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There are five different pairs of boots in the box. Find the

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VP
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There are five different pairs of boots in the box. Find the [#permalink] New post 02 Oct 2007, 10:10
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A
B
C
D
E

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There are five different pairs of boots in the box. Find the probability of two boots being pulled of the box at random are of the same pair?

1*1/9 = 1/9

first - any boot 10/10 = 1

second - his pair out of nine boots = 1/9

the answer is (D)

:)
VP
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 [#permalink] New post 02 Oct 2007, 10:15
What is the probability og two books being standing in adjacent places in a row of ten different books?

(9!*2)/10! = 1/5

9!*2 = ways to arrange ten books when two books together AB & BA hence multiply by two.

10! = ways to arrange ten books.

the answer is (B)

:)
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Re: Probability PS [#permalink] New post 02 Oct 2007, 12:11
OlgaN wrote:
1.There are five different pairs of boots in the box. Find the probability of two boots being pulled of the box at random are of the same pair?

A 1\10
B 1\45
C 1\90
D 1\9
E 3\20



Total No of boots = 2*5 = 10
No of ways of selecting 1 pair out of 5 pairs = 5C1 = 5
Total No of ways of selecting 2 boots out of 10 boots = 10C2 = 45

So Probability of selecting two boots of same pair = 5/45 = 1/9

Hence D.

- Brajesh
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Re: Probability PS [#permalink] New post 02 Oct 2007, 12:13
OlgaN wrote:
1.There are five different pairs of boots in the box. Find the probability of two boots being pulled of the box at random are of the same pair?

A 1\10
B 1\45
C 1\90
D 1\9
E 3\20


Take any one boot out. Now out of 9 boots, there is only one which will form pair.
Prob = 1/9
Director
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Re: Probability PS [#permalink] New post 02 Oct 2007, 12:15
OlgaN wrote:
2.What is the probability og two books being standing in adjacent places in a row of ten different books?

A 1\10
B 1\5
C 1\100
D 1\9
E 2\45

Thanks.


There are 10 books.
We can arrange total of 10 books in 10! ways.

Now, 2 books adjacent to each other can be arranged in 2 ways.
2 books together and rest of 8 books can be arranged in 9! ways.

Therefore total no of favourable ways of arrangement = 2 * 9!
Total ways of arrangement = 10!

So Required probability = 2*9!/10! = 2/10 = 1/5

Hence, B.

- Brajesh
Re: Probability PS   [#permalink] 02 Oct 2007, 12:15
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