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Probability and distributions / LESSON 06

Condition on the right group · AP Statistics

10 min + practice · Original foundation lesson

Probabilities describe long-run relative frequencies under a specified chance process. A condition changes which outcomes remain possible and therefore changes the denominator.

A reliable method

For two events, add their probabilities and subtract their overlap to find their union. For a conditional probability, divide the overlap by the probability of the conditioning event.

The common trap

Independent events do not change each other’s probability. Mutually exclusive events cannot occur together; if both have positive probability, they cannot also be independent.

Make the connection

Use the complement for at least one when calculating none is easier. First describe what the outcomes mean before multiplying probabilities.

A union versus a condition

If P(A)=.4, P(B)=.3 and P(A and B)=.1, the union is .6 because the overlap was counted twice. In contrast, P(A given B)=.1/.3=1/3 because only B is now the reference group. Drawing the same table can support both answers, but the denominators and event meanings differ.

A complement avoids case omissions

With two independent devices each failing with probability .2, no failures has probability .8²=.64. At least one failure is .36. Exactly one failure is 2(.2)(.8)=.32; the missing .04 is both failures. “At least one” and “exactly one” should not share an answer.

Transfer check

Explain why mutually exclusive events with positive probabilities are not independent: conditioning on one makes the other probability zero. Then construct independent events using separate coin tosses, and identify a possible outcome where both happen. Independence permits overlap; exclusivity forbids it. For the earlier events, independence would require overlap .4 times .3, or .12. The supplied overlap .10 differs, so multiplication without the independence condition would give the wrong joint probability.

WORKED EXAMPLE

P(A) = 3/10, P(B) = 3/10, and P(A and B) = 1/10. Find P(A or B).

  1. Add the two event probabilities.
  2. Subtract the overlap counted twice.
  3. P(A or B) = 1/2.
Answer: 1/2
Compare the other answer choices
2/5

Remove the overlap twice, giving exactly one event.

1/10

Report the intersection instead of the union.

9/100

Use an unjustified independence product.

Try it, then explain it.

Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.

Practice this skill

Sources & authorship

Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.

AP Statistics 2026–27 course framework