2.7 Independence and the Union of Events

Does Knowing One Result Change the Other Chance?

A display shows a coupon on 25% of visits. Separately, an independent spinner lands on blue with probability 0.40. Let $C$ mean coupon and $B$ mean blue. The stated independence means learning that a coupon appeared does not change the blue chance, and learning the spinner result does not change the coupon chance. Thus

For independent events the joint probability becomes

The resulting four outcome categories are a useful check:

	Coupon	No coupon	Total
Blue	0.10	0.30	0.40
Not blue	0.15	0.45	0.60
Total	0.25	0.75	1.00

From the table, $P(B\mid C)=0.10/0.25=0.40$, confirming the first conditional identity. The product shortcut is justified by the model’s independence, not by the fact that the events have different names.

“Or” Includes Both

The union $B\cup C$ means blue or coupon or both. Adding 0.40 and 0.25 counts the blue-with-coupon outcomes twice. Subtract their overlap once:

In word notation this is $P(B\text{ union }C)=P(B)+P(C)-P(B\text{ intersect }C)$. The addition rule works for dependent events too; only the way we find the overlap may change. The three union cells in the table, 0.10, 0.30 and 0.15, also sum to 0.55.

Disjoint Does Not Usually Mean Independent

Imagine selecting one ticket labeled either Red or Green, with each label having positive probability. Those events are disjoint: their joint probability is 0. But if the ticket is Red, the conditional chance that the same ticket is Green drops to 0. Learning one event has changed the other’s probability. They are dependent.

For positive-probability disjoint events, the product of the individual probabilities is positive while the joint probability is zero, so the independence identity fails. Independence permits both events to happen; disjointness forbids it. To verify a claimed independent model from a table, check whether the conditional probability equals the corresponding marginal probability, or whether the joint equals the product. Keep the event labels and denominator consistent throughout the check.

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.

Two Chance Stations and an Overlapping Survey

Station 1 rolls a fair six-sided die; event A is a result of 5 or 6. Station 2 tosses a fair coin independently; event B is heads. In a separate survey population, 60% visited a science museum (S), 50% visited a history museum (H), and 40% visited both. These survey percentages are exact for the stated population.

Part a

For the stations, calculate P(A), P(B), P(A and B), P(A given B), and P(B given A). Use your results to verify independence. Then calculate P(A or B or both), showing the union rule.

Your reasoning

Part b

Calculate the museum union probability in both word and symbol notation. Verify whether S and H are independent, and justify why simply adding 60% and 50% would be wrong.

Your reasoning

Part c

A separate spinner has two red sectors, two blue sectors and two green sectors, all equally likely. On one spin, calculate the joint and union probabilities for Red and Blue. Use a conditional probability to explain why these disjoint events are dependent.

Your reasoning

Worked answer and rubric