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

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Diana is going on a school trip along with her two brothers, [#permalink] New post 09 Jul 2010, 16:16
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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

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

Please comment and explain
[Reveal] Spoiler: OA
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Re: Probability Qs from Princeton Review [#permalink] New post 09 Jul 2010, 16:56
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rpgmat2010 wrote:
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

[Reveal] Spoiler:
E


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

Please comment and explain


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.
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Re: Probability Qs from Princeton Review [#permalink] New post 13 Jul 2010, 11:47
+1 Bunuel . Nice explanation.
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Re: Probability Qs from Princeton Review [#permalink] New post 09 Apr 2011, 02:49
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?
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Re: Probability Qs from Princeton Review [#permalink] New post 09 Apr 2011, 23:14
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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
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Re: Probability Qs from Princeton Review [#permalink] New post 10 Apr 2011, 05:41
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

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Re: Probability Qs from Princeton Review [#permalink] New post 25 Aug 2011, 04:37
Bunuel wrote:
rpgmat2010 wrote:
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

[Reveal] Spoiler:
E


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

Please comment and explain


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.


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.
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Re: Diana is going on a school trip along with her two brothers, [#permalink] New post 09 Jul 2013, 09:56
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Re: Diana is going on a school trip along with her two brothers, [#permalink] New post 04 Aug 2013, 06:44
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.
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Re: Diana is going on a school trip along with her two brothers, [#permalink] New post 12 Nov 2014, 07:22
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Re: Diana is going on a school trip along with her two brothers, [#permalink] New post 24 Jun 2015, 05:47
Bunuel wrote:
Bumping for review and further discussion.


hi bunnuel,

am struggling with probability and combinations questions can you please suggest me some material which can provide some clarity on the topic.
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Re: Diana is going on a school trip along with her two brothers, [#permalink] New post 24 Jun 2015, 06:11
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SunthoshiTejaswi wrote:
Bunuel wrote:
Bumping for review and further discussion.


hi bunnuel,

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


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.
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Re: Diana is going on a school trip along with her two brothers, [#permalink] New post 26 Jun 2015, 03:20
Bunuel wrote:
rpgmat2010 wrote:
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

[Reveal] Spoiler:
E


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

Please comment and explain


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.


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

Thanks
Re: Diana is going on a school trip along with her two brothers,   [#permalink] 26 Jun 2015, 03:20
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