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

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.

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

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