AQA · Mathematics · 8300
GCSE Maths: Probability: listing outcomes
Probability · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths textbook exercise and practice questions for Probability: listing outcomes (Corbettmaths Video 253), and every card links back to the exact page it came from. Practice tests are drawn from this deck’s own cards. Source material © Corbettmaths — used with attribution.
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What are all the possible outcomes when flipping a coin twice?
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HH, HT, TH, TT (where H = heads, T = tails)
A team plays two matches and can win (W), draw (D), or lose (L) each match. How many possible outcomes are there?
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9 outcomes: WW, WD, WL, DW, DD, DL, LW, LD, LL
When listing outcomes for a bead chosen from a bag containing green (G), white (W), and purple (P) beads, and a coin flip (H or T), how many outcomes are there?
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6 outcomes: GT, GH, WT, WH, PT, PH
What is the probability of getting two counters of the same colour when picking one from bag 1 (red, green, pink) and one from bag 2 (blue, yellow, red)?
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1/9 (only red-red combination out of 9 total outcomes)
What is the probability of flipping heads and rolling a 3 when flipping a fair coin once and rolling a dice once?
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1/12 (one outcome H3 out of 12 total possible outcomes)
When choosing 2 different vegetables from a list of 4 (Peas, Carrots, Turnip, Spinach), how many possible combinations are there?
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6 combinations: Peas-Carrots, Peas-Turnip, Peas-Spinach, Carrots-Turnip, Carrots-Spinach, Turnip-Spinach
What is the probability of getting a tail and a prime number when flipping a coin and rolling a dice?
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1/4 or 3/12 (outcomes T2, T3, T5 out of 12 total outcomes)
How many permutations exist for visiting 3 distinct friends in different orders?
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6 permutations (for example: ALT, ATL, LAT, LTA, TAL, TLA).
What are the positive square numbers below 20?
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1, 4, 9, and 16.
If a code consists of one square number below 20 paired with one of 3 colours, how many possible codes exist?
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12 codes (4 square numbers × 3 colours = 12).
If choosing 2 destinations from 4 options where destinations can repeat, how many possible pairs exist?
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16 possible pairs (4 × 4 = 16, because each choice has 4 options and repetition is allowed).
If one item in a sequence of 4 items is fixed in last position, how many different orderings of the remaining 3 items are possible?
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6 orderings (3! = 3 × 2 × 1 = 6).
When selecting one counter from box 1 (with 3 counters) and one from box 2 (with 2 counters), how many possible combinations exist?
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6 combinations (3 × 2 = 6).
For a 4-digit code where the first digit has 3 options, second and third digits are the same with 6 options, and fourth digit has 2 options, how many codes are possible?
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36 codes (3 × 6 × 1 × 2 = 36). The third digit has only 1 option once the second is chosen.
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