Choosing Balls Randomly 
4 balls are put randomly into $3$ cells
  what is the probability that:
  
  1. there is an empty cell?
  
  2. there is one cell with precisely $2$ balls



*

*if we know that one cell is empty there are $\frac{2^4}{3^4}$ options that a cell will be empty, there are $3$ cells so it is $3*\frac{2^4}{3^4}=\frac{42}{81}$, but the answer is $\frac{45}{81}$ doesn't my answer include that option that $2$ cells will be empty?

*Why the answer is $$\frac{{3\choose 1}{4\choose 2}*2!+{3 \choose 2}{4\choose 2}}{3^4}$$ ?
 A: Question 1. Your approach is generally correct, but just multiplying the answer by three is incorrect, not because you haven't included the possibility that two cells are empty, but because you've counted them twice.  That has been compounded by an arithmetical error ($3 \cdot 2^4 = 3 \cdot 16 = 48$, not $42$).  The correct number of ways to have (at least) one empty cell is
$$
3 \cdot 2^4 - 3 = 48 - 3 = 45
$$
and then the probability of having at least one empty cell is $45/81 = 5/9$.
Revised Question 2. There can be a cell with exactly two balls in two different ways.  One is to have a cell with two balls, and two cells with one ball each.  The number of ways to do this (and the associated probability) are analyzed in "Old Question 2", below.
The other way is to have two balls in each of two cells.  We choose two of the cells ($C(3, 2) = 3$ pairs of cells) and choose two of the balls to go in the "first" of that pair of cells ($C(4, 2) = 6$ pairs of balls), for a total of $3 \cdot 6 = 18$ different ways.  The probability is $18/81 = 2/9$, and if you add those together, you end up with the $2/3$ from the expression you gave.
Old Question 2. The original wording of this question was, "What is the probability that there is exactly one cell with two balls?"
The only way of having exactly one cell with two balls is to have one ball in each of the other two cells.  We choose two of the four balls ($C(4, 2) = 6$ different pairs), to share one of the cells ($3$ different cells), and the other two balls to split the two remaining cells ($2! = 2$ different ways).  The product is $6 \cdot 3 \cdot 2 = 36$ different ways, so the probability is $36/81 = 4/9$.
The expression you gave evaluates to $54/81 = 2/3$.  I don't think that can be right, if I understand the problem correctly (but I might not understand the question correctly).
It makes sense that the answers to Question 1 and Old Question 2 add up to $1$, since they are mutually disjoint (if there is one empty cell, there cannot be exactly one cell with exactly two balls), and cover all cases (one of those properties must be true).
A: For part one:  We ask the question "what is the probability that there is at least one empty cell."
You correctly begin by noting that there are $3^4$ possible ways to distribute the balls.  We have a number of ways of approaching this problem, but the easiest perhaps is via inclusion-exclusion.  We count instead: how many ways can $\underline{no}$ cells be empty?
Since there are four balls and three cells, that means that one cell gets two balls, and the other two cells each get one ball.  Pick which cell gets two balls (3 choices).  Pick which two balls went into the cell with two balls ($\binom{4}{2} = 6$ choices).  Pick which ball went into the leftmost cell with one ball ($2$ remaining choices).  Pick which ball went into the final cell ($1$ choice).  For a total of $3\cdot 6\cdot 2\cdot 1 = 36$ ways of distributing balls to not have an empty cell.  Thus there are $3^4-36=81-36=45$ ways of having at least one empty cell.  The probability is therefore $\frac{45}{81}$.
Your answer of $3\cdot \frac{2^4}{3^4}$ made two mistakes.  First was arithmetically.  $3\cdot 2^4 = 3\cdot 16 = 48$, not $42$.  The second mistake was that you had not properly applied inclusion-exclusion.  You have accidentally double-counted each scenario where 2 of the three cells are empty.  To fix this, you should subtract the number of ways that two cells are empty.  There are three ways this can occur.  Indeed, $48-3 = 45$ which gives the correct probability of $\frac{45}{81}$.
Read more at Inclusion-Exclusion Principle

For part two, I will interpret it as "There is at least one cell with exactly 2 balls."  There are two ways this can happen:


*

*There is one cell with 2 balls, and two cells with 1 ball each

*There are two cells with 2 balls each (and one cell which is empty)


The first case was calculated previously to occur $3\cdot\binom{4}{2}\cdot 2 = 36$ different ways.
To count the number of ways the second case can occur, apply multiplication principle.  Pick which of the cells is empty (equivalently pick which two cells are being used) (there are $\binom{3}{1}=\binom{3}{2}=3$ choices).  Pick which two balls go in the leftmost cell that will have two balls in it ($\binom{4}{2} = 6$ choices).  The remaining balls go in the remaining cell that will have two balls in it.  For a total of $\binom{3}{1}\binom{4}{2} = 18$ different ways to fall into the second case.
By the addition principle (since there is no overlap between these cases), there are a total of $36+18=54$ different ways to have "at least one cell has exactly two balls."  Thus the probability is $\frac{54}{81}$
