Question

According to the United States Centers for Disease Control and Prevention (CDC), among individuals 18–25 years old in the USA, it is reported1 that:

• 20% have used cannabis in the past month,
• 60% have consumed alcoholic beverages in the past month, • 10% have smoked cigars in the last month.

Assume using cannabis, consuming alcohol, and smoking cigars are independent but not mutually exclusive events. What is the probability that an individual 18–25 years old either used cannabis or consumed alcohol or smoked cigars in the past month?

Answer 1

20% have used cannabis in the last month.

60% have consumed alcoholic beverages last month.

10% have smoked cigars in the last month.

Let us denote the events of using cannabis, consuming alcoholic and that of smoking cigars, by C, A, and S respectively.

So,

P(C)=0.2

P(A)=0.6

P(S)=0.1

Given that, these 3 events are mutually independent, but not exclusive.

So,

P(A and S)=P(A)P(S)=0.06

P(A and C)=P(A)P(C)=0.12

P(S and C)=P(S)P(C)=0.02

P(A and S and C)=P(A)P(S)P(C)=0.012

Now, we have to find that an individual either used cannabis, or consumed alcoholic beverages, or smoked cigars in the last month.

ie. To find

P( A or S or C )

By inclusion exclusion property, this becomes

=P(A)+P(S)+P(C)-P(A and S)-P(S and C)-P(A and C)+P(A and S and C)

=0.6+0.1+0.2-0.06-0.02-0.12+0.012

=0.712.

So, the required probability is 0.712.