# Balls and Boxes (list) ```Determine the probability that each box contains exactly two balls
Since we are assuming that each ball cannot land anywhere else other than in the boxes, there are
only two positions that each ball can go; in Box 1 or Box 2. From this we can make a few inferences.
As there are 2 places for each ball to land, and there are 4 balls, so the total number of outcomes is 2
to the power of 4, which is 16. These combinations are:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0 represents the ball landing in the Box 1, and 1 represents the ball landing in Box 2. There are 6
permutations where there are 2 balls in each box, so the probability that two balls land in each box
is 6 over 16, which is 37.5%.
To what extent can you generalise?
Let’s say we are trying to find the probability that m out of n ball land in Box 1. We must first find the
total number of permutations, which is 2n. Furthermore, we must find the number of outcomes of n
out of m balls landing in the box. For this, we can use the Binomial Coefficient nm, which literally
reads ‘n choose m’ and finds how many of ways are there of picking m outcomes from n possibilities.
This Binomial Coefficient equates to:
n!m!n-m!
Therefore probability that m out of n ball land in Box 1 is:
n!m!n-m!&divide;2n
We can apply this to our original problem to work out the probability that 2 balls land in each box:
4!2!4-2!&divide;24
=37.5%
Probability and Pascal’s Triangle
Row 0
Row 4
We could also use Pascal’s Triangle to find the Binomial Coefficient. To find nm, we start from the 1
in Row 0, move down the left side n places and m places across. For instance, to find 42, you move
down the side 4 places and across 2 places to get 6, like before.
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