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? 
 A: 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.
A: It depends if you're looking for the odds to get exactly one job, or at least one job. 
To get at least one job you have to substract the probability to get none of the jobs from 1.
Pr(at least one job) = 1 - Pr(zero jobs) = 1 - $(\frac{9}{10})^7 * \frac{43}{50} * \frac{47}{50} * \frac{19}{20}$ 
