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

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Kim has 5 pairs of shoes; each pair is a different color  [#permalink]

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New post 20 Dec 2015, 19:40
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A
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C
D
E

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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]

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New post 20 Dec 2015, 19:41
1
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]

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New post 20 Dec 2015, 19:54
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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]

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New post 07 Dec 2017, 11:17
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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)

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Re: Kim has 5 pairs of shoes; each pair is a different color  [#permalink]

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New post 07 Dec 2017, 11:19
Top Contributor
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

RELATED VIDEO

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Re: Kim has 5 pairs of shoes; each pair is a different color &nbs [#permalink] 07 Dec 2017, 11:19
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