Suppose we throw 5 labelled balls randomly into 3 labelled bins. a) What is the probability that the first bin is empty? b) What is the probability that at least two bins are empty?
I know that the sample space is k^n, in this case being 3^5, but I can't figure out how I can solve these 2.
On
a) the first bin is empty:
you can fill the other two bins in 6 ways (we will consider only bin 2 because after choosing balls for bin 2, we have to put the rest of the balls in bin 3):
i) 5 balls in bin 2 and 0 balls in bin 3, you can do it in $^5C_5 = 1$ way
ii) 4 balls in bin 2 and 1 balls in bin 3,you can do it in $^5C_4 = 5$ way
iii) 3 balls in bin 2 and 2 balls in bin 3,you can do it in $^5C_3 = 10$ way
iv) 2 balls in bin 2 and 3 balls in bin 3,you can do it in $^5C_2 = 10$ way
v) 1 balls in bin 2 and 4 balls in bin 3,you can do it in $^5C_1 = 5$ way
vi) 0 balls in bin 2 and 5 balls in bin 3,you can do it in $^5C_0 = 1$ way
In total = (1+5+10+10+5+1) = 32 ways to keep the 1st bin empty.
Now the probability will be $$\frac{32}{3^5}$$.
(b) at least two bins are empty it has two cases:
case-1 : only two bins are empty
case-2 : more than two bins (three here) are empty.
case-1: two bins are empty that means one bin is full. We can choose one bin from the 3 bins in $^3C_2 = 3$ way.
case -2: All the three bins are empty this can be done by only $1$ way.
so the probability of at least two bins empty $$=\frac{3+1}{3^5}=\frac{4}{3^5}$$
Label the bins A,B,C
example a/ grants P(No balls land in A)=P(All balls land in B or C)=$(2/3)^5$
example b/ is a little more tricky, P(at least two bins are empty) = P(A is full) + P(B is full) + P(C is full) $= 3 \cdot (1/3)^5$