Question: In a 7 day week 1 student will drop out of school. What is the Probability that 3 students will drop out in exactly 2 of the next five days? * that is 3 students drop on each of these 2 days, (so 3 on one of the two and three more on the second) * Dropping out is independent Solution Given X~Poisson(=5/7=0.71 in five days) P(X=x)=(0.71)^x*exp(-0.71)/x! So the probability is P(X=3)=(0.71)^3*exp(-0.71)/6= 0.02932748.

1. Question: In how many ways can 5 people be seated on 5 chairs around a round table if
a) only their positions relative to each other count (that is, the arrangements obtained from each
other by rotation of all people are considered to be the same)
b)only who sits next to whom counts, but not on which side (rotations and relections do not
change the arrangement)
The answers for a is 5 factorial / 5 = 24
The answer for b is 5 factorial /10 = 12
Im just curious how my teacher got that answer. Can someone explain this to me?
Solution
(a) only their positions relative to each other count: The 1st place can be filled in 5
ways(total 5 people) once 1st seat is filled 2nd is filled in 4 ways (4 people left) once 2nd seat is
filled 3rd is filled in 3 ways Thus all the seats can be filled in 5*4*3*2*1 = 5! ways But if we
shift all people by 1 seat to left(equivalent to 4 to right) or 2 to left all(equivalent to 3 to right),
the arrangements only their positions relative to each other count are the same But since 5 seats
are in circle, so there are 5 arragements which are the same Thus, net no. of ways = 5! /5 = 24
(b) only who sits next to whom counts, but not on which side (rotations and reflections do not
change the arrangement): like last time in this case rotation doesn't change arrangement. But
now reflections do not change the arrangement hence, clock wise and anticlock wise are same
UNLIKE last case now for every 2 sets (felated by reflection i.e one Clockwise & one anticlock
wise) Net number of ways = (5! /5)/2 = 5! /10 = 12