Probability of one or more events occurring all with different probabilities For example imagine if a horse is racing in 5 different races, how would you calculate the probability of that horse winning 3 out of the 5 races. With the probabilities of  $\frac{1}{2.2}, \frac{1}{4}, \frac{1}{3.3}, \frac{1}{10}$ and $\frac{1}{5}$ respectively.
I am not a math's expert, so explanation of the how to use any formulas would be appreciated.
Currently to get an approximation I simply multiply the odds of the two favorites then multiply that again with the sum of the other probabilities. $(\frac{1}{2.2} \times \frac{1}{3.3}) \times (\frac{1}{10} + \frac{1}{4} + \frac{1}{5}$)
Thanks for the help.
 A: There is a possible point of ambiguity.  To a mathematician, $2.2/1$ (for example) translates into a win factor of $\displaystyle \frac{1}{3.2}$.
This corresponds to a payout of $2.2$ to $1$.  However, it is not uncommon for a casino (for example) to use the syntax $2.2/1$ to represent a payout of $2.2$ for $1$, which implies that if you win, you get back $2.2$, instead of $(2.2 + 1 = 3.2)$.
In the response below, I am going to assume that $2.2/1$ signifies a payout of $2.2$ to $1$, rather than $2.2$ for $1$.

Easiest to use your example for illustration, for an exact answer.
First, you have to convert each odds into a probability $p$ such that $0 < p < 1.$
In horse racing parlance, $X/1$ translates into
$\displaystyle p = \frac{1}{X + 1}.$
So, letting $p_k$ denote the probability of the horse winning race $k$, you have that
$p_1, p_2, p_3, p_4, p_5~~$ equal
$\displaystyle \frac{1}{3.2}, \frac{1}{5}, \frac{1}{4.3}, \frac{1}{11}, \frac{1}{6},~~$ respectively.
Set $q_1, q_2, q_3, q_4, q_5~~$ equal to
$(1 - p_1), (1 - p_2), (1 - p_3), (1 - p_4), (1 - p_5)~~$ respectively.
Per your example, I am assuming that you want the probability of exactly $3$ wins for the horse, rather than (for example) at least $3$ wins.
Unfortunately, there isn't much of a shortcut.  First, I will show the brute force method, and then discuss possible shortcuts.
There are $\displaystyle \binom{5}{3}$ different possibilities for which of the $3$ races that the horse wins.  The brute force method is to add together the following $10$ terms:
$p_1 \times p_2 \times p_3 \times q_4 \times q_5$. 
$p_1 \times p_2 \times q_3 \times p_4 \times q_5$. 
$p_1 \times p_2 \times q_3 \times q_4 \times p_5$. 
$p_1 \times q_2 \times p_3 \times p_4 \times q_5$. 
$p_1 \times q_2 \times p_3 \times q_4 \times p_5$. 
$p_1 \times q_2 \times q_3 \times p_4 \times p_5$. 
$q_1 \times q_2 \times p_3 \times p_4 \times p_5$. 
$q_1 \times p_2 \times q_3 \times p_4 \times p_5$. 
$q_1 \times p_2 \times p_3 \times q_4 \times p_5$. 
$q_1 \times p_2 \times p_3 \times p_4 \times q_5$.
A shortcut of sorts is to notice that the first $6$ terms all have the horse winning the 1st race.  This corresponds to $\displaystyle \binom{4}{2} = 6$, which represents that there are $6$ ways of choosing which of the other $2$ races the horse wins.
So, you could ignore the $p_1$ factor, brute force compute the sum of the 1st 6 terms, using only $4$ factors each (i.e. where the $p_1$ factor is excluded in each of the $6$ computations), and then multiply the sum of these $6$ terms by $p_1$.  This would get you the actual sum of the first $6$ terms.
Similarly, you could notice that the last $4$ terms all have a factor of $q_1$.  So, you could brute force compute these last $4$ terms, excluding the $q_1$ factor in each term, and then collectively apply the $q_1$ factor.
A (quick) estimate is to notice that the probability of the horse winning the $4th$ race is very small $\left(
\text{i.e.} ~ \displaystyle \frac{1}{11}\right)$.  So, if you assume that the horse loses that race, then instead of having to sum $\displaystyle \binom{5}{3}$ terms, you only have to sum $\displaystyle \binom{4}{3}$ terms.
There are certainly other approximation games that can be played.  For example, instead of simply ignoring the $6$ possibilities that correspond to the horse winning the $4$th race, you could reason as follows:
In exactly $3$ of the $10$ races, the horse wins race $3$ and loses race $4$.  Call this group A.  In exactly $3$ of the races, the horse (instead) wins race $4$ and loses race $3$.  Call this group B.
Consider $\displaystyle p_4 \times q_3 = \frac{1}{11} \times \frac{3.3}{4.3}$. 
Denote this as $R$.
Consider $\displaystyle q_4 \times p_3 = \frac{10}{11} \times \frac{1}{4.3}$. 
Denote this as $S$.
As a shortcut to computing the collective sum of the $3$ races in group B, you can take the collective sum of the $3$ races in group A, and multiply it by $\displaystyle \frac{R}{S} = \frac{3.3}{10}.$
