.

sm176811 · Answer

This is what I had. Not sure whether this is correct:

P(RRGB) x 4! = 3/6 * 2/6 * 2/5 * 2/4 * 4! = 1/35 * 4! = 24/35 

Answer: 24/35

Professor · Answer

dude, your solution has two flaws... 

first, 4! is not correct cuz its not permutation and the order doesnot matter. it is a combination. even if order matters, it is not 4!.

second, 3/6 * 2/6 * 2/5 * 2/4 * 4! is not equal to 24/35.

M8 · Answer

Professor please elaborate this one. 

I see the denominator 7!/4!3!, but what is in the numerator?

Professor · Answer

= (2 red from 3 red in 3 ways)(1 green from 2 green in 2 ways)(1 blue from 2 red in 2 ways) = 3x2x2=12

M8 · Answer

Why do you use combination formulae for the red ones (3c2)?
What do you mean 'in 2 ways'? First or second of the same colour?

Professor · Answer

this is not a permutation problem because here order doesnot matter. for example: there are three people as under: 
M8 (M), professor (P) and sm176811 (S).

how many ways can these three people be in a que? 

= 3! = 6 ways (they are .... MPS, MSP, PMS, PSM, SMP, SPM) 

in this question, order matters because MPS and PMS are different ques (arrangements).

ok, lets say this problem in different way i.e how many ways can we select 2 from these three people M, P and S. 

= 3c2 = 3 ways becasue it doesnot matter if we select M first and P second or P first and M second.

similarly how you chose red doesnot matter. so use combination....

gmat_crack · Answer

Professor why you have used 7C4 in denominator.

According to the question "balls are drawn at random without replacement".

It means order matters here. If we draw a balls according to 

RRBG

P = 3/7 * 2/6 * 2/5*2/4

Similarly if we draw balls RBGR

p = 3/7* 2/6* 2/5* 2/4

or BGRR
2/7*2/6*3/5*2/4

I think here order matters because we are not putting balls back. Can you elaborate on this.

Professor · Answer

hmm.... i overlooked..... i should revise my posts. do tomorrow... late to bed :sleep

swaroh · Answer

I think the professors is correct, this is a combination. Combination does not mean there is replacement. It just means order does not matter. The order in which we pick the colors do not matter so it is still a combination problem. So what is the OA?

thanks

gmatacer · Answer

Professor - Can you please break your solutions in steps and explain each step clearly. I did'nt uderstand this (3c2 x 2 x 2)/(7c4)=12/35 I am sorry but I am just not getting the solutions

GmatAcer

swaroh · Answer

I will try the explanation,

Denominator= Total space= 7C4 (7=total items of which 4 are being chosen)

Numerator= Out of 3 red we need 2 of them = 3C2
= Out of 2 green we need 1 = 2C1
= Out of 2 blue we need 1 = 2C1

Now we want the first condition, second and third not first or second or third, so we need to multiply.

3C2*2C1*2C1/7C4

Hope this helps.

sm176811 · Answer

Sorry I do not have the OA... I copied the qn from another post!

conocieur · Answer

Professor is right, this is an hypergeometric distribution problem.
and the key word for this problems is without replacement every time you hear that word you know you should apply hypergeometric.

so 7 c 4 is the number of ways you can pick 4 balls out of 7, that is the size of the sample

the number of ways you can get 2 red balls out of 3 red balls is 3c2 and same procedure for the rest.

use multiplication rule so (3*2*2) = 12

12/35

Professor · Answer

yah swaroh well explained in his post. btw, i replied your PM but for some reason its not going. PM me your email address.....

gmatacer · Answer

Got it. Thanks alot swaroh for you explanation.

I'll send you my email add professor.

GmatAcer!

This is a repaste of a problem posted by gmatacer: There

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