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 0.064
The probability that the student neither smokes nor eats between meals is 0.322
Step-by-step explanation:
Consider the provided information.
Suppose that in a senior college class of 500 students, it is found that 215smoke, 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) = 124/500
Thus,
![P(A\cap B')=P(A)-P(A\cap B)](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%27%29%3DP%28A%29-P%28A%5Ccap%20B%29)
![P(A\cap B')=\frac{215}{500}-\frac{124}{500}=0.43-0.248=0.182](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%27%29%3D%5Cfrac%7B215%7D%7B500%7D-%5Cfrac%7B124%7D%7B500%7D%3D0.43-0.248%3D0.182)
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:
![P(C\cap B \cap A')=P(B\cap C)-P(A\cap B \cap C)](https://tex.z-dn.net/?f=P%28C%5Ccap%20B%20%5Ccap%20A%27%29%3DP%28B%5Ccap%20C%29-P%28A%5Ccap%20B%20%5Ccap%20C%29)
From the given information.
P(B ∩ C) = 78/500 and P(A ∩ B ∩ C)= 46/500
Thus,
![P(C\cap B \cap A')=\frac{78}{500}-\frac{46}{500}=0.156-0.092=0.064](https://tex.z-dn.net/?f=P%28C%5Ccap%20B%20%5Ccap%20A%27%29%3D%5Cfrac%7B78%7D%7B500%7D-%5Cfrac%7B46%7D%7B500%7D%3D0.156-0.092%3D0.064)
The probability that the student eats between meals and drinks alcoholic beverages but does not smoke is 0.064
Part (C) Neither smokes nor eats between meals.
Neither smokes nor eats between meals can be written as:
![P((A\cup C)')=1-P(A\cup C)](https://tex.z-dn.net/?f=P%28%28A%5Ccup%20C%29%27%29%3D1-P%28A%5Ccup%20C%29)
From the given information.
P(A) = 215/500 and P(C)= 215/500 and P(A∩C)=91/500
P(A∪C)=P(A)+P(C)-P(A∩C)
P(A∪C)=![\frac{215}{500}+\frac{215}{500}-\frac{91}{500}=\frac{339}{500}](https://tex.z-dn.net/?f=%5Cfrac%7B215%7D%7B500%7D%2B%5Cfrac%7B215%7D%7B500%7D-%5Cfrac%7B91%7D%7B500%7D%3D%5Cfrac%7B339%7D%7B500%7D)
Substitute the respective values in the above formula.
![P((A\cup C)')=1-P(A\cup C)=1-\frac{339}{500} \\1-0.678=0.322](https://tex.z-dn.net/?f=P%28%28A%5Ccup%20C%29%27%29%3D1-P%28A%5Ccup%20C%29%3D1-%5Cfrac%7B339%7D%7B500%7D%20%5C%5C1-0.678%3D0.322)
The probability that the student neither smokes nor eats between meals is 0.322