probabilty of getting one of each character Kelloggs (the breakfast-foods company) and Nintendo (which
owns the Pokemon franchise) have agreed to distribute Pokemon toys for free
inside specially-marked boxes of Fruit Loops. Upon opening a single box of
Fruit Loops, you will have an equal chance of getting either a Pikachu, a
Bulbasaur, or a Jigglypuff. What is the probability that you will get at least
one of each character after opening 5 boxes?
Does this mean I would have equally likely outcomes?
seems to me (1/3)^5, however i am not sure 
1/3 because we either one character out of the three 
And raised it to the 5 because we are opening 5 boxes 
Any corrections or simplifications would be highly appreciated 
Thanks
 A: Let the objects be called A, B, and C. We assume that on any purchase, each of A, B, C is equally likely. Then there are $3^5$ equally likely sequences of $5$ objects. 
We need to count the favourables, the number of ways to get all $3$ objects. There are many ways to do this, some "fancier" than others. Since our numbers are small, how we do it does not matter much, as long as we are very careful.  We do it three ways. The last way uses the Principle of Inclusion/Exclusion. Once we get used to it, it is easier than the others. 
Let us for example count the bad sequences, the sequences where we do not get all three objects. There are $3$ very bad sequences, in which we get the same object $5$ times. 
Next we count the number of sequences where we get exactly two different objects. Which two? These can be selected in $\binom{3}{2}$ ways. Now let us count, for example, the number of ways to get A's and/or B's, at least one of each. 
So we can get $1$ A, or $2$, or $3$, or $4$. There are $\binom{5}{1}$ patterns with $1$ A and $4$ B, $\binom{5}{2}$ patterns with $2$ A and $3$ B, and so on for a total of $30$.  Multiply by $\binom{3}{2}$. There are $90$ patterns with $2$ different toys.
Thus the number of bad patterns is $93$. It follows that the number of favourables is $3^5-93$, which is $150$.
Our required probability is therefore $\frac{150}{3^5}$.  
Another way: We can count the favourables directly, but there is more opportunity for error. We get all three prizes in two basic patterns: (i) one prize $3$ times, and the other two once each or (ii) two prizes twice each, and the other prize once.
We count (i). The prize that occurs $3$ times can be chosen in $3$ ways. When we get it can be chosen in $\binom{5}{3}$ ways. Then the remaining two slots can be filled in $2$ ways, for a total of $60$.
We count (ii). The prize we only get one of can be chosen in $3$ ways. When we get it can be chosen in $5$ ways. Now consider the two remaining prizes, and order them alphabetically. The two places where the prize higher in the alphabet goes can be chosen in $\binom{4}{2}$ ways, for a total of $90$. 
So the number of favourables is $150$. 
Inclusion/Exclusion: There are $3^5$ possible strings made up of the letters A and/or B and/or C. 
There are $2^5$ strings over the alphabet A, B, and the same with B, C, and the same with $A, C$. We should take them away from the $3^5$. But when we remove $3\cdot 2^5$ from $3^5$, we are removing once too many times the $3$ single letter strings. Thus the number of favourables is
$$3^5-3\cdot 2^5+3.$$
A: Directed graphs/finite automata can solve many problems in combinatorics automatically (easy if you do it with a CAS)
The problem's adjacency matrix can be written as
$$\displaystyle 
A=
\left(\begin{array}{ccccccccc}
 & I & x & y & z & xy & xz & yz & xyz \\
I & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
x & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\
y & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\
z & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 \\
xy & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 1 \\
xz & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 1 \\
yz & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 \\
xyz & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3
\end{array}\right)
$$
'I' is the initial state.
x,y and z represents the three different characters.
Absence of any character indicates it hasn't seen that character yet.
To explain one of the entries, e.g., from yz it can't go to xz, since it has already seen y. It can stay in yz in two ways, on encountering character y or z, and it can go to state xyz on encountering z, i.e one way.
For your question of opening 5 boxes, you check the entry
$$(A^5)[0,7]=150$$
You can do more, like getting a generating function and hence obtaining the $n^{th}$ term:
$$\displaystyle
(I-x\, A)^{-1}
$$
gives the matrix of generating functions.
Our required g.f. is first row, eigth column in that, which is:
$$\displaystyle
G(x)=-\frac{6 \, x^{3}}{{\left(3 \, x - 1\right)} {\left(2 \, x - 1\right)} {\left(x - 1\right)}}
$$
and the $n^{th}$ coefficient can be obtained on partial fractions, 
$$ \displaystyle
[x^n]G(x)=3^n-3\cdot 2^n+3
$$
From here, obtaining the probability is easy!
(We may also do it using exponential g.f., anyway, this is more general)
See also: a pdf on g.f. I stumbled upon
A: Alternative route.
Let $p_{n,k}$ denote the probability that exactly $k$ characters
show up by opening of $n$ boxes. 
Then $p_{0,0}=1$, $p_{n,0}=0$ if $n>0$, $p_{0,k}=0$ if $k>0$
and: $$p_{n,k}=\frac{k}{3}\times p_{n-1,k}+\frac{4-k}{3}\times p_{n-1,k-1}$$
if $n>0$ and $k>0$. 
If you would like an explanation of this recurrence relation then let me know in a comment. 
This leads to:
$p_{1,1}=1$
$p_{2,1}=\frac{1}{3}$; $p_{2,2}=\frac{2}{3}$
$p_{3,1}=\frac{1}{9}$; $p_{3,2}=\frac{6}{9}$; $p_{3,3}=\frac{2}{9}$
$p_{4,1}=\frac{1}{27}$; $p_{4,2}=\frac{14}{27}$; $p_{4,3}=\frac{12}{27}$
$p_{5,1}=\frac{1}{81}$; $p_{5,2}=\frac{30}{81}$; $p_{5,3}=\frac{50}{81}$
You are searching for $p_{5,3}=\frac{50}{81}$.

addendum
Denote the number of characters that show up by opening of $n$ boxes
by $K_{n}$ and denote the event that by the opening of the $n$-th
box a new character shows up by $E_{n}$. To be found here is $P\left\{ K_{5}=3\right\} $.
It is evident that $P\left\{ K_{n}=k\right\} =1$ if $k=n=0$ and
$P\left\{ K_{n}=k\right\} =0$ if $n=0\wedge k=1$ or $n=1\wedge k=0$.
If $n>0\wedge k\in\left\{ 1,2,3\right\} $ then:
$$P\left\{ K_{n}=k\right\} =P\left\{ K_{n-1}=k\wedge E_{n}^{c}\right\} +P\left\{ K_{n-1}=k-1\wedge E_{n}\right\} =P\left\{ E_{n}^{c}\mid K_{n-1}=k\right\} P\left\{ K_{n-1}=k\right\} +P\left\{ E_{n}\mid K_{n-1}=k-1\right\} P\left\{ K_{n-1}=k-1\right\} $$
resulting in:
$$P\left\{ K_{n}=k\right\} =\frac{k}{3}\times P\left\{ K_{n-1}=k\right\} +\frac{4-k}{3}\times P\left\{ K_{n-1}=k-1\right\} $$
