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Probability

by hey_thr67 » Tue Jun 19, 2012 2:11 am

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If the exam centre of 4 students can be any one of the 7 cities, then calculate the probability that all the 4 students get any one of exactly 2 centres.

A: 3/49
B: 6/49
C: 7/48
D: 12/39
E: 7/32

OA is B.

I want to know why the no. of ways of centres can be allocated to students is not 4^7 but 7^4. I always do mistake at this concept.
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by GMATGuruNY » Tue Jun 19, 2012 4:03 am
hey_thr67 wrote:If the exam centre of 4 students can be any one of the 7 cities, then calculate the probability that all the 4 students get any one of exactly 2 centres.

A: 3/49
B: 6/49
C: 7/48
D: 12/39
E: 7/32

OA is B.
Two cities in total must be chosen.
Let the 4 students be A, B, C and D.
Student A can choose any city.

Case 1: A and B choose the same city.
P(B chooses the same city as A) = 1/7.
Since only one city has been chosen thus far, C can choose any of the 7 cities.
D must choose one of two cities already chosen.
P(D chooses one of the two cities already chosen) = 2/7.
Since we want both of the probabilities above to happen, we multiply the fractions:
1/7 * 2/7 = 2/49.

Case 2: A and B choose different cities.
Here, B can choose any of the 7 cities.
C and D must choose one of the two cities chosen by A and B.
P(C chooses one of the 2 cities already chosen) = 2/7.
P(D chooses one of the 2 cities already chosen) = 2/7.
Since we want both of these probabilities to happen, we multiply the fractions:
2/7 * 2/7 = 4/49.

Since either Case 1 or Case 2 will yield a favorable outcome, we add the results above:
2/49 + 4/49 = 6/49.

The correct answer is B.
I want to know why the no. of ways of centres can be allocated to students is not 4^7 but 7^4. I always do mistake at this concept
There are fewer students than cities.
The result: while not every city will get a student, EVERY STUDENT MUST CHOOSE A CITY.
Thus, we count from the perspective of the STUDENTS.
Each of the 4 students has 7 choices:
7*7*7*7 = 7^4.
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by shekhar.kataria » Tue Jun 19, 2012 6:35 am
HI MItch

I used the Combination formula in this problem and somehow got lost in the question. Please correct em where i went wrong.

4 students, 7 cities, students to placed in any one of the exactly two chosen cities.

1. Total number of ways :- basically any student can be pleaced in any of the cities. therefore,7*7*7*7 no of ways.

2. possible no of ways :- first chose exactly two cities, which can be chosen in 7C2 ways.

Now 4 students can be placed in any of the chosed two cities, each student has 2 options i.e 2^4.
total required ways == 7C2 * 2^4

but here dividing the possible no of ways by the total number of ways, i didnt get the exact answer. Help please

Thanks in advance.
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by GMATGuruNY » Tue Jun 19, 2012 7:01 am
shekhar.kataria wrote:HI MItch

I used the Combination formula in this problem and somehow got lost in the question. Please correct em where i went wrong.

4 students, 7 cities, students to placed in any one of the exactly two chosen cities.

1. Total number of ways :- basically any student can be pleaced in any of the cities. therefore,7*7*7*7 no of ways.

2. possible no of ways :- first chose exactly two cities, which can be chosen in 7C2 ways.

Now 4 students can be placed in any of the chosed two cities, each student has 2 options i.e 2^4.
total required ways == 7C2 * 2^4

but here dividing the possible no of ways by the total number of ways, i didnt get the exact answer. Help please

Thanks in advance.
Once two cities have been chosen (7C2 = 21), we must make sure that each of the two cities is occupied by at least one student.
Thus, the portion in red above includes two distributions that are NOT viable:
All 4 students in the first city and all 4 students in the second city.
These two BAD distributions must be subtracted from the total:
(2^4) - 2 = 14.

Alternatively, we could count directly the number of ways that the students could be distributed.
Case 1: 3 students in the first city, 1 student in the other city
4C3 * 1C1 = 4.
Case 2: 2 students in each city
4C2 * 2C2 = 6.
Case 3: 1 student in the first city, 3 students in the other city
4C1 * 3C3 = 4.
Total ways to distribute the 4 students = 4+6+4 = 14.

Thus, there are 21 pairs of cities that could be chosen, and -- within the chosen pair -- 14 ways to distribute the 4 students.
To combine these options, we multiply:
Total good options for the 4 students = 21*14.

