Answer: a) 0.16, b) 0.058, and c) 0.856.
Step-by-step explanation:
Since we have given that
Number of students = 500
Number of students smoke = 179
Number of students drink alcohol = 228
Number of students eat between meals = 119
Number of students eat between meals and drink alcohol = 59
Number of students eat between meals and smoke = 72
Number of students engage in all three = 30
a) Probability that the student smokes but does not drink alcohol is given by
![P(S-A)=P(S)-P(S\cap A)\\\\P(S-A)=\dfrac{179}{500}-\dfrac{99}{500}\\\\P(S-A)=\dfrac{179-99}{500}\\\\P(S-A)=\dfrac{80}{500}\\\\P(S-A)=0.16](https://tex.z-dn.net/?f=P%28S-A%29%3DP%28S%29-P%28S%5Ccap%20A%29%5C%5C%5C%5CP%28S-A%29%3D%5Cdfrac%7B179%7D%7B500%7D-%5Cdfrac%7B99%7D%7B500%7D%5C%5C%5C%5CP%28S-A%29%3D%5Cdfrac%7B179-99%7D%7B500%7D%5C%5C%5C%5CP%28S-A%29%3D%5Cdfrac%7B80%7D%7B500%7D%5C%5C%5C%5CP%28S-A%29%3D0.16)
b) eats between meals and drink alcohol but does not smoke.
![P((M\cap A)-S)=P(M\cap A)-P(M\cap S\cap A)\\\\P((M\cap A)-S)=\dfrac{59}{500}-\dfrac{30}{500}\\\\P((M\cap A)-S)=\dfrac{59-30}{500}\\\\P((M\cap A)-S)=\dfrac{29}{500}\\\\P((M\cap A)-S)=0.058](https://tex.z-dn.net/?f=P%28%28M%5Ccap%20A%29-S%29%3DP%28M%5Ccap%20A%29-P%28M%5Ccap%20S%5Ccap%20A%29%5C%5C%5C%5CP%28%28M%5Ccap%20A%29-S%29%3D%5Cdfrac%7B59%7D%7B500%7D-%5Cdfrac%7B30%7D%7B500%7D%5C%5C%5C%5CP%28%28M%5Ccap%20A%29-S%29%3D%5Cdfrac%7B59-30%7D%7B500%7D%5C%5C%5C%5CP%28%28M%5Ccap%20A%29-S%29%3D%5Cdfrac%7B29%7D%7B500%7D%5C%5C%5C%5CP%28%28M%5Ccap%20A%29-S%29%3D0.058)
c) neither smokes nor eats between meals.
![P(S'\cap M')=1-P(S\cup M)\\\\P(S'\cap M')=1-\dfrac{72}{500}\\\\P(S'\cap M')=\dfrac{500-72}{500}\\\\P(S'\cap M')=\dfrac{428}{500}=0.856](https://tex.z-dn.net/?f=P%28S%27%5Ccap%20M%27%29%3D1-P%28S%5Ccup%20M%29%5C%5C%5C%5CP%28S%27%5Ccap%20M%27%29%3D1-%5Cdfrac%7B72%7D%7B500%7D%5C%5C%5C%5CP%28S%27%5Ccap%20M%27%29%3D%5Cdfrac%7B500-72%7D%7B500%7D%5C%5C%5C%5CP%28S%27%5Ccap%20M%27%29%3D%5Cdfrac%7B428%7D%7B500%7D%3D0.856)
Hence, a) 0.16, b) 0.058, and c) 0.856.