问题及解答

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, $60\%$ can be repaired whereas the other $40\%$ must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

[中文]

$20\%$的某種類型的電話在保修期間需要維修。其中$60\%$可以修好，其餘$40\%$必須更換新裝置。如果一家公司購買了其中十部手機，那麼在保修期內，正好有兩部電話被替換的可能性是多少？

[法语]

Vingt pour cent de tous les téléphones d'un certain type sont soumis au service pendant la garantie. De ce nombre, $60\%$ peuvent être réparés alors que les $40\%$ restants doivent être remplacés par de nouvelles unités. Si une entreprise achète dix de ces téléphones, quelle est la probabilité que exactement deux finissent par être remplacés sous garantie?
 

Let $A$ denote the event that the telephone under warranty is need service. Then $P(A)=0.2$ by assumption.

Let $B$ denote the event that the telephone underwarranty can be repaired. Then $P(B|A)=0.6$. Thus, $P(B^c|A)=0.4$, which is the probability of the telephone among the event $A$ that must be replaced with new units.

Then, $B^c\cap A$ is the event $\{\text{telephone is need service and is ERROR}\}$. Here ERROR means it must be replaced its units. Therefore, the probability of $B^c\cap A$ is 

\[
P(B^c\cap A)=P(B^c|A)\cdot P(A)=0.4\times 0.2=0.08
\]

Let $X$ be the number of the telephones whose units are replaced during the service. Then, for a company purchased ten of these telephones, the probability that exactly two will end up being replaced under waranty is calculated as follows:

\[
\begin{split}
&C_{10}^{2}\cdot P(B^c\cap A)^2\cdot(1-P(B^c\cap A))^{10-2}\\
=&45\times 0.08^2\times 0.92^8\\
=&0.14780703546361774080
\end{split}
\]