Good options/total options = (21*14)/(7*7*7*7) = 6/49.
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by dhonu121 » Tue Jun 19, 2012 9:00 am
GMATGuruNY wrote: Once two cities have been chosen (7C2 = 21), we must make sure that each of the two cities is occupied by at least one student.
This is not mentioned in the question. Why are we assuming this ?
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by GMATGuruNY » Tue Jun 19, 2012 10:20 am
dhonu121 wrote:
GMATGuruNY wrote: Once two cities have been chosen (7C2 = 21), we must make sure that each of the two cities is occupied by at least one student.
This is not mentioned in the question. Why are we assuming this ?
Calculate the probability that all the 4 students get any one of exactly 2 centres.
While the wording here is awkward, the implication seems to be that exactly 2 of the 7 centers must be occupied.
Thus, once two centers have been chosen, each must be occupied by at least one student.
If all 4 students occupy the same center, then only one of the 7 centers -- not exactly 2 -- will be occupied.
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by shekhar.kataria » Tue Jun 19, 2012 11:04 am
MItch,

thank you the quick response. I had the same issue of understanding the question correctly.

i also do doubt such questions, which does not clearly clarify or are ambiguous in meaning, will appear in GMAT. Moreover, i tried googling it and this question has never been discussed before.





GMATGuruNY wrote:
dhonu121 wrote:
GMATGuruNY wrote: Once two cities have been chosen (7C2 = 21), we must make sure that each of the two cities is occupied by at least one student.
This is not mentioned in the question. Why are we assuming this ?
Calculate the probability that all the 4 students get any one of exactly 2 centres.
While the wording here is awkward, the implication seems to be that exactly 2 of the 7 centers must be occupied.
Thus, once two centers have been chosen, each must be occupied by at least one student.
If all 4 students occupy the same center, then only one of the 7 centers -- not exactly 2 -- will be occupied.
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by karthik1239 » Tue Jun 19, 2012 6:50 pm
Hello GMATGURUNY,
Case 1: A and B choose the same city.
Case 2: A and B choose different cities.
What if
Case 3:A and B and C choose the same city and D chooses a different city.
That will add 1/49 to the probability.
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by GMATGuruNY » Wed Jun 20, 2012 2:51 am
karthik1239 wrote:Hello GMATGURUNY,
Case 1: A and B choose the same city.
Case 2: A and B choose different cities.
What if
Case 3:A and B and C choose the same city and D chooses a different city.
That will add 1/49 to the probability.
This possibility is covered by Case 1:
Case 1: A and B choose the same city.
P(B chooses the same city as A) = 1/7.
Since only one city has been chosen thus far, C can choose any of the 7 cities.
The portion in red enables C to choose the same city chosen by A and B.
If C chooses the same city as A and B, then D must choose a different city, since it is required that 2 cities be occupied.
The result: A, B and C in the same city, with D in a different city.
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Re: Probability

by Scott@TargetTestPrep » Fri Apr 03, 2020 7:35 am
hey_thr67 wrote: ↑
Tue Jun 19, 2012 2:11 am
 If the exam centre of 4 students can be any one of the 7 cities, then calculate the probability that all the 4 students get any one of exactly 2 centres.

A: 3/49
B: 6/49
C: 7/48
D: 12/39
E: 7/32

OA is B.

I want to know why the no. of ways of centres can be allocated to students is not 4^7 but 7^4. I always do mistake at this concept.
Letting A, B, C, D, E, F and G denote the cities, we are looking for the probability that a selection of four cities (such as A-B-C-D or D-E-F-G) includes exactly two different cities.

Since there are 7 cities, the number of different selections is 7^4 (repetitions allowed).

Let’s first find the number of ways where the selection only includes A and B. Since there are two choices for each student, there are 2 x 2 x 2 x 2 = 16 selections which include only A and B. However, one of these selections is A-A-A-A and one is B-B-B-B, neither of which is possible. Therefore, there are 16 - 2 = 14 selections which include only A and B and at least one of each.

Next, we note that the requirement “all students get any one of exactly 2 centers” can also be satisfied through selections such as C-D-C-D or E-F-F-E. Thus, we need to determine the number of ways we can select two letters out of seven. There are 7C2 = (7 x 6)/2 = 21 ways to make such a selection.

Since there are 7^4 possibilities and 14 x 21 of them are favorable, the required probability is:

(14 x 21)/7^4 = (2 x 3)/7^2 = 6/49

Answer: B

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