Answer: Our required probability is 81%.
Explanation:
Since we have given that
Probability of defective calculators (C)= 10%
Probability of defective batteries (B) = 14%
Probability of both defective calculators and defective batteries = 5%
So, it becomes.
![P(C\cup B)=P(C)+P(B)-P(B\cap C)\\\\P(C\cup B)=10+14-5\\\\P(C\cup B)=19\%](https://img.qammunity.org/2020/formulas/mathematics/college/wwnvusgbywh8yor67ubjr5llt5924otdtd.png)
So, probability that the calculator has a good case and good batteries is given by
![100\%-19\%=81\%](https://img.qammunity.org/2020/formulas/mathematics/college/hovrjbmbnjubvyidcaovd0pv5n78kafnuu.png)
Hence, our required probability is 81%.