a PS from Gmat

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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 am; Thanked: 29 times; Followed by:3 members

a PS from Gmat

by diebeatsthegmat » Sun Dec 25, 2011 5:46 am

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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

i do not understand this problem because it doesnt tell me how many people in the group. i read the forum which solved this probability and it said " 3*3*3=27"
why is that, can you please explain????/

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Source: Beat The GMAT — Problem Solving | Sun Dec 25, 2011 5:46 am

user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pm; Thanked: 62 times; Followed by:6 members

by user123321 » Sun Dec 25, 2011 6:09 am

assigning 3 people to 3 groups is just an analogy of 3 letters going into 3 post boxes.
the possibilities are 3^3 = 27 ways

instead of calculating answer we will calculate possibilities with Diana not in any of her brothers' groups.
1) 2 brothers in one group & Diana in any of the other two groups
1/2 _ D
possibilities are 2*3 = 6ways
2) 2 brothers in different groups & Diana in third group
1 2 D
possibilities are 2*3 = 6 ways

=> answer 1- (6+6)/27 = 1-12/27 = 5/9

well there can be better way than this, but if we need to solve in 2 min the raw method of calculating possibilities is helpful sometimes.

user123321

Just started my preparation

Want to do it right the first time.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

Re: a PS from Gmat

by Scott@TargetTestPrep » Mon Nov 30, 2020 10:32 am

diebeatsthegmat wrote: ↑
Sun Dec 25, 2011 5:46 am
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

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 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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