91Ó°ÊÓ

$\text{An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were}9,886\text{African Americans,}2,745\text{Native Hawaiians,}12,831\text{Latinos,}8,378\text{Japanese Americans, and}7,650\text{Whites. Of the people smoking}11to20\text{cigarettes per day, there were}6,514\text{African Americans,}3,062\text{Native Hawaiians,}4,932\text{Latinos,}10,680\text{Japanese Americans, and}9,877\text{Whites. Of the people smoking}21to30\text{cigarettes per day, there were}1,671\text{African Americans,}1,419\text{Native Hawaiians,}1,406\text{Latinos,}4,715\text{Japanese Americans, and}6,062\text{Whites. Of the people smoking at least}31\text{cigarettes per day, there were}759\text{African Americans,}788\text{Native Hawaiians,}800\text{Latinos,}2,305\text{Japanese Americans, and}3,970\text{Whites}$

We know, according to the table

There were $19969$Latino people and $100450$total no. of people.

So, by using normal method of probability in which to get the probability we divide no. of favorable outcomes by total no. of events.

Therefore, the probability that the person was Latino is $19969dividedby100450$.

$$\begin{gathered} P(ThatthepersonwasLatino)=\frac{\text{No.of people who were Latino}}{\text{Total no.of people}} \\ P(ThatthepersonwasLatino)=\frac{19969}{100450}=0.1988 \end{gathered}$$

The probability is $0.1988.$