# 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%?

• The probability of missing on both is $.8^2=.64$ so the probability of getting at least one offer is $1-.64=.36$
– lulu
Oct 4 '20 at 10:01

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 $$36\%$$.

• Thanks a lot! I understood it! :)
– Leah
Oct 4 '20 at 10:07
• You're welcome :) Oct 4 '20 at 10:08

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.

• Thanks a lot! Got it! :)
– Leah
Oct 4 '20 at 10:08
• @N.F.Taussig that's the same thing as not getting offers from both companies, right? Oct 4 '20 at 10:10
• @N.F.Taussig I revised the answer according to your suggestions. Oct 4 '20 at 10:12

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 $$0.5$$.