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# In a race of 6 horses A, B, C, D, E and F, what is the probability of

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In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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01 Apr 2018, 01:13
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85% (hard)

Question Stats:

40% (02:07) correct 60% (02:17) wrong based on 82 sessions

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In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.

A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15

Source: www.GMATinsight.com

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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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01 Apr 2018, 01:40
Only 2 out of 6 scenarios work: ABC, ACB
Total number of ways to arrange 3 horses ABC within themselves = 3! = 6 ways. Probability of fulfiling question stem requirements = 1/3.
Total number of arrangements possible = 6! ---> Only 1/3 of the time, do the cases fulfil the requirements.
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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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16 Apr 2018, 08:24
GMATinsight, Bunuel, chetan2u

Kindly help with solving this?
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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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16 Apr 2018, 09:49
RLokesh wrote:
GMATinsight, Bunuel, chetan2u

Kindly help with solving this?

Hi...
6 horses out of which we are interested in placing three horses A,B and C..
Rest all can take any of the three..

When these three are placed, A should be ahead of B and C...
So either ABC or ACB...so 2 ways
Total ways these three can be placed =3*2*1=6

Probability=2/6=1/3
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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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16 Apr 2018, 11:12
1
GMATinsight wrote:
In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.

A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15

Source: http://www.GMATinsight.com

OA: B
Attachment:

horse probability.PNG [ 6.13 KiB | Viewed 884 times ]

Case 1 is self explainatory.
In case 2 , as A is in 2nd position,also B and C are behind it. We have only 3 [6(total number of horse)- 3 ( A,B and C)] horses to fill 1st position.
Similarily, cases 3 and 4 can be explained .

We will have 240 cases when A will be ahead of B and C.

Total number of cases ( no condition of any particular horse being in front of another) =6! = 6*5*4*3*2*1 = 720

Probability of A finishing in front of B and C = $$\frac{240}{720}$$= $$\frac{1}{3}$$
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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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20 Apr 2018, 00:49
1
1
GMATinsight wrote:
In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.

A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15

Source: http://www.GMATinsight.com

METHOD-1:

If A comes at 1st position i.e. Positions are A - - - - - then B can take position in 5 ways and C can take position in 4 ways and rest in 3! ways = 5*4*3!

If A comes at 2nd position i.e. Positions are - A - - - - then B can take position in 4 ways and C can take position in 3 ways and rest in 3! ways = 4*3*3!

If A comes at 3rd position i.e. Positions are - - A - - - then B can take position in 3 ways and C can take position in 2 ways and rest in 3! ways = 3*2*3!

If A comes at 4th position i.e. Positions are - - - A - - then B can take position in 2 ways and C can take position in 1 ways and rest in 3! ways = 2*1*3!

Total Favourable ways of arrangements = 5*4*3! +4*3*3! +3*2*3! + 2*1*3!= 3! (20+12+6+2) = 6*40 = 240
Total Arrangements of 6 horses = 6! = 720

Probability = 240/720 =1/3

METHOD-2:

Select 3 places for A B and C out of 6 positions in 6C3 = 20 ways

Arrange A, B and C on selected position in 1 (For A at first of three selected positions) *2*1(for B and C) = 2 ways

Arrange remaining three on three remaining places in 3! = 6 ways

Total Favourable ways of arrangements of 6 horses = 6C3* (1*2*1)*(3!) = 20*2*6 = 240
Total Arrangements of 6 horses = 6! = 720

Probability = 240/720 =1/3

METHOD-3:
A has as much probability of being ahead of B and C as B has to be ahead of A and C as C has to be ahead of A and B therefore each one (of A, B and C) has equal probability to be winner among themselves

Hence probability of each horse (of A, B and C) to be ahead of other two = 1/3

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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of  [#permalink]

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23 May 2019, 17:20
GMATinsight wrote:
In a race of 6 horses A, B, C, D, E and F, what is the probability of A finishing ahead of B and C both? There are no ties between any two horses.

A) 1/2
B) 1/3
C) 1/6
D) 1/12
E) 1/15

Source: http://www.GMATinsight.com

Since we only care about A finishing ahead of B and C, we can ignore the other horses. If we only have three horses A, B and C, the ordering of the 3 horses can be:

ABC, ACB, BAC, BCA, CAB, CBA

We see that of the 6 orderings, only (the first) 2 have A ahead of B and C, therefore, the probability is 2/6 = 1/3.

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Re: In a race of 6 horses A, B, C, D, E and F, what is the probability of   [#permalink] 23 May 2019, 17:20
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