Question

the table gives the proportion of adults in a certain country in each age group, as well as the proportion in each group who smoke cigarettes. find the probability that a randomly selected adult who smokes is between 18 - 24 years of age (inclusive).
the probability is
(type an integer or decimal rounded to four decimal places as needed )

Explanation:

Step1: Calculate the number of smokers in each age - group

Let \(P(A_i)\) be the proportion of the population in age - group \(i\) and \(P(S|A_i)\) be the proportion of smokers in age - group \(i\). The number of smokers in age - group \(i\) is \(P(A_i)\times P(S|A_i)\).
For \(18 - 24\) years: \(P(A_1)P(S|A_1)=0.121\times0.071 = 0.008591\)
For \(25 - 44\) years: \(P(A_2)P(S|A_2)=0.348\times0.168 = 0.058464\)
For \(45 - 64\) years: \(P(A_3)P(S|A_3)=0.342\times0.167 = 0.057114\)
For \(65\) years and over: \(P(A_4)P(S|A_4)=0.189\times0.086 = 0.016254\)

Step2: Calculate the total proportion of smokers

\(P(S)=P(A_1)P(S|A_1)+P(A_2)P(S|A_2)+P(A_3)P(S|A_3)+P(A_4)P(S|A_4)\)
\(P(S)=0.008591 + 0.058464+0.057114 + 0.016254=0.140423\)

Step3: Calculate the conditional probability

We want to find \(P(A_1|S)\) using Bayes' theorem \(P(A_1|S)=\frac{P(A_1)P(S|A_1)}{P(S)}\)
\(P(A_1|S)=\frac{0.008591}{0.140423}\approx0.0612\)

Answer:

\(0.0612\)