Homework SourseSuppose that 10 women including Alice and Betty are arranged
Suppose that 10 women, including Alice and Betty, are arranged at random in a line. For each value of k between k=0 and k=8, find the probablitly that exactly k women are between Alice and Betty.
Solution
total exhaustive number iof ways = 10P2= 10*9
= 90
if k=8 then 8 women can arraged between alice and betty in only 2 way (if alice in first positon and betty at 10 th position andif betty in first positon and alice at 10 th position )
if k= 8 then p
if k=7 then n= 4 ( 1) if alice in 1st postion and betty 9 th postion, 2) if batty in 1st postion and alice 9 th postion 3) if alice in 2nd postion and betty 10 th postion 4) if batty in 2nd postion and allice 10 th postion
thus probability are calculated in the table shown below
(1, 8) (2,9) (3,10) (8,1) (9,2) (10,3)
| k | position of (alice,betty) | probability |
| k=8 | (1,10) (1, 10) | 2/90 |
| k=7 | (1,9) (2,10) (9,1) (10,2) | 4/90 |
| k=6 | (1, 8) (2,9) (3,10) (8,1) (9,2) (10,3) | 6/90 |
| k=5 | (1, 7) (2,8) (3,9) (4,10) (7,1) (8,2) (9,3) (10,4) | 8/90 |
| k=4 | (1,6) (2,7) (3,8) (4.9) (5,10) (6,1) (7,2) (8,3) (9,4) (10,5) | 10/90 |
| k=3 | (1,5) (2,6) (3, 7) (4,8) (5,9) (6,10) (5,1) (6,2) (7,3) (8,4) (9,5) (10,6) | 12/90 |
| k=2 | (1,4) (2,5) (3, 6) (4, 7) (5,8) (6,9) (7, 10) (4,1) (5, 2) (6,3) (7,4) (8,5) (9,6) (10,7) | 14/90 |
| k=1 | (1,3) (2,4) (3,5) (4,6) (5,7) (6,8) (7,9) (8,10) (3,1) (4,2) (5,3) (6,4) (7,5) (8,6) (9,7) (10,8) | 16/90 |
