Probability of recieving an offer Your friend tells you that he had two job interviews last week. He says that based on how the interviews went, he thinks he has a 20% chance of receiving an offer from each of the companies he interviewed with. Nevertheless, since he interviewed with two companies, he is 50% sure that he will receive at least one offer. Is he right?
My Approach:
since he had 2 interviews, 20% each, the total probability of getting should be 40%?
 A: He is not right.
The probability of getting at least one offer is 1 minus the probability of getting an offer from neither company. This probability is $P = \frac{4}{5} \times \frac{4}{5} = \frac{16}{25}$, which comes out to $64\%$. Thus, there is a $36\%$ chance of getting at least one offer.
A: Hello and welcome to Math.SE.
Your intuition is not serving you correctly here.
If we assume that the events of getting offers from the different companies are independent we find
$$
\begin{align}
\Pr(\text{Friend receives at least 1 offer})
&= 1 - \Pr(\text{Friend receives no offer}) \\
&= 1 - \Pr(\text{No offer from C1 and no offer from C2}) \\
&= 1 - \Pr(\text{No offer from C1}) \cdot \Pr(\text{No offer from C2}) \\
&= 1 - 0.8^2 \\
&= 0.36
\end{align}
$$
Hence the chance of receiving at least one offer would be $36\%$.
A: I just want to point out that we needn't assume the independence of events. In that case we can use union bound to get that $P( \text{at least one offer}) \leq P(\text{offer from A}) + P(\text{offer from B}) =2 \cdot 1/5 =0.4$, which is still less than $0.5$.
