2.5 Joint Probability and Disjoint Events

Can Both Things Be True?

A selected transit ticket can be a student ticket and a weekday ticket at the same time. The intersection $A\cap B$ contains those shared outcomes. Its probability, $P(A\cap B)$, is the joint probability—the chance that both events occur. The notation $P(A\text{ intersect }B)$ means the same thing.

Consider a complete set of 200 tickets, each equally likely to be selected. Each ticket has one student-status label and one weekday-status label.

	Weekday B	Not weekday	Total
Student A	60	20	80
Not student	80	40	120
Total	140	60	200

The overlap contains 60 tickets, so $P(A\cap B)=60/200=0.30$. This is not $P(A)$ or $P(B)$: the numerator must satisfy both descriptions. Thirty percent of this ticket population has both labels.

Worked Contrast: One Assigned Route

Now let $R$ mean “a ticket is assigned the Red route” and $G$ mean “it is assigned the Green route.” Suppose each ticket has exactly one route assignment. No ticket belongs to both route categories. Thus $R\cap G=\varnothing$, and $P(R\cap G)=0$. The events are mutually exclusive, also called disjoint: they cannot happen together on the same selected ticket.

Compare that with student and weekday status. Their positive joint probability proves they are not disjoint. A useful justification names the observational trial, checks whether shared outcomes exist, and connects that overlap to joint probability. Simply calling the labels “different” is insufficient; different attributes can overlap.

For finite equally likely ticket outcomes, zero shared tickets means zero joint probability. The set statement $R\cap G=\varnothing$ and the numerical statement $P(R\cap G)=0$ describe different objects. Keep the set and probability notation distinct. Also keep disjoint separate from independent: whether events can occur together is a different question from whether learning one event changes the probability of the other.

When reading a table or diagram, point to the overlap explicitly. In the table, it is the 60 in the student/weekday cell. Adding row and column totals would double-count that overlap rather than calculate a joint probability.

Try It Yourself

The following original practice applies the ideas above to a fresh setting. Show the reasoning requested in each part. Answers and task-specific rubrics are on the separate solutions page.

Original generated practice for the Fall 2026 course.

Which Locker-Code Events Are Disjoint?

A locker-code generator selects one integer uniformly from 1 through 10. Let A={1,2,3}, B={8,9,10}, C={2,4,6,8,10}, and D={8,9,10}. A trial is one selected code; every code has positive probability 1/10.

Part a

Calculate P(A intersect B), also writing the probability with the intersection symbol. Justify whether A and B are mutually exclusive using both possible joint outcomes and joint probability.

Your reasoning

Part b

Calculate P(C intersect D). Explain why this is a joint probability and justify whether C and D are disjoint. Distinguish the intersection set from its numerical probability.

Your reasoning

Worked answer and rubric