Questions: Suppose that in a senior college class of 500 students, it is found that 197 smoke, 245 drink alcoholic beverages, 215 eat between meals, 121 smoke and drink alcoholic beverages, 81 eat between meals and drink alcoholic beverages, 93 smoke and eat between meals, and 53 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. (a) P( smokes but does not drink alcoholic beverages )= (Type an integer or a decimal. Do not round.)

Solution Steps

To solve this problem, we will use the principle of inclusion-exclusion to find the number of students who meet each specific condition. Then, we will calculate the probabilities by dividing the number of students meeting each condition by the total number of students (500).

1. Calculate the number of students who smoke.
2. Subtract the number of students who both smoke and drink alcoholic beverages.
3. Divide by the total number of students to get the probability.

1. Calculate the number of students who eat between meals and drink alcoholic beverages.
2. Subtract the number of students who engage in all three practices.
3. Divide by the total number of students to get the probability.

1. Calculate the number of students who smoke or eat between meals using the principle of inclusion-exclusion.
2. Subtract this number from the total number of students to get the number of students who neither smoke nor eat between meals.
3. Divide by the total number of students to get the probability.

To find the number of students who smoke but do not drink alcoholic beverages, we subtract the number of students who both smoke and drink from the total number of students who smoke: \[ \text{smoke only} = \text{smoke} - \text{smoke and drink} = 197 - 121 = 76 \]

The probability is given by the ratio of the number of students who smoke but do not drink to the total number of students: \[ \mathrm{P}(\text{smokes but does not drink}) = \frac{\text{smoke only}}{\text{total students}} = \frac{76}{500} = 0.152 \]

To find the number of students who eat between meals and drink alcoholic beverages but do not smoke, we subtract the number of students who engage in all three practices from the number of students who eat between meals and drink: \[ \text{eat and drink only} = \text{eat and drink} - \text{all three} = 81 - 53 = 28 \]

The probability is given by the ratio of the number of students who eat between meals and drink but do not smoke to the total number of students: \[ \mathrm{P}(\text{eats and drinks but does not smoke}) = \frac{\text{eat and drink only}}{\text{total students}} = \frac{28}{500} = 0.056 \]

Using the principle of inclusion-exclusion, we first find the number of students who smoke or eat between meals: \[ \text{smoke or eat} = \text{smoke} + \text{eat} - \text{smoke and eat} = 197 + 215 - 93 = 319 \] Then, we subtract this number from the total number of students to find the number of students who neither smoke nor eat between meals: \[ \text{neither smoke nor eat} = \text{total students} - \text{smoke or eat} = 500 - 319 = 181 \]

The probability is given by the ratio of the number of students who neither smoke nor eat between meals to the total number of students: \[ \mathrm{P}(\text{neither smokes nor eats}) = \frac{\text{neither smoke nor eat}}{\text{total students}} = \frac{181}{500} = 0.362 \]

Final Answer