Decide whether two events are independent and explain conditional probability in everyday situations

Learning goal: Understand that two events A and B are independent if and only if the probability of A and B occurring together is the product of their probabilities, use this characterization to determine whether events are independent, interpret independence as P(A | B) = P(A) and P(B | A) = P(B) (each when the conditioning event has positive probability), and recognize and explain conditional probability and independence in everyday language and everyday situations.

The definition. Two events A and B are independent if and only if

P(A and B) = P(A) · P(B).

This is a definition, not a theorem, and it is a statement about numbers in the model. It does not say the events "have nothing to do with each other" in some physical sense, and it is not decided by intuition. To determine whether two events are independent you compute three probabilities, P(A), P(B) and P(A and B), and check whether the third equals the product of the first two. If it does, they are independent; if it does not, they are dependent.

The same idea through conditional probability. Since P(A | B) = P(A and B) / P(B), dividing both sides of the product equation by P(B), when P(B) > 0, gives P(A | B) = P(A). Dividing by P(A) instead, when P(A) > 0, gives P(B | A) = P(B). So independence means: the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. In words, learning that B happened does not change how likely A is, and learning that A happened does not change how likely B is. Either conditional statement is equivalent to the product statement (when the events have positive probability), so you may test independence whichever way is convenient.

A surprising independent pair. Roll a red and a blue die. Let B be "red die is even", P(B) = 18/36 = 1/2, and A be "sum is 7", P(A) = 6/36 = 1/6. From the previous lessons, P(A and B) = 3/36 = 1/12. Is 1/12 equal to (1/2)(1/6)? Yes. So A and B are independent. Check through conditionals: P(A | B) = 3/18 = 1/6 = P(A), and P(B | A) = 3/6 = 1/2 = P(B). Knowing the red die is even tells you nothing about whether the sum is 7. Many students expect the sum and the red die to be related, and in general they are; this particular pair of events happens to satisfy the definition, which is why the definition, not intuition, must decide.

A dependent pair from the same dice. Keep B, red die even, and let E be "sum is 8". E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, so P(E) = 5/36. E and B share (2, 6), (4, 4), (6, 2), so P(E and B) = 3/36 = 1/12. The product is (1/2)(5/36) = 5/72. Since 1/12 = 6/72 is not 5/72, E and B are dependent. Through conditionals: P(E | B) = 3/18 = 1/6, which is larger than P(E) = 5/36; knowing the red die is even raises the chance of a sum of 8, because a sum of 8 requires the two dice to have the same parity, and an even red die then needs an even blue die. The two tests always agree, because they are algebraically the same statement.

Independence by construction. Sometimes independence is part of the model rather than something to check. If a coin is tossed and, separately, a die is rolled, we model the 12 outcomes as equally likely; then "heads" (6 of 12) and "rolls a 5" (2 of 12) have P(heads and 5) = 1/12 = (1/2)(1/6), independent. Physically separate trials are usually modeled as independent, but that is an assumption justified by the setup, not a consequence of separation alone: separate trials that share a changing condition, such as free throws by a player who is tiring, may be dependent. When the assumption is justified, the product rule lets us build the joint probabilities. The check and the construction go in opposite directions: in one you have the joint probability and test it; in the other you assume independence and compute the joint probability.

Everyday language: what "given" means. In everyday situations, conditional probability hides inside words like "if", "among", "of those who", "when", and "given that". "The chance of rain if the sky is grey at breakfast" is P(rain | grey sky). "Among students who eat breakfast, the fraction who pass" is P(pass | breakfast). Recognizing the given event is the whole skill: the sentence names a group, and the probability is a share of that group.

