91影视

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

We know, according to the table,

Here,

"One person who is Japanese American and smokes $21to30$cigarettes per day" means a person who is Japanese American and also the same person who smokes $21to30$cigarettes.

Since, there is $4715$person who is Japanese American and smokes $21to30$cigarettes.

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, probability is $4715$divided by localid="1650550306235" $100450.$

$P(ThatapersonisJapaneseAmericanandsmokes21to30cigarettesperday)=\frac{4715}{100450}=0.047.$

The required probability is$0.047.$