Given:
The number of students who drank alcohol is 836 students.
[tex]n(A)=836[/tex]
The number of students who smoked cigarettes is 624 students.
[tex]n(C)=624[/tex]
The number of students who used illegal drugs is 176 students.
[tex]n(D)=176[/tex]
The number of students who drank alcohol and used smoked cigarettes is 395.
[tex]n(A\cap C)=395[/tex]
The number of students who drank alcohol and used illegal drugs is 101
[tex]n(A\cap D)=101[/tex]
The number of students who smoked cigarettes and used illegal drugs is 106
[tex]n(C\cap D)=106[/tex]
The number of students engaged in all three is 85.
[tex]n(A\cap C\cap D)=85[/tex]
The total number of students is 1250 students.
[tex]n(A\cup C\cup D)=1250-131=1119[/tex]
131 engaged in none of the behavior.
The number of students who drank alcohol or used illegal drugs is 911
[tex]n(AUD)=911[/tex]
Required:
We need to find the number of students who drank alcohol only.
Explanation:
Let set A represents the number of students who drank alcohol.
Let set C represents the number of students who smoked cigarettes.
Let set D represents the number of students who used illegal drugs.
Consider the formula.
[tex]n(A\cup C\cup D)=n(A)+n(C)+n(D)-n(A\cap C)-n(A\cap D)-n(C\cap D)-n(A\cap C\cap D)[/tex]
[tex]n(AUD)=911[/tex]
[tex]n(A\text{ only\rparen=n\lparen A\rparen-}n(A\cap D)-n(A\cap C)+n(A\cap C\cap D)[/tex]
Substitute know values.
[tex]n(A\text{ only\rparen=836-395-101+85}[/tex][tex]n(A\text{ only\rparen=425}[/tex]
Final answer:
The number of students who drank alcohol only is 425 students.
