This set has eleven original questions, ordered from easier to harder, covering all five lessons in probability reasoning. Questions 1 to 3 are warm-ups on outcomes, 4 to 8 need a full method, and 9 to 11 mix skills and ask for a second-route check.
Give fractions in their simplest form and decimals exactly. Draw a tree for every question that has two stages, and finish each answer with a total check. The probability tree and counting board and the non-calculator working trainer are helpful companions, and the mistake log and retest queue is the place to record what went wrong.
Questions
1. Ten cards are numbered 1 to 10. One is picked at random. Find P(a multiple of 4 or a prime number).
Show answer
Multiples of 4: {4, 8}. Primes: {2, 3, 5, 7}. No number is in both lists, so the events are mutually exclusive. Six cards fit, so P = 6/10 = 3/5.
2. A bag has red, blue, green and yellow counters. P(red) = 0.15, P(blue) = 0.4 and P(green) = 0.25. Find P(yellow) and P(not blue).
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Known total: 0.15 + 0.4 + 0.25 = 0.8, so P(yellow) = 1 − 0.8 = 0.2. P(not blue) = 1 − 0.4 = 0.6. Check: 0.15 + 0.25 + 0.2 = 0.6.
3. A fair die is rolled. Find P(an even number or a factor of 12).
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Even: {2, 4, 6}. Factors of 12 on the die: {1, 2, 3, 4, 6}. The two sets overlap, so list the distinct outcomes: {1, 2, 3, 4, 6}. That is 5 outcomes, so P = 5/6. Adding 3/6 + 5/6 would give more than 1, which is impossible.
4. A footballer scores a penalty with probability 0.7, independently each time. She takes two penalties. Find P(both scored), P(exactly one scored) and P(at least one scored).
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Both: 0.7 × 0.7 = 0.49. One scored: 0.7 × 0.3 + 0.3 × 0.7 = 0.21 + 0.21 = 0.42. Neither: 0.3 × 0.3 = 0.09, so at least one = 1 − 0.09 = 0.91. Check: 0.49 + 0.42 + 0.09 = 1.
5. Of 60 students, 36 cycle to school (C), 25 wear glasses (G), and 18 do both. Are C and G independent? Show your working.
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P(C) = 36/60 = 3/5 and P(G) = 25/60 = 5/12. Product: 3/5 × 5/12 = 15/60 = 1/4 = 0.25. P(C and G) = 18/60 = 3/10 = 0.3. Since 0.25 ≠ 0.3, the events are not independent.
6. A box holds 4 blue and 2 black pens. Two are taken without replacement. Find P(both blue) and P(one of each colour).
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Both blue: 4/6 × 3/5 = 12/30 = 2/5. Both black: 2/6 × 1/5 = 2/30 = 1/15. One of each = 1 − 2/5 − 1/15 = 15/15 − 6/15 − 1/15 = 8/15. Second route: 4/6 × 2/5 + 2/6 × 4/5 = 8/30 + 8/30 = 16/30 = 8/15.
7. An outcome model has four outcomes with probabilities x, 2x, 0.1 and 0.3. Find x and the probability of the outcome 2x.
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x + 2x + 0.1 + 0.3 = 1, so 3x + 0.4 = 1 and 3x = 0.6, giving x = 0.2. The outcome 2x has probability 0.4. Check: 0.2 + 0.4 + 0.1 + 0.3 = 1.0.
8. A spinner lands on blue with probability 2/5. A student writes P(blue, then blue) = 2/5 + 2/5 = 4/5. Explain the error and find the correct value.
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“Blue then blue” needs both spins to be blue, so the probabilities are multiplied, not added. Also, 4/5 is larger than either single probability, which signals an error. Correct value: 2/5 × 2/5 = 4/25 (= 0.16), assuming the spins are independent.
9. A bag holds 3 red, 4 green and 3 yellow sweets. Two are taken without replacement. Find P(both the same colour) and P(at least one green).
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Same colour: RR = 3/10 × 2/9 = 6/90, GG = 4/10 × 3/9 = 12/90, YY = 3/10 × 2/9 = 6/90. Total 24/90 = 4/15. At least one green = 1 − P(no green). No green: 6/10 × 5/9 = 30/90. So 1 − 30/90 = 60/90 = 2/3. Check directly: green first 36/90 plus non-green then green 24/90 = 60/90.
10. The probability of rain today is 0.3. If it rains today, the probability of rain tomorrow is 0.6. If it does not rain today, the probability of rain tomorrow is 0.2. Find P(rain both days) and P(rain tomorrow).
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Rain both days: 0.3 × 0.6 = 0.18. Rain tomorrow can happen two ways: rain today then rain (0.18) or no rain today then rain (0.7 × 0.2 = 0.14). Add: 0.18 + 0.14 = 0.32. Check: P(no rain tomorrow) = 0.3 × 0.4 + 0.7 × 0.8 = 0.12 + 0.56 = 0.68, and 0.32 + 0.68 = 1. The events are dependent, so the second stage uses different branch values.
11. An archer hits the target with probability 0.6 on each independent shot. She takes three shots. Find P(at least one hit) and P(exactly one hit).
Show answer
No hits: 0.4 × 0.4 × 0.4 = 0.064, so at least one hit = 1 − 0.064 = 0.936. Exactly one hit can occur in 3 ways (first, second or third shot): each has probability 0.6 × 0.4 × 0.4 = 0.096. Add: 3 × 0.096 = 0.288. Check with the rest: two hits 3 × 0.6 × 0.6 × 0.4 = 0.432 and three hits 0.216. Then 0.064 + 0.288 + 0.432 + 0.216 = 1.000.
If you got these wrong
| What went wrong | Where to go back |
|---|---|
| Added probabilities that overlap, or missed an outcome (1 to 3) | Represent mutually exclusive outcomes |
| Added along a branch, or multiplied between alternatives (4, 8, 11) | Use a two-stage probability tree |
| Multiplied without checking independence, or could not justify it (5, 10) | Explain when multiplication needs independence |
| Kept the same fractions on the second draw (6, 9) | Update probabilities after an item is not replaced |
| Answer did not total 1, or a missing value was wrong (2, 7) | Check that an outcome model totals one |
Fix the lesson first, then retry a fresh question of the same type a few days later. Record the error type in the mistake log and retest queue so the repeat is deliberate.
If an error returns even after you have read the lesson, online one-to-one Mathematics tuition lets a teacher read your written working and find which step keeps slipping.