Question

a company is expanding globally and sending its employees overseas for business meetings. the travel department is booking flights on a number of different airlines. the frequency table below shows the distribution of airline tickets purchased over the course of the last year. assume that each employee attended only one overseas meeting that year.
what is the probability of traveling to beijing or flying on skylark airlines?
rows represent the different airlines with which the employees have flown.
columns represent the different cities to which the employees have traveled.

	singapore	johannesburg	beijing	total
cloud crew	6	7	7	20
skylark	9	6	8	23
total	20	23	17	60

event a: traveling to beijing
event b: flying on skylark airlines

use the frequency table to compute the following:
probability of traveling to beijing:
p(a) =

probability of flying on skylark airlines:
p(b) =

probability of traveling to beijing and flying on skylark airlines:
p(a and b) =

probability of traveling to beijing or flying on skylark airlines:
p(a or b) =

Explanation:

Step1: Calculate probability of traveling to Beijing

Probability is number of favorable outcomes over total outcomes. Number of employees traveling to Beijing is 17 and total number of employees (total tickets) is 60. So $P(A)=\frac{17}{60}$.

Step2: Calculate probability of flying on Sky - Lark airlines

Number of employees flying on Sky - Lark airlines is 23 and total number of employees is 60. So $P(B)=\frac{23}{60}$.

Step3: Calculate probability of traveling to Beijing and flying on Sky - Lark airlines

Number of employees traveling to Beijing and flying on Sky - Lark airlines is 8 and total number of employees is 60. So $P(A\ and\ B)=\frac{8}{60}=\frac{2}{15}$.

Step4: Calculate probability of traveling to Beijing or flying on Sky - Lark airlines

Use the formula $P(A\ or\ B)=P(A)+P(B)-P(A\ and\ B)$. Substitute the values: $P(A\ or\ B)=\frac{17}{60}+\frac{23}{60}-\frac{8}{60}=\frac{17 + 23-8}{60}=\frac{32}{60}=\frac{8}{15}$.

Answer:

$P(A)=\frac{17}{60}$
$P(B)=\frac{23}{60}$
$P(A\ and\ B)=\frac{2}{15}$
$P(A\ or\ B)=\frac{8}{15}$