Probability

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sud21: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sat Nov 12, 2011 3:21 amFollowed by: 2 members; Thanked: 1 times

by sud21 » Mon Sep 28, 2015 5:00 pm

Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

4/7

1/2

27/70

2/7

9/35

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Source: Beat The GMAT — Problem Solving | Mon Sep 28, 2015 5:00 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Sep 28, 2015 5:29 pm

sud21 wrote:Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

4/7

1/2

27/70

2/7

9/35

P = (good combinations)/(all possible combinations).

All possible combinations:
From 8 shirts, the number of ways to choose 4 = 8C4 = (8*7*6*5)/(4*3*2*1) = 70.

Good combinations:
In a good combination, 3 non-blue shirts are packed with the pink shirt.
From the 6 non-pink and non-blue shirts, the number of ways to choose 3 to be packed with the pink shirt = 6C3 = (6*5*4)/(3*2*1) = 20.

Thus:
(good combinations)/(all possible combinations) = 20/70 = 2/7.

The correct answer is D.

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vishalwin: Master | Next Rank: 500 Posts; Posts: 274; Joined: Fri Sep 18, 2015 10:58 pmFollowed by: 1 member; Thanked: 12 times; GMAT Score:530

by vishalwin » Wed Sep 30, 2015 1:44 am

Hi Mitch,

I have a doubt here.

I got that out 6 non- blue and non-pink we are selecting the 3 shirts.

But now we have to select 1 pink shirt as well. So 2 left shirts are blue and pink so either of them can be selected i.e. either pink or blue.

Can you please clear what mistake I am making here.

Thanks & Regards,
Vishal Chopra

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed Sep 30, 2015 3:10 am

vishalwin wrote:Hi Mitch,

I have a doubt here.

I got that out 6 non- blue and non-pink we are selecting the 3 shirts.

But now we have to select 1 pink shirt as well. So 2 left shirts are blue and pink so either of them can be selected i.e. either pink or blue.

Can you please clear what mistake I am making here.

Thanks & Regards,
Vishal Chopra

A good combination is composed to two different types of shirts:
1 PINK SHIRT and 3 NON-PINK, NON-BLUE SHIRTS.

From the 8 shirts, the number of ways to choose 1 pink shirt = 1. (Since there is only 1 pink shirt.)
From the 6 non-pink and non-blue shirts, the number of ways to choose 3 to be packed with the pink shirt = 6C3 = 20.
To combine these options, we multiply:
1*20 = 20.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8188; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Mon Jul 23, 2018 5:13 pm

sud21 wrote:Before leaving for his business trip, Chad asks his assistant to choose and pack four shirts from his closet, which currently contains eight shirts. If each shirt is a different color, including one blue shirt and one pink shirt, and the assistant chooses the shirts at random, what is the probability that the pink shirt will be one of the four packed but the blue shirt will not?

4/7

1/2

27/70

2/7

9/35

The number of ways to select the pink shirt but not blue shirt is 1C1 x 1C0 x 6C3 = (6 x 5 x 4)/3! = 20. (Notice that 1C1 is for the pink shirt to be selected, 1C0 is for the blue shirt not to be selected, and 6C3 is the number of ways to select the 3 shirts from the remaining 6 shirts.)

The number of ways to select 4 shirts from 8 shirts is:
8C4 = 8!/(4! x 4!) = (8 x 7 x 6 x 5)/4! = (8 x 7 x 6 x 5)/(4 x 3 x 2) = 7 x 2 x 5 = 70

Thus, the probability that the pink shirt will be one of the four packed but the blue shirt will not is 20/70 = 2/7.

Answer: D

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