Sure this is a university level question?
(i)
You know 9 urns have a ball in, and there are 12 urns total. the probability you want is 9/12
(ii)
Work out
P(at least 2 empty) = 1 - P(0 empty) - P(1 empty)
P(0 empty)=
P(1 empty) =
Suppose 9 balls are distributed amongst 12 urns in such a way that no more than one ball is put in any one urn, and that all such arrangements are equally likely.
(i) What is the probability that a specified urn will contain a ball?
(ii) If the urns are placed in a row before the balls are distributed, what is the probability that at least two of the first four urns will be empty?
Thanks for any help.
Sure this is a university level question?
(i)
You know 9 urns have a ball in, and there are 12 urns total. the probability you want is 9/12
(ii)
Work out
P(at least 2 empty) = 1 - P(0 empty) - P(1 empty)
P(0 empty)=
P(1 empty) =