Thread: What is the probability that Jimmy gets to stand next to Sally?

Jimmy has a crush on Sally, a girl in his swimming class. The class contains 10 students, including Jimmy and Sally. On the last day of class, the teacher randomly lines the students up for a picture. What is the probability that Jimmy gets to stand next to Sally?

let "A" be the event in which jimmy sits near to sally
n(S)=10 ........... (sample space)
n(A)=1

P(A)= n(A)/n(S) ........(P means probabilty)
therefore P(A)=1/10


(I m an indian, this is how i we do the sum in my school)
(sry... for my bad english)

There's a 1/10 chance that Jimmy will be on one end of the line, and a 1/9 chance that Sally will then be next to him.

1/10 * 1/9

There's also a 1/10 chance that Jimmy will be on the other end of the line, with a 1/9 chance that Sally will be next to him there.

1/10 * 1/9

So there's an 8/10 chance that Jimmy will be somewhere other than the ends of the line, and a 2/9 chance that Sally will be next to him on one side or the other.

8/10 * 2/9

The overall probability is the sum of these mutually exclusive situations:

1/10 * 1/9 + 1/10 * 1/9 + 8/10 * 2/9 =
1/90 + 1/90 + 16/90 =
18/90 =
1/5

Answer: 1/5

The probability that Jimmy stands on one end is 2/10.
Given that Jimmy is standing on one end, the probability that Sally is standing next to him is 1/9.
The probability that Jimmy stands on the end with Sally next to him is 2/10 * 1/9 = 1/45.

The probability that Jimmy is not on one end is 8/10.
Given that Jimmy is not on one end, the probability that Sally is standing next to him is 2/9.
The probability that Jimmy is not one end but Sally is next to him is 8/10 * 2/9 = 8/45.

The total probability is 1/45 + 8/45 = 9/45 = 1/5.

1/10 is correct. Don't lisnten to that stupid
Kid who said 1/5z he sucks.

JSXXXXXXXX = 8!
XJSXXXXXXX = 8!
XXJSXXXXXX = 8!
XXXJSXXXXX = 8!
XXXXJSXXXX = 8!
XXXXXJSXXX = 8!
XXXXXXJSXX = 8!
XXXXXXXJSX = 8!
XXXXXXXXJS = 8!
OR
SJXXXXXXXX = 8!
XSJXXXXXXX = 8!
XXSJXXXXXX = 8!
XXXSJXXXXX = 8!
XXXXSJXXXX = 8!
XXXXXSJXXX = 8!
XXXXXXSJXX = 8!
XXXXXXXSJX = 8!
XXXXXXXXSJ = 8!

There are 18 ways that they can stand next to each other
There are 10! ways that they can line up
(18 * 8!) / 10!

.2

20% probability

There are 10! ways to arrange the students.

Thinking of Jimmy and Sally standing together as a group of 2 people. So the number of ways to arrange the students so that Jimmy and Sally stand together is the number of ways of permuting the 9 groups (8 singles and a group of two) times the number of ways to permute the students within the group of two, which equals 9!*2!.

So P(Jimmy & Sally stand together) = 9!*2!/10! = 2/10 = 1/5.

Lord bless you today!