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

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NoHalfMeasures
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Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  20 Dec 2015, 18:40
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Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly selects 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
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  20 Dec 2015, 18:54
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NoHalfMeasures wrote:
Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly selects 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


Can be tackled in this way as well: Probability of selecting any 1 out of 10 shoes = 10/10 = 1

Probability of selecting the next shoe (out of 9 available) having the same color = 1/9 (as after selecting the 1st one, there is only 1 another shoe left with the same color).

Thus the total probability = 1*1/9=1/9.

C is the correct answer.
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  07 Dec 2017, 10:19
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NoHalfMeasures wrote:
Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly selects 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. (see below)

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

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NoHalfMeasures
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  20 Dec 2015, 18:41
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Total pairs = 10C2 = 45; Same Color Pairs= 5C1*1C1=5; Prob = 1/9
Or 2/10 * 1/9 * 5 = 1/9

Ans C
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  07 Dec 2017, 10:17
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NoHalfMeasures wrote:
Kim has 5 pairs of shoes; each pair is a different color. If Kim randomly selects 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
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karan12345
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  01 Jul 2019, 09:30
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Probability of selecting 1 out of 10 shoes = 10/10 = 1
Probability of selecting the next shoe having the same color = 1/9 (as only 1 shoe is left of the same color and 9 shoes are left after selecting one)
So, Probability that she will select 2 shoes of the same color = 1 * 1/9 = 1/9
Hence, Answer choice is (C).
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  30 Sep 2020, 02:46
Can someone please help me with what is wrong with my approach:

Picking up first shoe: 1/5 (we can pick up any one of the shoe from a pair)
Picking up the 2nd shoe: 1/9 (since only one shoe of the pair is left to be picked up from the remaining 9 shoes)

Reqd Probability = 1/5 * 1/9 = 1/45
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BrentGMATPrepNow
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  30 Sep 2020, 04:56
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ProfChaos wrote:
Can someone please help me with what is wrong with my approach:

Picking up first shoe: 1/5 (we can pick up any one of the shoe from a pair)
Picking up the 2nd shoe: 1/9 (since only one shoe of the pair is left to be picked up from the remaining 9 shoes)

Reqd Probability = 1/5 * 1/9 = 1/45


We can pick ANY shoe for the first shoe. So the P(selecting the first shoe) = 1 (not 1/5)

If we say P(selecting the first shoe) = 1/5, then we're also saying that P(NOT selecting the first shoe) = 4/5, which makes no sense.
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Re: Kim has 5 pairs of shoes; each pair is a different color [#permalink] New post  30 Sep 2020, 06:46
BrentGMATPrepNow wrote:
ProfChaos wrote:
Can someone please help me with what is wrong with my approach:

Picking up first shoe: 1/5 (we can pick up any one of the shoe from a pair)
Picking up the 2nd shoe: 1/9 (since only one shoe of the pair is left to be picked up from the remaining 9 shoes)

Reqd Probability = 1/5 * 1/9 = 1/45


We can pick ANY shoe for the first shoe. So the P(selecting the first shoe) = 1 (not 1/5)

If we say P(selecting the first shoe) = 1/5, then we're also saying that P(NOT selecting the first shoe) = 4/5, which makes no sense.


Oh yes!
I understood what you are saying.

So I can pick any shoe in the first pick and thus prob is 1
But the second pick has to be the the other shoe of the pair, hence prob of picking up a second shoe so that both form a pair is 1/9

I had read your explanation before as well but I dont know why, at that time it didnt click with me.
Anyway, thank you so much for your reply BrentGMATPrepNow
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