Answer: [tex]Pr(Nissau\text{ or Paris\rparen= }\frac{1}{5}\text{ \lparen option B\rparen}[/tex]
Explanation:
Given:
list of locations as a giveaway
To find:
the probability that a trip will be either to Nissua or Paris
the probability that a trip will be either to Nissua or Paris = Pr(Nassau) + Pr(Paris)
[tex]\begin{gathered} Pr(Nissau)\text{ = }\frac{number\text{ of times Nissau appeared}}{total\text{ trips}} \\ \\ Nissau\text{ = once} \\ Total\text{ trip = 10} \\ \\ Pr(Nissau)\text{ = }\frac{1}{10} \end{gathered}[/tex][tex]\begin{gathered} Pr(Paris)\text{ = }\frac{number\text{ of times Paris appearance}}{total} \\ \\ Pr(Paris)\text{ = }\frac{1}{10} \end{gathered}[/tex][tex]\begin{gathered} Pr(Nissau\text{ or Paris\rparen = }\frac{1}{10}\text{ + }\frac{1}{10} \\ \\ Pr(Nissau\text{ or Paris\rparen= }\frac{2}{10}\text{ } \\ \\ Pr(Nissau\text{ or Paris\rparen= }\frac{1}{5}\text{ \lparen option B\rparen} \end{gathered}[/tex]
