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

Question

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: https://www.GMATinsight.com

tinytiger · Answer

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.
Answer: 1/3 (B)

RLokesh · Answer

GMATinsight ,  Bunuel ,  chetan2u

Kindly help with solving this?

chetan2u · Answer

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

Princ · Answer

OA: B horse probability.PNG 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 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}

GMATinsight · Answer

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

Answer: option B

ScottTargetTestPrep · Answer

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.

Answer: B

bumpbot · Answer

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards