mathewmithun wrote:I am getting stumped by this question-I wrote the different possibilities and I was getting 3/5. But clearly my approach was wrong and I am certain that my answer is also wrong. So pls help me solve this question.

Diana is going on a school trip along with her two brothers bruce and clark. 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 one of her brothers?

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

OA is E

rephrased, this question asks about probability for Diana to be placed in the same group with either of her brothers. Since we are not given the number of students, our pool must be a factor of 3 (we divide all people into three-person groups). Also note, the condition "at least one brother" implies one and more (two, here) brothers selected for the same group with Diana.
here, I have to exclaim, this question is missing something in it - either number of students or some other data ...

mathewmithun wrote:I am getting stumped by this question-I wrote the different possibilities and I was getting 3/5. But clearly my approach was wrong and I am certain that my answer is also wrong. So pls help me solve this question.

Diana is going on a school trip along with her two brothers bruce and clark. 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 one of her brothers?

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

OA is E

P(at least 1 brother is assigned to Diana's group) = 1 - P(neither brother is assigned to Diana's group).

It is given that Diana is assigned to one of the 3 groups.
What we need to determine is how Diana's brothers can be assigned RELATIVE to Diana's group.
Since the students are RANDOMLY assigned to the 3 groups, the probability of being assigned to Diana's group = 1/3.

P(neither brother is assigned to Diana's group):
P(the 1st brother is to NOT ASSIGNED to Diana's group) = 2/3. (Of the 3 groups, 2 do not include Diana.)
P(the 2nd brother is NOT ASSIGNED to Diana's group) = 2/3. (Of the 3 groups, 2 do not include Diana).
Since we want both events to happen, we multiply the fractions:
2/3 * 2/3 = 4/9.

P(at least 1 brother is assigned to Diana's group):
1 - 4/9 = 5/9.

The correct answer is E.

mathewmithun wrote:I am getting stumped by this question-I wrote the different possibilities and I was getting 3/5. But clearly my approach was wrong and I am certain that my answer is also wrong. So pls help me solve this question.

Diana is going on a school trip along with her two brothers bruce and clark. 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 one of her brothers?

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

OA is E

Of course GMAT Guru's reply is most elegant way of solving this problem.
If you want to break down the problem and take a lengthier way here is more detail -
How are B and C placed relative to D, into the three trips T1, T2 and T3, so that they are not along with D?
Fix D in T1 and see how many scenarios are possible
D B C
D C B
D BC *
D * BC

Similarly there are 4 placements each with D alone in T2 and T3

Total ways in which brothers are not in same trip 4*3 = 12

Probability that brothers are not in same trip = 12/27

Probability that Diana is in the same trip as at least one of her brothers = 1- 12/27=15/27=5/9

Here 27 is the total number of combinations in which the three can go on the three trips T1, T2, T3.
I calculated 27 by actually counting the scenarios.