Tridip C. answered • 08/13/20
2+ years of Experience in Tutoring Discrete Math
n balls are tossed into b (≥ 2) bins so that each ball is equally likely to fall into any of the bins and that the tosses are independent.
Ej be the event that bin j is not empty, where j ∈ {1, 2, ..., b}.
Number of different ways those n balls can be put into the b bins, are bn
Number of different ways n balls can be put into the (b-1) bins but the 1st bin, are (b-1)n
∴ There are (b-1)n different ways in which the 1st bin can remain empty.
p(bin 1 is empty) = (b-1)n/bn = (b-1/b)n = (1-1/b)n
∴ p(E1) = p(bin 1 is not empty) = 1- p(bin 1 is empty) = 1 - (1-1/b)n
Similar argument can be used to show that p(E2) = p(bin 2 is not empty) = 1- p(bin 2 is empty) = 1 - (1-1/b)n
∴ p(E1) = p(E2) = 1 - (1-1/b)n
The event E1∩E2 denotes that, bin 1 and bin 2 are both non-empty.
There are:
(b-1)n different ways in which both bin 1 remain empty
(b-1)n different ways in which both bin 2 remain empty
(b-2)n different ways in which both bin 1 and bin 2 remain empty
∴ Number of ways in which both are non-empty = bn - (b-1)n - (b-1)n + (b-2)n = bn - 2⋅(b-1)n + (b-2)n
∴ p(E1∩E2) = (bn - 2⋅(b-1)n + (b-2)n) / bn = 1 - 2⋅(1-1/b)n + (1-2/b)n
But, p(E1)⋅p(E2) = (1 - (1-1/b)n)⋅(1 - (1-1/b)n) = 1 - 2⋅(1-1/b)n + (1-1/b)2n
∴ p(E1∩E2) ≠ p(E1)⋅p(E2)
Thus, E1, E2 are not independent.
