Diana is going on a school trip along with her two brothers,

Question

Diana is going on a school trip along with her two brothers, Bruce and Clerk. The students are to be randomly assigned into 3 groups, with each group leaving at a different time. What is the probability that DIana leaves at the same time as AT LEAST on her bothers?

A. 1/27
B. 4/27
C. 5/27
D. 4/9
E. 5/9

A. 1/27
B. 4/27
C. 5/27
D. 4/9
E. 5/9

Friends,
the explanation is given in the princeton review book, but i am not able to follow their explanation.

Please comment and explain

Bunuel · Accepted Answer

Diana and her two brothers can be assigned to one of the 3 groups, so each has 3 choices, so total # of different assignments of Diana and her 2 brothers to 3 groups is  3*3*3=3^3=27 .
 
Now, "Diana leaves at the same time as AT LEAST one her brothers" means that Diana is in the same group as at least one her brothers.

Let's find the opposite probability of such event and subtract it from 1. Opposite probability would be the probability that  Diana is not in the group with any of her brothers .

In how many ways we can assign Diana and her two brothers to 3 groups so that Diana is not in the group with any of her brothers? If Diana is in the first group, then each of her two brothers will have 2 choices (either the second group or the third) and thus can be assigned in  2*2=2^2=4  ways to other two groups. As there are 3 groups, then total # of ways to assign Diana and her 2 brothers to these groups so that Diana is not in the group with any her bothers is  3*4=12  (For each of Diana's choices her brother can be assigned in 4 ways, as Diana has 3 choices: first, second or the third group, then total  3*4=12 ). So probability of this event is  \frac{12}{27} .

Probability that Diana is in the same group as at least one her brothers would be  1-\frac{12}{27}=\frac{15}{27}=\frac{5}{9} .

Answer: E.

Hope it's clear.

ravsg · Answer

For "atleast" problems, its easier to do (1 - opposite of what is being asked). Here 1 - different groups

# of ways in which Diana can be assigned a group = 3
# of ways in which Bruce can be assigned a group = 2 (since bruce cannot be assigned the same group as diana)
# of ways in which Clerk can be assigned a group = 2 (since clerk cannot be assigned the same group as diana)

=> 1 - (3/3)(2/3)(2/3)
=> 5/9

bhandariavi · Answer

+1 Bunuel . Nice explanation.

sdas · Answer

bunuel, need your help. u r marvellous in ur approach. i went through all of them.i think my problem areas are permutation/probability. i understand though ur approaches. could you explain me, the basic way to approach such problems?the way to understand the independent, mutually exclusive, both combined, permutation/combination approach?meaning hw cn a problem be broken down to understand which approach to apply?

gmat1220 · Answer

I think forward approach also works well. Total ways are 3*3*3=27

The number of cases Diana will be with "one" of her brother= 3*2+3*2

The number of ways all three will be in one team=3*1*1=3

Total fav cases = 6+6+3=15

Probability = fav cases / total cases
= 15/27= 5/9

Posted from my mobile device

mpqr · Answer

Bunuel! 
I got the same answer but I cannot really identify if my approach was a different way to get the same result or just a matter of luck.

I also substract the opposite probability: 1- P(ALL DIFERENT GROUPS)= 1-3C1x3/3x2/3x1/3x3!=5/9

My line of thought: 
3C1: which group will be chose first
3/3: Any of the 3 groups cann be selected
2/3: prob of selecting any of the remainder 2
1/3: prob of selecting the remainder group
3! Number of ways the first selection can be made

Where I am wrong?

Thanks not just for your reply but for your amazing willingness to help.

PiyushK · Answer

why we did 3x3x3 for assignment or distribution ?

why we have not considered cases of 0,1,2,3 ways of distribution of three people in three groups.

n+r-1Cr-1 looks much appropriate to find all possible cases.

5!/2! = 60 ways to distribute them in three groups.

G1 G2 G3
D D D
B B B
C C C

3X3X3 covers cases like DDD or DBD or BBB which is not a possible distribution.

Either I am not able to understand this question properly or answer choices are not correct.

SunthoshiTejaswi · Answer

hi bunnuel,

am struggling with probability and combinations questions can you please suggest me some material which can provide some clarity on the topic.

Bunuel · Answer

Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html Theory on probability problems: math-probability-87244.html All DS probability problems to practice: search.php?search_id=tag&tag_id=33 All PS probability problems to practice: search.php?search_id=tag&tag_id=54 Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html Also, check articles on these topics in our Important Topics Directory . Hope it helps.

