Probability of getting at least one job offer based on interview odds I have recently interviewed for a number of jobs, and am wondering what the odds are of getting accepted for one based on the odds of each interview and the total number of interviews. 


*

*I had $7$ interviews which had a $1$ in $10$ chance of securing a job ($10$
interviewees for every applicant).

*I had another interview with a $6\%$ chance, another with a $5\%$ chance, and another with a $14\%$ chance.


So in total, $10$ places with odds of $10\%$, $10\%$, $10\%$, $10\%$, $10\%$, $10\%$, $10\%$, $14\%$, $6\%$, $5\%$. 
I know that I cannot simply add the numbers, but I am also stumped in that I am trying to find the odds of not one job offer, but a minimum of one offer....so the odds of either one positive return, two positive return, etc. vs the odds of all $10$ interviews coming up negative. 
Anyone know how to calculate this problem? 

This is a problem in which the complementary approach will be the most fruitful - let's instead consider how likely you are to not get a job. We know that, for an event $A$, then
$$P(A) = 1 - P(\text{not} \; A)$$
That is to say, more relevant to your case,
$$P(\text{getting at least one job offer}) = 1 - P(\text{getting no job offers})$$
Since the odds of getting a job doesn't affect that for any other job, we know
$$\begin{align}
P(\text{getting no offers}) &= (1 - P(\text{getting job #1})) \\
&\times (1 - P(\text{getting job #2})) \\
&\times (1 - P(\text{getting job #3})) \\
&... \\
&\times (1 - P(\text{getting job #10}))
\end{align}$$
With these two facts in mind you should find it easy to complete.
