Total=12 balls

Blue ball can be selected in 5C1 =5 ways

Probability of drawing 1 blue ball from 12 balls=5/12

Green ball can be selected in 4C1 =4 ways

Probability of drawing 1 green ball from 11 balls=4/11


Yellow ball can be selected in 3C1 =3 ways

Probability of drawing 1 yellow ball from 10 balls=3/10

5/12*4/11*3/10

=1/22

Hope this clarify

paridhi wrote:Three balls aee drawn (w/o replacement) from an urn containing 5 Blue balls, 4 Green balls and 3 yellow balls. What is the probability of drawing a blue ball, green ball and a yellow ball in that order?

A. 1/220

B. 1/60

C. 5/144

D. 1/22

E. 47/60

Can someone please explain....

We can also just use simple probability to solve - no need for combinations.

Going to our basic formula:

Probability = (# of desired outcomes)/(total # of possibilities)

and applying another important rule:

when you calculate the probability of MULTIPLE events occurring, you MULTIPLY the individual probabilities.

So, since we want the 3 balls in order:

Prob blue, then green, then yellow = (5/12) * (4/11) * (3/10)

(bunch of cancelling out on top and bottom)

= 1/22

thanks..

i mistakenly took each probability as 5/12, 4/12 & 3/12 but since each step will reduce the number of balls this is incorrect.

thanks for explaining

5/12*4/11*3/10=(5*4*3)/(12*11*10)=1/22

I wish all the questions on GMAT were this easy

Its a D , the important thing being that one must reduce the no of balls at every juncture.
1/22.
rgds

Aditya

the answer to this question is 1/22 option D.i found out the probability of drawing 1 blue ball from 12 that is 5/12 then 1 green ball from 11 balls left that is 4/11 and then 1 yellow ball from 10 left 1/10 adding all the probabilities gives the order of drawing in which the balls should be drawn

Opening this old thread...

Is the answer to this question 1/22 or 1/(6*22)
1. 1/22 will be the probability of drawing 1 blue, 1 green and 1 yellow ball, but they can be drawn in any order BGY, BYG, YBG, YGB, GBY or GYB.
2. The probability of drawing blue, green and yellow ball in this order will be 1/(6*22). And this is what I think the question asked.
Please correct me if I misunderstood something

I don't think its asks for the 3 colors in any order, it does specify a particular order (Blue/Green and then Yellow)

If it asked for all 3 in any order, we would need to do them individually for the 6 cases you have mentioned and add the six cases. And also probability of getting them in any order will be higher, don't see why you divide it by 6, which infact makes it much lower than 1/22 which is Prob for a particular order.

gmatclubmember wrote:Opening this old thread...

Is the answer to this question 1/22 or 1/(6*22)
1. 1/22 will be the probability of drawing 1 blue, 1 green and 1 yellow ball, but they can be drawn in any order BGY, BYG, YBG, YGB, GBY or GYB.
2. The probability of drawing blue, green and yellow ball in this order will be 1/(6*22). And this is what I think the question asked.
Please correct me if I misunderstood something

shankar.ashwin wrote:I don't think its asks for the 3 colors in any order, it does specify a particular order (Blue/Green and then Yellow)

If it asked for all 3 in any order, we would need to do them individually for the 6 cases you have mentioned and add the six cases. And also probability of getting them in any order will be higher, don't see why you divide it by 6, which infact makes it much lower than 1/22 which is Prob for a particular order.

gmatclubmember wrote:Opening this old thread...

Is the answer to this question 1/22 or 1/(6*22)
1. 1/22 will be the probability of drawing 1 blue, 1 green and 1 yellow ball, but they can be drawn in any order BGY, BYG, YBG, YGB, GBY or GYB.
2. The probability of drawing blue, green and yellow ball in this order will be 1/(6*22). And this is what I think the question asked.
Please correct me if I misunderstood something

I am sorry ashwin, I didnt really understood your point
But anyways the question says... "What is the probability of drawing a blue ball, green ball and a yellow ball in that order? " So the order of the balls should be BGY. Now since there can be 6 possibilities of drawing 1 B, 1 G and 1 Y balls, so probability of B then G and then Y should be (probability of drawing 1B, 1G and 1Y), divided by 6.

Maybe I am messing up something big time

Total balls - 12.

1st pick ( I need a blue) P(Blue) = No of blue balls/ Total = 5/12

At this point I have a blue ball for sure as my first pick.

2nd Pick : P(Green) = N(Green)/Total-1 = 4/11

Again, now I have a blue first and the green second.

3rd pick: P(yellow) = N(yellow)/Total - 2 = 3/10

So, P(BGY) in that order = 5/12 * 4/11 * 3/10 = 1/22.

So the order of the balls should be BGY. Now since there can be 6 possibilities of drawing 1 B, 1 G and 1 Y balls, so probability of B then G and then Y should be (probability of drawing 1B, 1G and 1Y), divided by 6.

Since we consider only a particular order here, we won't have 6 ways of getting BGY, we have only one way of getting that order, BGY can be arranged among themselves in 3! = 6 ways, but that would consider all arrangements you have mentioned. (BYG,YGB etc)

The difference is we already account for the order by doing this. If that was not the case, the first ball we pick up would be any one of the 3 colors, so probability should be 1 because it will certainly be a B,G or a Y. But we do 5/12 for the first because we only want a blue ball as our first pick.

Get it? I am not sure if I explain it well here.

Thanks bro... Got it.
Vacation hangover .. you see:))

5*4*3/12*11*10 => 1/22
Option D

The probability problem is without replacement type and also the order of selection is specified.That makes it easey.

5/12*4/11*3/10= 1/22