Mo2men · Answer

Hi Bunnel,
I'm little struggle to understand why it is 2^2=4  ways, What does the first (2) and the (2) represent? Also why does 3*4 represent?

Thanks

WilDThiNg · Answer

If Diana is in the first group, then each of her two brothers will have 2 choices (either the second group or the third) and thus can be assigned in \(2*2=2^2=4\) ways to other two groups. As there are 3 groups, then total # of ways to assign Diana and her 2 brothers to these groups so that Diana is not in the group with any her bothers is \(3*4=12\) (For each of Diana's choices her brother can be assigned in 4 ways, as Diana has 3 choices: first, second or the third group, then total \(3*4=12\)). So probability of this event is \(\frac{12}{27}\).

Hi - please could you help me understand why do we make such an assumption, and can we take the other assumption i.e. her brothers have 3 choices?

Thanks

Hi - please could you help me understand why do we make such an assumption, and can we take the other assumption i.e. her brothers have 3 choices?

Thanks

Mahmud6 · Answer

Should be AT LEAST  one  her bothers

sobby · Answer

Nice question

My Approach:
total number of way we can assign children to group is 3*3*3=27
timing must be same for Diana with her brother(atleast one)
we have Diana(D),Bruce(B),Clerk(C).
we have 3 possibility D and B will be in one team,D and C will be in one team or all three in a single team
let us assume D and B in a team ..we can assume DB as a single unit --so 3P2 ways.---6 ways
same we assume D and C as a single unit ---so 3P2 ways--6 ways
all in one team-we will assume all as single unit -3P1--3 ways

total =15 ways .
p=15/27=5/9 .

sillyboy · Answer

Good question.

got D.

understood where I went wrong...Nice explanations!

LGIX · Answer

In case it helps someone, I did it in a way that took me less than a minute and no options calculation. Please correct me if it was luck!

First: understand what is being asked. Diana can leave at the same time as one of her brothers if they are both (or all 3) in the same group. So the question is: what is the probability of Diana NOT being alone in her group?

Second: its easier to do 1-P(Diana is alone). P(Diana is alone)= the brothers are in different groups.

Till here everything is the same.
 
 But now, there is no need to calculate number of options or groups or anything like that:
 
Diana is in a group (any) with a probability of 1.

Now the brothers: brother 1 has a probability of 2/3 of NOT being with Diana. Brother 2 has a probability of 2/3 of NOT bieng with Diana. This is so because they can be in 3 groups and Diana is in one of them, so the probability of them NOT being with her is the 2 empty ones out of 3.

\frac{2}{3}*\frac{2}{3}=\frac{4}{9}  is the probability of both brothers NOT being in the same group as Diana.

1-\frac{4}{9}=\frac{5}{9}  is the probability of Diana not being alone.

Please correct me if I did something wrong, but doing  \frac{2}{3}*\frac{2}{3}  seemed a lot faster.

ScottTargetTestPrep · Answer

Solution:

Since Diana and her two brothers can be assigned to any of the 3 groups, each has 3 choices, and therefore, overall there are 3 * 3 * 3 = 27 assignments. If we can determine the number of assignments in which Diana leaves at a different time than both of her brothers, we can determine the number of assignments in which she leaves at the same time as at least one of her brothers.

Diana leaves at a different time than both of her brothers if each of the 3 people is in a different group, or if both of her brothers are in one group and she is in a different group.

Option 1 : Each of the 3 people is in a different group 
 
If we let 1, 2, and 3 be the 3 groups, we could have, for example:

1 | 2 | 3
D | B1 | B2

However, we can rearrange the 3 people in a different group in 3! = 6 ways. Therefore, there are 6 assignments in option 1.

Option 2 : Both of her brothers are in one group, and she is in a different group
 
If we let 1, 2, and 3 be the 3 groups, we could have, for example:

1 | 2 | 3
D | B1B2 | --

However, we can choose 2 groups out of 3 in 3C2 = 3 ways, and for each pair of groups we choose, we can rearrange the 3 people in 2! = 2 ways (for example, because Diana is in group 1 and her two brothers are in group 2, we could have Diana in group 2 and her two brothers in group 1). Therefore, there are 3 x 2 = 6 assignments in option 2.

In total, there are 6 + 6 = 12 assignments in which Diana is leaving at a different time than her brothers. Therefore, there are 27 - 12 = 15 assignments in which Diana is leaving at the same time as at least one of her brothers, and the probability of this happening is 15/27 = 5/9.

Answer: E

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Diana is going on a school trip along with her two brothers,

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