Three houses, labeled in order A, B, and C, are equally spaced on a st

Question

Three houses, labeled in order A, B, and C, are equally spaced on a street in which five children live. There is at least one child in each house. The children plan to meet for a play-date and decide to meet in the house that minimizes the total distance traveled by the five children. What is the probability that the children meet at house B?

A. 1/2

B. 3/5

C. 2/3

D. 3/4

E. 4/5

Show Answer

Official Answer

C

A. 1/2

B. 3/5

C. 2/3

D. 3/4

E. 4/5

gmatophobia · Answer

HarryGeorge - Not sure if this is a GMAT question as the number of options is only 4. Nevertheless, it's an interesting question. Given : A total of five children live in three houses. Each house has at least 1 child. The houses are in order A, B, and C and are equally spaced . Inference : The possible arrangement can be 2 - 2 - 1 (i.e. two houses with two children each and one house with one child) OR 3 - 1 - 1 (i.e. two houses with one child each and one house with three children). Let's assume that each house is one unit apart. So the placement of the houses is as follows - ------ A -<--1 unit-->- B -<--1 unit-->- C ------ Case 1 : 3 - 1 - 1 (Two houses have one child and one house has three children) In this arrangement, to minimize the distance the play date has to be arranged in the house which has three children. Hence, for the play date to occur at house B, that house should have three children. The probability that three children are at house B = 1/3 (i.e. 3 children can be at house A, house B, or house C) Total Probability for this case to occur = 1/2 * 1/3 = 1/6 We have multiplied by 1/2 as this arrangement ( 3 - 1 - 1) is one of the two possible arrangements. Case 2 : 2 - 2 - 1 (Two houses have one child and one house has three children) In this arrangement, either house B has two children or has one child. Let's assume that house A and house B have two children each and house C has one child. In that case, the arrangement would be as follows - ------ A (2) -<--1 unit-->- B (2) -<--1 unit-->- C (1) ------ If the play date was arranged in house A - Two children from house B will have to travel to house 'A': Total distance traveled by children of house B = 2 units One child from house C will have to travel to house 'A': Total distance traveled by the child of house C = 2 units Total distance traveled by three children = 4 units The distance traveled will be the same if the play date was arranged in house C instead of house A. If the play date was arranged in house B - Two children from house A will have to travel to house 'B': Total distance traveled by children of house A = 2 units One child from house C will have to travel to house 'B': Total distance traveled by the child of house C = 1 unit Total distance traveled by three children = 3 units So if house B has two children, the play date will be arranged in house B. Let's assume that house A and house C have two children each and house B has one child. In that case, the arrangement would be as follows - ------ A (2) -<--1 unit-->- B (1) -<--1 unit-->- C (2) ------ If the play date was arranged in house A - Two children from house C will have to travel to house 'A': Total distance traveled by children of house B = 2*2 = 4 units One child from house B will have to travel to house 'A': Total distance traveled by the child of house B = 1 unit Total distance traveled by three children = 5 units The distance traveled will be the same if the play date was arranged in house C instead of house A. If the play date was arranged in house B - Two children from house A will have to travel to house 'B': Total distance traveled by children of house A = 2 units Two children from house A will have to travel to house 'B': Total distance traveled by the child of house C = 2 unit Total distance traveled by three children = 4 units Hence, even in this case to minimize the distance, the play date should be arranged in house B. In a nutshell in a case when two houses have two children each and one house has one child, it makes sense to have the play date in house B. Probability of this occurring = 1/2 * (1) = 1/2 Total Probability = \frac{1}{2} +\frac{ 1}{6} = \frac{3 + 1 }{ 6} = \frac{4}{6} = \frac{2}{3} Option C

Kinshook · Answer

Given:  
Three houses, labeled in order A, B, and C, are equally spaced on a street in which five children live. 
There is at least one child in each house.
The children plan to meet for a play-date and decide to meet in the house that minimizes the total distance traveled by the five children.

Asked:  What is the probability that the children meet at house B?

Total cases = 3C2 + 3C1 = 6

Case 1: - 
A(3), B(1), C(1) 
Meet at A.

Case 2 & 3:-
A(2),B(2),C(1) & A(1), B(2), C(2)
Meet at B

Case 4: -
A(2),B(1),C(2) 
Meet at B

Case 5: 
A(1), B(1), C(3)
Meet at C

Case 6:
A(1), B(3), C(1)
Meet at B

The probability that the children meet at house B = 4/6 = 2/3

IMO C

HarshaBujji · Answer

Over all combinations. 6, As we know how to divide m similar items in n boxes with at least one. .

There are 2 possibilities where meeting can happen at A or C (Extreme ends)

So P = (6-2)/6 = 2/3

DelademAnku · Answer

how is the total combinatons 6 and how do you have 4/6?

sdp4 · Answer

We define the question mathematically:
Pr(Playdate at B) = (Cases of playdate at B) / (All possible cases)
We solve for the numerator:
(Cases of playdate at B) = 2c1 (2 children to choose from, one place they can be (at B)) = 2

We solve for the denominator:
(All possible cases) = 3c2 (3 houses to choose from, 2 children to place) = 3

Hence, Pr(Playdate at B) = (Cases of playdate at B) / (All possible cases) = 2/3.

C

hr1212 · Answer

Each house requires at least one child, so after distributing 3 children out of 5, we have 2 remaining children.

Total ways to distribute 2 children among 5 houses => a + b + c = 2

Can be calculated via (n + r - 1)C(r - 1) for x1 + x2 + ... + xr = n

(2 + 3 - 1)C(3 - 1) = 4C2 = 6

To minimize the distance, both the children cannot be either in 'a' or 'c' ie. (2 0 0 or 0 0 2), rest of the 4 options will make 'b' as the optimal choice.

P = Favorable outcomes/Total outcomes = 4/6 = 2/3

IMO: C

Shaunak3004 · Answer

Not sure if this approach is right, but I got the right answer somehow:

We arrange 3 children in 3 houses out of 5 in 5C3 ways = 10.

2 children remain, Probability of the meet happening in house B:
1. 1 child in house A & B each= 2C1*1 = 2
2. 1 child in house A & C each= 2C1*1 = 2
3. 1 Child in house B & C each = 2C1*1 = 2
4. Both children in house A = 1
5. Both children in house B = 1
6 Both children in house C = 1.

Total Possibilities for 5 children = 5!

Therefore, (10*2*2*2*1*1*1)/(5*4*3*2*1) = 2/3

Correct me if I've made a mistake anywhere

Dooperman · Answer

Children can be divided in:
2 1 2 = 3 ways (3!/2) 
or 
3 1 1 = 3 ways

Total 6 ways.
In 1st case, children will meet at B in all 3 cases 212; 122; 221
In 2nd cade children will meet at B in only 1 case ie. 313.

p= 4/6 = 2/3

Three houses, labeled in order A, B, and C, are equally spaced on a st

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

Three houses, labeled in order A, B, and C, are equally spaced on a st

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