kim has 5 pairs of shoes; each pair is a different color...

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

kim has 5 pairs of shoes; each pair is a different color...

by AAPL » Thu Dec 07, 2017 10:05 am

Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly select 2 shoes without replacement from the 10 shoes, what is the probability that she will select 2 shoes of the same color?

(A) 2/5
(B) 1/5
(C) 1/9
(D) 1/10
(E) 1/25

The OA is C.

I don't understand why that is the correct answer, I'm a little bit confused.

The first time that she select any shoes, the probability should be 2/10? I'm confused. Then in the second time that she select any shoes the probability should be 1/9? I'm not sure.

Experts, I need your help with this PS question. Thanks.

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Source: Beat The GMAT — Problem Solving | Thu Dec 07, 2017 10:05 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

kim has 5 pairs of shoes; each pair is a different color...

by Brent@GMATPrepNow » Thu Dec 07, 2017 10:10 am

AAPL wrote:Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly select 2 shoes without replacement from the 10 shoes, what is the probability that she will select 2 shoes of the same color?

(A) 2/5
(B) 1/5
(C) 1/9
(D) 1/10
(E) 1/25

P(matching pair) = P(select ANY shoe for 1st selection AND select matching shoe for 2nd selection)
= P(select ANY shoe for 1st selection) x P(select matching shoe for 2nd selection)
= 1 x 1/9
= 1/9
= C

ASIDE: Once we have selected ANY shoe as the 1st selection, there are 9 shoes remaining. Of those 9 remaining shoes, only 1 matches the first shoe (thus the 1/9)

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Thu Dec 07, 2017 10:17 am

AAPL wrote:Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly select 2 shoes without replacement from the 10 shoes, what is the probability that she will select 2 shoes of the same color?

(A) 2/5
(B) 1/5
(C) 1/9
(D) 1/10
(E) 1/25

Here's an approach that uses counting techniques

P(matching pair) = (number of ways to get a matching pair)/(TOTAL number of ways to select 2 shoes)

number of ways to get a matching pair
There are 5 different colors.
So, there are 5 different ways to get a matching pair

TOTAL number of ways to select 2 shoes
There are 10 shoes altogether.
Since the order in which we select the 2 shoes does not matter, we can use combinations.
We can select 2 shoes from 10 shoes in 10C2 ways
10C2 = 45

Aside: If anyone is interested, we have a video on calculating combinations (like 10C2) in your head: [url] https://www.gmatprepnow.com/module/gmat- ... /video/789

So, P(matching pair) = (5)/(45)
= 1/9
= C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Fri Oct 11, 2019 2:39 pm

AAPL wrote:Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly select 2 shoes without replacement from the 10 shoes, what is the probability that she will select 2 shoes of the same color?

(A) 2/5
(B) 1/5
(C) 1/9
(D) 1/10
(E) 1/25

The OA is C.

The probability that Kim will select two shoes of the same color is:

2/10 x 1/9 x 5 = 1/45 x 5 = 1/9

Alternate solution:

Since the first shoe she picks does not matter, then the probability of picking the first shoe is 10/10 = 1. However, since only 1 shoe is left to match the color of that shoe, the probability of picking the second shoe matching the color of the first one is 1/9. Therefore, the probability that Kim will pick two shoes of the same color is 1 x 1/9 = 1/9.

Answer: C

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