Question

Suppose that in a senior college class of 500 ​students, it is found that 179 ​smoke, 241 drink alcoholic​ beverages, 192 eat between​ meals, 103 smoke and drink alcoholic​ beverages, 66 eat between meals and drink alcoholic​ beverages, 73 smoke and eat between​ meals, and 33 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 1

(a) 179 students smoke and 103 out of them also drink alcoholic beverages. Thus 179-103 = 76 many students smoke but do not drink alcoholic beverages. Thus, the required probability is: 76/500=0.152

(b) We know that 33 students engage in all the bad health practices. Thus, 33 of them eat between meals, smoke as well as drink alcoholic beverages. Now, it is given that 66 eat between meals and drink alcoholic​ beverages. Thus 66-33=33 many eats between meals, drinks alcoholic bevrages but doesn't smoke. Hence the required probability is:33/500=0.066

(c) We know that 179 students smoke, 192 eat between meals and 73 does both. Thus, using inclusion exclusion principle, numeber of students who smoke or eat between meals is 179+192-73=298. Thus 500-298=202 many of them doesn't smoke noe do they eat between meals. Hence, the required probability is: 202/500=0.404