Question

a travel agency is giving away a free trip as part of a grand opening. the trip will be to one of the ten locations listed.

Answer 1

Answer:
`Pr(Nissau\text{ or Paris\rparen= }(1)/(5)\text{ \lparen option B\rparen}`

Step-by-step 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)

`\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{ = }(1)/(10) \end{gathered}`
`\begin{gathered} Pr(Paris)\text{ = }\frac{number\text{ of times Paris appearance}}{total} \\ \\ Pr(Paris)\text{ = }(1)/(10) \end{gathered}`
`\begin{gathered} Pr(Nissau\text{ or Paris\rparen = }(1)/(10)\text{ + }(1)/(10) \\ \\ Pr(Nissau\text{ or Paris\rparen= }(2)/(10)\text{ } \\ \\ Pr(Nissau\text{ or Paris\rparen= }(1)/(5)\text{ \lparen option B\rparen} \end{gathered}`