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D unions intersections and complements
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Easy


In a group of students, $25%$ smoke cigarettes, $60%$ drink alcohol, and $15%$ do both. What fraction of students have at least one of these bad habits?

Answer:

$70%$

Solution:

Let $C$ denote the event that a randomly picked student in the group smokes cigarettes, and let $A$ be the event that the student drinks alcohol. From the information given, we see that $P(C)=0.25$, $P(A)=0.6$, and $P(C\cap A)=0.15$.

The event that a student has at least one of these bad habits is $C\cup A$. And,\[\begin{align}P(C\cup A)&=P(C)+P(A)-P(C\cap A)\text{, addition rule }\\
&=0.25+0.6-0.15\\
&=0.7.\end{align}
\]Therefore, 70% of students have at one of these bad habits.