Log inGet started
DrillsLessonsSectionsPractice testsPricingTry 5 questions
← All lessonsMATH · TECHNIQUE

Probability traps

Three setups cause most lost points: “at least one”, drawing without replacement, and “given that”.

When to use it

  • The wording says “at least one”.
  • Items are drawn one after another without being put back.
  • The question says “given that…”, which shrinks the list of possible outcomes.

How it works

  1. “At least one” → find the probability of none and subtract from 1.
  2. “Without replacement” → the second probability uses the smaller total and the changed count. Multiply the stages; add the ways.
  3. “Given that” → list only the outcomes that satisfy the condition; that count is the new denominator.
  4. Before choosing, ask whether the result is between 0 and 1 and whether it makes sense (a complement should be big).
Watch out: Using the full sample space when a condition has already removed outcomes, or forgetting that a different order is a second way to get the same result.

Time: Two lines of setup are worth writing. Probability errors are almost always setup errors, not arithmetic errors.

Worked examples

Example 1

“At least one” is the complement of “none”.

A fair coin is flipped 3 times. What is the probability of getting at least one head?

  1. A1/8
  2. B1/2
  3. C3/4
  4. D7/8Correct

Using the technique

  1. The only way to get no heads in three flips is three tails: (1/2)³ = 1/8.

  2. At least one head = 1 − 1/8 = 7/8.

  3. Listing one head, two heads and three heads works too, but takes longer and invites a miss.

Practice this set

Example 2

Without replacement, and “different colors” can happen in two orders.

An urn has 6 red and 4 blue marbles. Two are drawn without replacement. What is the probability that they are different colors?

  1. A4/15
  2. B2/5
  3. C12/25
  4. D8/15Correct

Using the technique

  1. There are 6 red and 4 blue marbles, 10 in all. Red then blue: 6/10 × 4/9 = 24/90. Blue then red: 4/10 × 6/9 = 24/90.

  2. Add the two orders: 48/90 = 8/15.

  3. The second fraction uses 9, not 10, because a marble is already out. Forgetting the second order gives the trap 4/15.

Practice this set

Example 3

“Given that” changes the denominator.

Two fair dice are rolled. Given that at least one die shows a 5, what is the probability that the sum is 9?

  1. A1/9
  2. B1/6
  3. C2/9
  4. D2/11Correct

Using the technique

  1. Outcomes with at least one 5: 36 − 25 (no 5 on either die) = 11.

  2. Of those, the ones with sum 9 are (4, 5) and (5, 4): 2 outcomes.

  3. The probability is 2/11. Dividing by 36 would ignore the condition and give 1/9, the trap.

Practice this set

Practice with a custom Math set

Related lessons: Compound and conditional probability · Counting · Mean, median and mode