Please explain me the solutions for this problem?

Question

Suppose that in a senior college class of 500 students it is found that 210 smoke, 258 drink alcoholic beverages, 216 east between meals, 122 smoke and drink alcoholic beverages, 83 east between meals and drink alcoholic beverages, 97 smoke and eat between meals, and 52 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

At first let's sum up:  class consists of 500 students and

52 (eat, drink and smoke)

45 (smoke and eat but does not drink)

31 (eat and drink but does not smoke)

70 (smoke and drink but does not eat)

88 (just eat)

105 (just drink)

43 (just smoke)

66 (are healthy)

a) smokes but does not drink 45+43=88 probability= 88/500

b) eats and drinks but does not smoke 31 probability=31/500

c) neither smokes nor eats 105+66= 171 probability=171/500

Answer 2

This problem is more easily solved with Venn diagrams.  You should look in your book for this topic.

Start with the group that engages in all three---52 seniors.

Next consider the groups that engage in two of these activities.

122 smoke and drink.  Since 52 do all three, 70 more must smoke and drink.

83 eat and drink.  Since 52 do all three, 31 more must drink and eat.

97 smoke and eat; since 52 do all three, 45 more must smoke and drink.

Now consider each activity alone.

210 smoke.  Since 52 do all three, 70 smoke and drink, and 45 smoke and eat, then 43 more must smoke only.

Similarly, 105 must drink only and 88 must eat only.

This leaves 66 seniors who do not smoke nor drink nor eat between meals.  That is the entire universe---500

a)  P(smokes but does not drink) = # who smoke or smoke and eat/# who do not drink alcohol = 88/242 = 4/11

You should be able to do the rest now that you have the counts for all the possible groups.  A Venn diagram would do wonders...