Calculate the likelihood of an event from a sample space, including independent and mutually exclusive events — probability is really just counting, dressed up as a fraction. Every question comes with a written walkthrough of exactly what was counted.
Once an item is removed from a sample space and not replaced, the total for the next draw is smaller — reusing the original total treats a dependent event as if it were independent.
If two events can happen at the same time, adding their probabilities double-counts the overlap — that case needs P(A)+P(B)−P(A and B) instead.
The complement rule works on probabilities, which sum to 1 — mixing it up with a count of outcomes produces a number that isn't a valid probability at all.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. What is the probability of drawing a blue marble?
A — 3/10. There are 10 marbles total, 3 of them blue: P(blue) = 3/10. 3/8 (B) comes from dividing by only the non-blue marbles (5+3=8) instead of the full sample space of 10.
A standard deck of 52 cards has 13 cards of each suit. Two cards are drawn without replacement. What is the probability both cards are hearts?
A — 1/17. The first draw has probability 13/52 = 1/4. Since the card isn't replaced, the second draw has only 51 cards left and 12 hearts remaining: 12/51. Multiplying: (13/52)(12/51) = 156/2652 = 1/17. 1/16 (B) comes from treating the two draws as independent — using 1/4 for both draws instead of adjusting the second fraction to reflect the card removed: (1/4)(1/4) = 1/16.
A jar contains 4 red, 3 blue, and 5 green balls. Two balls are drawn without replacement. What is the probability that the two balls are NOT the same color?
A — 47/66. It's faster to find the complement: P(same color) = P(RR)+P(BB)+P(GG) = (4/12)(3/11) + (3/12)(2/11) + (5/12)(4/11) = 12/132 + 6/132 + 20/132 = 38/132 = 19/66. The probability the two balls are not the same color is 1 − 19/66 = 47/66. 19/66 (B) is the probability the balls are the same color — the complement itself, left unsubtracted from 1.
Probability problems are counting problems wearing a fraction — the fastest gains come from correctly identifying the sample space before any arithmetic starts.
Name which category a problem falls into before choosing whether to multiply, adjust the denominator, or add. Untimed practice is where that classification gets fast.
Move to timed sessions once classification is automatic. Multi-step probability problems run longer since there's often more than one event to track.
Pair basic probability with counting problems in a mock — many probability questions need a combination or permutation just to find the sample space size.
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