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by manik11 » Mon Dec 21, 2015 6:34 am
A bag contains 15 wool scarves, exactly one of which is red and exactly one of which is green. If Deborah reaches in and draws three scarves, simultaneously and at random, what is the probability that she selects the red scarf but not the green scarf?

A) 2/35

B) 1/15

C) 6/35

D) 13/70

E) 1/5

OA : C
Source : Veritas Prep

My approach : If we remove the green scarf from the bag, it would leave 14 scarves. Now Deborah has 14 scarves to draw from.
Lets pick 1 RED scarf out (to satisfy the condition in the question). This leaves 13 scarves in the bag and she has to draw 2 (13 C 2)
So Probability would be (13 C 2) / (15 C 3)

Experts..Could you guys please help me figure out the problem with my approach here. I think I'm missing something fundamental here.
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by GMATGuruNY » Mon Dec 21, 2015 7:22 am
manik11 wrote:A bag contains 15 wool scarves, exactly one of which is red and exactly one of which is green. If Deborah reaches in and draws three scarves, simultaneously and at random, what is the probability that she selects the red scarf but not the green scarf?

A) 2/35

B) 1/15

C) 6/35

D) 13/70

E) 1/5

OA : C
Source : Veritas Prep

My approach : If we remove the green scarf from the bag, it would leave 14 scarves. Now Deborah has 14 scarves to draw from.
Lets pick 1 RED scarf out (to satisfy the condition in the question). This leaves 13 scarves in the bag and she has to draw 2 (13 C 2)
So Probability would be (13 C 2) / (15 C 3)

Experts..Could you guys please help me figure out the problem with my approach here. I think I'm missing something fundamental here.
Your solution is perfect.
13C2 = (13*12)/(2*1) = 13*6.
15C3 = (15*14*13)/(3*2*1) = 5*7*13.
Thus:
(13C2)/(15C3) = (13*6)/(5*7*13) = 6/35.

The correct answer is C.
Last edited by GMATGuruNY on Tue Dec 22, 2015 5:47 am, edited 1 time in total.
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by [email protected] » Mon Dec 21, 2015 3:17 pm
Hi manik11,

Many permutation, combination and probability questions can be solved using slightly different versions of the same type of math. Thus, you should select whichever method is most comfortable for you.

Beyond the approach that you used, you can also solve this prompt with a slightly longer permutation-based approach:

According to the prompt, we have 1 red scarf, 1 green scarf and 13 other scarfs. We're asked for the probability of pulling 3 scarves that include the red scarf but NOT the green scarf. Selecting one scarf at a time, there would be 3 'options' that would fit this goal:

1st red, 2nd non-green, 3rd non-green = (1/15)(13/14)(12/13)

1st non-red & non-green, 2nd red, 3rd non-green = (13/15)(1/14)(12/13)

1st non-red & non-green, 2nd non-red & non-green, 3rd red = (13/15)(12/14)(1/13)

You'll notice that all 3 calculations are the SAME calculation, so we just have to calculate it once, then multiply it by 3...

(1/15)(13/14)(12/13) = (13)(12)/(15)(14)(13) = 12/(15)(14) = 6/(15)(7) = 2/(5)(7) = 2/35

3[2/35] = 6/35

Final Answer: C

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by Shazi1711 » Tue Dec 22, 2015 5:43 am
13C2 = (13*12)/(3*2) = 13*6.


Mitch,

Shouldn't this be (13*12)/(2*1)
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by GMATGuruNY » Tue Dec 22, 2015 5:49 am
Shazi1711 wrote:13C2 = (13*12)/(3*2) = 13*6.


Mitch,

Shouldn't this be (13*12)/(2*1)
Good catch.
I've corrected the typo.
Thanks for pointing it out.
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