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Diana is going on a school trip along with her two brothers, [#permalink]

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09 Jul 2010, 17:16

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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.

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?

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 \(1-\frac{12}{27}=\frac{15}{27}=\frac{5}{9}\).

Re: Probability Qs from Princeton Review [#permalink]

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09 Apr 2011, 03:49

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?
_________________

"When the going gets tough, the tough gets going!"

Re: Probability Qs from Princeton Review [#permalink]

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10 Apr 2011, 00:14

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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)

Re: Probability Qs from Princeton Review [#permalink]

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25 Aug 2011, 05:37

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?

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.

Re: Diana is going on a school trip along with her two brothers, [#permalink]

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04 Aug 2013, 07:44

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]

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12 Nov 2014, 08:22

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Re: Diana is going on a school trip along with her two brothers, [#permalink]

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26 Jun 2015, 04:20

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?

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?

Re: Diana is going on a school trip along with her two brothers, [#permalink]

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20 May 2016, 23:39

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?

Re: Diana is going on a school trip along with her two brothers, [#permalink]

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05 Feb 2017, 19:00

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

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

Please comment and explain

Should be AT LEAST one her bothers
_________________

Re: Diana is going on a school trip along with her two brothers, [#permalink]

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05 Feb 2017, 23:20

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

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

Please comment and explain

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

Diana is going on a school trip along with her two brothers, [#permalink]

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30 Apr 2017, 00:33

Bunuel wrote:

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 - getting my thought slightly fuzzy about the highlighted portion. Assuming that Diana is not in the first group, each of her brothers has 3 options (since they can be together in the same group too), in which case the probability of the event in question increases. Please can you let me know where I am going wrong?

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