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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%?

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

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

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  • $\begingroup$ Thanks a lot! Got it! :) $\endgroup$
    – Leah
    Oct 4 '20 at 10:08
  • $\begingroup$ @N.F.Taussig that's the same thing as not getting offers from both companies, right? $\endgroup$ Oct 4 '20 at 10:10
  • $\begingroup$ @N.F.Taussig I revised the answer according to your suggestions. $\endgroup$ Oct 4 '20 at 10:12
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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 $0.5$.

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