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by him1985 » Mon Mar 12, 2012 3:31 am
1/22 is the right answer...

5/12*4/11*3/10 = 1/22
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by ronnie1985 » Fri Mar 30, 2012 7:21 am
(5/12)*(4/11)*(3/10) = 1/22
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by Lifetron » Fri Jul 27, 2012 9:57 pm
(5/12)*(4/11)*(3/10)=1/22

D !
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by Ganesh hatwar » Thu Aug 16, 2012 11:14 pm
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

OA is D
Can someone please explain....
P = Favorable number of ways/ Total number of ways

Here order of selection matters


So,

P= 5c1/12c1 * 4c1/11c1 * 3c1/10c1 = 1/22
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by chris558 » Thu Aug 23, 2012 7:30 am
5 blue
4 green
3 yellow

= 12 total

probability of drawing a blue ball first= 5/12
without replacement- there are now 11 total balls

probability of drawing a green ball next= 4/11
without replacement- there are now 10 total balls

probability of drawing a yellow ball last= 3/10

now, we multiply these probabilities together because we want these to know what happens when all circumstances are fulfilled.

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

ANSWER IS D
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by coolhabhi » Thu Aug 23, 2012 10:19 pm
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

OA is D
Can someone please explain....
Why isnt the answer : (5C1 * 4C1 * 3C1)/12C3 ??
Experts please help..
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by chris558 » Fri Aug 24, 2012 5:57 am
coolhabhi wrote:
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

OA is D
Can someone please explain....
Why isnt the answer : (5C1 * 4C1 * 3C1)/12C3 ??
Experts please help..
I'm certainly no expert, but here goes...

Remember this RULE:
To determine the probability that event X AND event Y will both occur, MULTIPLY the two probabilities together.

What is the probability of each event happening without replacement?
Well, probability= number of successful outcomes/total number of possible outcomes.

There are a TOTAL number of 12 balls.
1) blue ball: 5/12
There are 5 blue balls (therefore 5 chances of picking one of them) and 12 total balls.
2) green: 4/11
There are now a pool of 11 balls because we had already taken one out of the original pool of balls
3) yellow: 3/10

Multiplying these together... 5/12*4/11*3/10=1/22

Notice that it doesn't matter what order you pick the balls in. Your outcome will still be the same
For example: yellow green blue
3/12*4*11*5/10=1/22
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by mparakala » Tue Dec 11, 2012 7:32 pm
1/22

D
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by Kshitij Sharma » Thu Jan 10, 2013 12:09 am
[5/12]*[4/11]*[3/10]=1/22

Ans. D
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by rajeshsinghgmat » Mon Mar 04, 2013 12:12 am
(D) 1/22
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by ganeshrkamath » Thu Aug 15, 2013 7:16 pm
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

OA is D
Can someone please explain....
5B 4G 3Y

Probability of drawing a blue ball first = 5/(5+4+3) = 5/12
Probability of drawing a green ball next = 5/12 * 4/(4+4+3) = 5/12 * 4/11
Probability of drawing a yellow ball next = 5/12 * 4/11 * 3/(4+3+3) = 5/12 * 4/11 * 3/10
= 1/22

Choose D

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by Java_85 » Fri Sep 27, 2013 8:24 am
IMO D
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by theCodeToGMAT » Fri Sep 27, 2013 8:59 am
I know the solution is (5/12)*(4/11)*(3/10) = 1/22

However, if i try to solve the question using : (5C1 x 4C1 x 3C1)/12C3 then the answer is different.. What is the logical mistake?
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by jaspreetsra » Wed Nov 26, 2014 11:43 am
Three balls are 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

P(blue ball) = 5/12
P(green ball) = 4/11
P(yellow ball) = 3/10
P(a blue ball, green ball and a yellow ball) = 5/12 * 4/11 * 3/10 = 1/22
Answer: D
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by GMATinsight » Thu Nov 27, 2014 5:08 am
theCodeToGMAT wrote:I know the solution is (5/12)*(4/11)*(3/10) = 1/22

However, if i try to solve the question using : (5C1 x 4C1 x 3C1)/12C3 then the answer is different.. What is the logical mistake?
Because the order is given BYG therefore selection method would require the denominator to be multiplied by 3! as in total outcomes, all the orders must be taken into account

The correct answer of the question should be
(5C1 x 4C1 x 3C1)/12C3x3! = 60 / 220X6 = 6/22 = 1/22

The answer Option D is CORRECT

The first method 5/12 * 4/11 * 3/10 = 1/22 Doesn't require multiplication of 3! in numerator for the same reason as the ordering of colors is fixed and the order of colors does NOT need to be taken into account {BGY, BYG, GBY, GYB, YBG, YGB} which can be arranged in 3! ways therefore alternate correct method would be

5/12 * 4/11 * 3/10 = 1/22
Last edited by GMATinsight on Fri Nov 28, 2014 5:43 am, edited 1 time in total.
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