Suppose that in a senior college class of 500500 ​students, it is found that 215215 ​smoke, 252252 drink alcoholic​ beverages, 215215 eat between​ meals, 124124 smoke and drink alcoholic​ beverages, 7878 eat between meals and drink alcoholic​ beverages, 9191 smoke and eat between​ meals, and 4646 engage in all three of these bad health practices. If a member of this senior class is selected at​ random, find the probability that the student​ (a) smokes but does not drink alcoholic​ beverages; (b) eats between meals and drinks alcoholic beverages but does not​ smoke; (c) neither smokes nor eats between meals.

Answer:The probability that the student​ smokes but does not drink alcoholic​ beverages is 0.182.The probability that the student​ eats between meals and drinks alcoholic beverages but does not​ smoke is 0.064The probability that the student​ neither smokes nor eats between meals is 0.322Step-by-step explanation:Consider the provided information.Suppose that in a senior college class of 500 ​students, it is found that 215​smoke, 252 drink alcoholic​ beverages, 215 eat between​ meals, 124 smoke and drink alcoholic​ beverages, 78 eat between meals and drink alcoholic​ beverages, 91 smoke and eat between​ meals, and 46 engage in all three of these bad health practices. Let A is the represents student smoke.B is the event, which represents student drink alcoholic beverage.Part (A) Smokes but does not drink alcoholic​ beverages.From the above information.P(A) = 215/500 and P(A ∩ B) = 124/500Thus,[tex]P(A\cap B')=P(A)-P(A\cap B)[/tex][tex]P(A\cap B')=\frac{215}{500}-\frac{124}{500}=0.43-0.248=0.182[/tex]The probability that the student​ smokes but does not drink alcoholic​ beverages is 0.182.Part (B) Eats between meals and drinks alcoholic beverages but does not​ smoke.Let C is the event, which represents student eat between meals.Eats between meals and drinks alcoholic beverages but does not​ smoke can be written as:[tex]P(C\cap B \cap A')=P(B\cap C)-P(A\cap B \cap C)[/tex]From the given information.P(B ∩ C) = 78/500 and P(A ∩ B ∩ C)= 46/500Thus, [tex]P(C\cap B \cap A')=\frac{78}{500}-\frac{46}{500}=0.156-0.092=0.064[/tex]The probability that the student​ eats between meals and drinks alcoholic beverages but does not​ smoke is 0.064Part (C) Neither smokes nor eats between meals.Neither smokes nor eats between meals can be written as:[tex]P((A\cup C)')=1-P(A\cup C)[/tex]From the given information.P(A) = 215/500 and P(C)= 215/500 and P(A∩C)=91/500P(A∪C)=P(A)+P(C)-P(A∩C)P(A∪C)=[tex]\frac{215}{500}+\frac{215}{500}-\frac{91}{500}=\frac{339}{500}[/tex]Substitute the respective values in the above formula.[tex]P((A\cup C)')=1-P(A\cup C)=1-\frac{339}{500} \\1-0.678=0.322[/tex]The probability that the student​ neither smokes nor eats between meals is 0.322