Mixing
What's the probability two students present consecutively? [closed]
During the last class period of the semester, each student in a graduate computer science class with 10 students is required to give a brief report on his or her class project. The professor randomly selects the order in which the reports are to be given. Two students have been working on similar projects and would like to give their reports consecutively. What is the probability that this will happen?
Here's what I have (pretty sure this is wrong): C(10,2)/10! (10 choose 2 divided by 10 factorial)
discrete-mathematics
asked Nov 21 '18 at 2:42
Smith Jones
93
closed as off-topic by Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo Nov 22 '18 at 1:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
During the last class period of the semester, each student in a graduate computer science class with 10 students is required to give a brief report on his or her class project. The professor randomly selects the order in which the reports are to be given. Two students have been working on similar projects and would like to give their reports consecutively. What is the probability that this will happen?
Here's what I have (pretty sure this is wrong): C(10,2)/10! (10 choose 2 divided by 10 factorial)
discrete-mathematics
asked Nov 21 '18 at 2:42
Smith Jones
93
closed as off-topic by Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo Nov 22 '18 at 1:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
What have you tried?
– Don Thousand
Nov 21 '18 at 2:47
I found the number of 2 combinations out of 10 and divided that by 10! (factorial)
– Smith Jones
Nov 21 '18 at 2:50
Not sure I'm on the right track...
– Smith Jones
Nov 21 '18 at 2:50
1
Can you write what you've done in the actual question?
– Don Thousand
Nov 21 '18 at 2:50
add a comment |
During the last class period of the semester, each student in a graduate computer science class with 10 students is required to give a brief report on his or her class project. The professor randomly selects the order in which the reports are to be given. Two students have been working on similar projects and would like to give their reports consecutively. What is the probability that this will happen?
Here's what I have (pretty sure this is wrong): C(10,2)/10! (10 choose 2 divided by 10 factorial)
discrete-mathematics
asked Nov 21 '18 at 2:42
Smith Jones
93
During the last class period of the semester, each student in a graduate computer science class with 10 students is required to give a brief report on his or her class project. The professor randomly selects the order in which the reports are to be given. Two students have been working on similar projects and would like to give their reports consecutively. What is the probability that this will happen?
Here's what I have (pretty sure this is wrong): C(10,2)/10! (10 choose 2 divided by 10 factorial)
discrete-mathematics
discrete-mathematics
asked Nov 21 '18 at 2:42
Smith Jones
93
asked Nov 21 '18 at 2:42
Smith Jones
93
edited Nov 21 '18 at 2:53
asked Nov 21 '18 at 2:42
Smith Jones
93
asked Nov 21 '18 at 2:42
Smith Jones
93
asked Nov 21 '18 at 2:42
Smith Jones
93
93
closed as off-topic by Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo Nov 22 '18 at 1:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo Nov 22 '18 at 1:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Don Thousand, José Carlos Santos, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
What have you tried?
– Don Thousand
Nov 21 '18 at 2:47
I found the number of 2 combinations out of 10 and divided that by 10! (factorial)
– Smith Jones
Nov 21 '18 at 2:50
Not sure I'm on the right track...
– Smith Jones
Nov 21 '18 at 2:50
1
Can you write what you've done in the actual question?
– Don Thousand
Nov 21 '18 at 2:50
add a comment |
What have you tried?
– Don Thousand
Nov 21 '18 at 2:47
I found the number of 2 combinations out of 10 and divided that by 10! (factorial)
– Smith Jones
Nov 21 '18 at 2:50
Not sure I'm on the right track...
– Smith Jones
Nov 21 '18 at 2:50
1
Can you write what you've done in the actual question?
– Don Thousand
Nov 21 '18 at 2:50
What have you tried?
– Don Thousand
Nov 21 '18 at 2:47
What have you tried?
– Don Thousand
Nov 21 '18 at 2:47
I found the number of 2 combinations out of 10 and divided that by 10! (factorial)
– Smith Jones
Nov 21 '18 at 2:50
I found the number of 2 combinations out of 10 and divided that by 10! (factorial)
– Smith Jones
Nov 21 '18 at 2:50
Not sure I'm on the right track...
– Smith Jones
Nov 21 '18 at 2:50
Not sure I'm on the right track...
– Smith Jones
Nov 21 '18 at 2:50
1
1
Can you write what you've done in the actual question?
– Don Thousand
Nov 21 '18 at 2:50
Can you write what you've done in the actual question?
– Don Thousand
Nov 21 '18 at 2:50
add a comment |
1 Answer
1
active
oldest
votes
There are $10!$ ways to order the students. If we consider the number of ways to order the other $8$ students and the pair of students, then there are $2cdot9!$ ways to order them where the two students present consecutively. Therefore the probability is $$
frac{2cdot 9!}{10!}=frac{2}{10}=frac{1}{5}
$$
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
2
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are $10!$ ways to order the students. If we consider the number of ways to order the other $8$ students and the pair of students, then there are $2cdot9!$ ways to order them where the two students present consecutively. Therefore the probability is $$
frac{2cdot 9!}{10!}=frac{2}{10}=frac{1}{5}
$$
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
2
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
add a comment |
There are $10!$ ways to order the students. If we consider the number of ways to order the other $8$ students and the pair of students, then there are $2cdot9!$ ways to order them where the two students present consecutively. Therefore the probability is $$
frac{2cdot 9!}{10!}=frac{2}{10}=frac{1}{5}
$$
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
2
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
add a comment |
There are $10!$ ways to order the students. If we consider the number of ways to order the other $8$ students and the pair of students, then there are $2cdot9!$ ways to order them where the two students present consecutively. Therefore the probability is $$
frac{2cdot 9!}{10!}=frac{2}{10}=frac{1}{5}
$$
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
There are $10!$ ways to order the students. If we consider the number of ways to order the other $8$ students and the pair of students, then there are $2cdot9!$ ways to order them where the two students present consecutively. Therefore the probability is $$
frac{2cdot 9!}{10!}=frac{2}{10}=frac{1}{5}
$$
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
answered Nov 21 '18 at 2:51
Joey Kilpatrick
1,181422
1,181422
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
2
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
add a comment |
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
2
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Consider the pair of students as one student. There are $9!$ ways to order the 8 students and the pair and $2$ ways to order the pair.
– Joey Kilpatrick
Nov 21 '18 at 2:55
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Could the downvoter explain?
– Joey Kilpatrick
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
Alternate way to count: there are 9 ways to choose where the two students give consecutive presentations, 2 ways to choose which goes first between them, and then $8!$ ways to order the remaining students.
– Daniel Schepler
Nov 21 '18 at 2:56
2
2
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
@DanielSchepler This gives the same answer.
– Joey Kilpatrick
Nov 21 '18 at 2:59
add a comment |
What have you tried?
– Don Thousand
Nov 21 '18 at 2:47
I found the number of 2 combinations out of 10 and divided that by 10! (factorial)
– Smith Jones
Nov 21 '18 at 2:50
Not sure I'm on the right track...
– Smith Jones
Nov 21 '18 at 2:50
1
Can you write what you've done in the actual question?
– Don Thousand
Nov 21 '18 at 2:50