Everyday language: reversing the condition changes the question. The chance of having a disease if you have a risk factor is not the chance of having the risk factor if you have the disease. Take the classic comparison: the chance of having lung cancer given that a person is a smoker, versus the chance that a person is a smoker given that they have lung cancer. These are P(cancer | smoker) and P(smoker | cancer), and they can be wildly different. Use an explicitly hypothetical population (invented numbers for arithmetic practice, not medical statistics): 10,000 adults, of whom 2,500 smoke; 150 of the smokers and 50 of the 7,500 non-smokers develop the disease. Then P(cancer | smoker) = 150/2,500 = 3/50 = 0.06, while P(smoker | cancer) = 150/200 = 3/4 = 0.75. The first is a share of smokers; the second is a share of patients. Both are much larger than they would be under independence, since P(cancer | non-smoker) = 50/7,500 = 1/150 ≈ 0.0067 is far below 0.06, so in this hypothetical population smoking and the disease are dependent. Notice also that a large P(smoker | cancer) does not by itself mean that most smokers get the disease; in the invented data 94% of smokers do not.

Everyday language: independence. Two everyday events are independent when knowing one happened does not change your probability for the other. Whether a randomly chosen person was born on a Tuesday and whether their surname begins with M is a plausible independent pair; whether it rained today and whether the sidewalk is wet is a dependent pair, since P(wet | rain) is far above P(wet). But "plausible" is not proof: for a claim of independence inside a data set, compute P(A and B) and P(A)P(B) and compare. When the probabilities are estimated from a sample, exact equality is not expected, and a small gap may be sampling variation; the next lesson treats that case.

The misconception to avoid. "Independent" is not the same as "mutually exclusive". Mutually exclusive events cannot both happen, so P(A and B) = 0; if both have positive probability the product P(A)P(B) is positive, so mutually exclusive events with positive probability are always dependent. Knowing that A happened tells you B did not, which is very strong information. "Sum is 7" and "doubles" are mutually exclusive, and P(doubles | sum 7) = 0 while P(doubles) = 1/6, so they are dependent.

Worked independence test: Draw one card from a standard 52-card deck. A = "heart", P(A) = 13/52 = 1/4. B = "king", P(B) = 4/52 = 1/13. A and B is "king of hearts", P(A and B) = 1/52. Product: (1/4)(1/13) = 1/52. Equal, so suit and rank are independent for this pair. Conditionally, P(king | heart) = 1/13 = P(king): knowing the suit does not change the chance of a king.

Worked dependent test with the class model: 30 students, 18 play an instrument, 12 are on a team, 8 do both. P(instrument) = 3/5, P(team) = 2/5, product = 6/25 = 0.24. P(both) = 8/30 = 4/15 ≈ 0.267. Not equal, so the events are dependent in this class. Conditionally: P(team | instrument) = 8/18 = 4/9 ≈ 0.444, above P(team) = 0.4. The two tests agree, as they must.

Worked everyday reversal: In a hypothetical school, 40% of students walk to school, and among walkers 30% are late at least once a month; among non-walkers 10% are. P(late | walker) = 0.30. To find P(walker | late), use a population of 1,000: 400 walkers, of whom 120 are late; 600 non-walkers, of whom 60 are late; 180 late in total. P(walker | late) = 120/180 = 2/3. So while only 30% of walkers are late, two thirds of late students are walkers. The sentence "most late students walk" and the sentence "most walkers are late" are different claims, and here the first is true and the second false.

Represent and record: Make a table with rows for four pairs of dice events: (sum 7, red even), (sum 8, red even), (sum 7, doubles), (red even, blue even). For each pair record P(A), P(B), P(A and B), the product P(A)P(B), P(A | B) and the verdict independent or dependent. Below the table, rewrite each verdict as an everyday sentence beginning "Knowing that … does / does not change the chance that …".

Independent performance: Write three everyday statements containing a hidden conditional probability (using words such as if, among, of those who) and for each identify the given event and the event whose probability is stated. Then take one statement and write its reversed conditional; invent an explicitly hypothetical population of 1,000 with counts that make the two conditionals differ, and compute both. Finally give one pair of events you claim are independent and one pair you claim are dependent, each with a numerical check of P(A and B) against P(A)P(B). Success criteria: every given event is named; the reversed conditional is computed from counts, not guessed; the product test is written out with all three probabilities; you state that a hypothetical population is hypothetical.