Module 29 of 30Lesson 1 of 12 · P-E0 Probability Foundations at Extended Level
Probability strand · Unit P-E0 · Syllabus E8.1, E8.2

Probability Foundations at Extended Level

🟠 Extended levelFrom the scale to expected frequency

1 Learn it

The probability scale. Every probability is a number from 0 (impossible) to 1 (certain). You may write it as a fraction, a decimal or a percentage — but never as a ratio like “1 : 6” and never bigger than 1.

0ImpossibleUnlikely0.5EvensLikely1CertainP(6 on a dice) ≈ 0.17P(not 6) ≈ 0.83

Equally likely outcomes:

\(P(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of equally likely outcomes}}\)

The “not” rule: an event either happens or it does not, so

\(P(\text{not } A) = 1 – P(A)\)

Tables of probabilities. If a table lists ALL the possible outcomes, the probabilities add up to 1. That gives you an equation for any unknown, such as x or 2x.

Relative frequency (experimental probability):

\(\text{relative frequency} = \dfrac{\text{number of times the event happened}}{\text{total number of trials}}\)

Relative frequency is an estimate of the probability. The more trials, the more reliable the estimate.

Expected frequency — how many times you expect an outcome in n trials:

\(\text{expected frequency} = P(A) \times n\)

🔑 Is it fair? Compare the relative frequency from MANY trials with the theoretical probability. A fair dice gives a six with probability \(\dfrac{1}{6} \approx 0.167\). If 200 rolls give 48 sixes (relative frequency 0.24, compared with about 33 expected), the dice is probably biased. Small differences are normal; big differences over many trials suggest bias.

2 Watch it work
EXAMPLE 1

A bag has 5 red, 7 blue and 3 green counters. (a) Find P(blue) and P(not green). (b) Some more green counters are added so that P(green) becomes \(\dfrac{1}{3}\). How many green counters were added?

(a) Total = 15. \(P(\text{blue}) = \dfrac{7}{15}\).   \(P(\text{not green}) = 1 – \dfrac{3}{15} = \dfrac{12}{15} = \dfrac{4}{5}\).

(b) Add x green counters. Green becomes 3 + x and the total becomes 15 + x:

\(\dfrac{3 + x}{15 + x} = \dfrac{1}{3} \;\Rightarrow\; 9 + 3x = 15 + x \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3\)

3 green counters were added. Check: \(\dfrac{6}{18} = \dfrac{1}{3}\) ✔. Remember: adding counters changes the TOTAL too.

EXAMPLE 2

A spinner can land on red, blue, green or yellow. P(red) = 0.25, P(blue) = 2x, P(green) = x, P(yellow) = 0.15. Find x, then the expected number of blues in 250 spins.

All four outcomes are listed, so the probabilities add up to 1:

\(0.25 + 2x + x + 0.15 = 1 \;\Rightarrow\; 3x = 0.6 \;\Rightarrow\; x = 0.2\)

P(blue) = 2 × 0.2 = 0.4. Expected blues = 0.4 × 250 = 100.

EXAMPLE 3

Lesedi rolls a dice 200 times. Results — 1: 29, 2: 31, 3: 30, 4: 34, 5: 28, 6: 48. (a) Find the relative frequency of a six. (b) Is the dice fair? (c) Estimate how many sixes she will get in her next 600 rolls.

(a) \(\dfrac{48}{200} = 0.24\).

(b) For a fair dice, P(six) = \(\dfrac{1}{6} \approx 0.167\), so we would expect about \(\dfrac{1}{6} \times 200 \approx 33\) sixes. 48 is much higher after a large number of rolls, so the dice is probably biased towards six.

(c) Use the relative frequency as the best estimate: 0.24 × 600 = 144 sixes. (Do not use \(\dfrac{1}{6}\) here — we have just decided the dice is not fair.)

EXAMPLE 4

Kago spins a spinner 20 times and gets red 7 times. Neo spins the same spinner 300 times and gets red 81 times. Whose result gives the better estimate of P(red)? Give the estimate.

Kago: \(\dfrac{7}{20} = 0.35\).   Neo: \(\dfrac{81}{300} = 0.27\).

Neo’s estimate, 0.27, is better because it is based on many more trials. Experimental results settle down towards the true probability as the number of trials grows.

3 Your turn — try first, then reveal
🚫 No calculator[1 mark]

Q1. The probability that it rains in Maun tomorrow is 0.35. Find the probability that it does not rain in Maun tomorrow.

👀 Show answer
✍️ How your working should look in the exam
1 − 0.35 = 0.65   ✓ B1
Why this works
Rain and no rain cover every possibility, so their probabilities add up to 1.

🚫 No calculator[3 marks]

Q2. The letters of the word BOTSWANA are written on 8 cards. One card is picked at random. Find (a) P(A), (b) P(vowel), (c) P(not a vowel). Give each answer as a fraction in its simplest form.

👀 Show answer
✍️ How your working should look in the exam
(a) 28 = 14   ✓ B1
(b) O, A, A → 38   ✓ B1
(c) 1 − 38 = 58   ✓ B1
Why this works
Count every card, including the repeated A. There are two A cards, so A is twice as likely as any other single letter. The vowels are O, A and A — three cards.

🚫 No calculator[3 marks]

Q3. A four-sided spinner is numbered 1 to 4. P(1) = 0.3, P(2) = x, P(3) = 2x and P(4) = 0.1. Find (a) x, (b) the probability of an even number.

👀 Show answer
✍️ How your working should look in the exam
(a) 0.3 + x + 2x + 0.1 = 1   ✓ M1
3x = 0.6 → x = 0.2   ✓ A1
(b) P(2) + P(4) = 0.2 + 0.1 = 0.3   ✓ B1
Why this works
The four outcomes are the only possibilities, so they add up to 1. “Even” means 2 or 4 — add those two probabilities, not 2 and 3.

🚫 No calculator[4 marks]

Q4. A biased spinner has four colours. P(red) = x, P(blue) = x + 0.1, P(green) = 2x − 0.05 and P(white) = 0.15. Find x, then the expected number of blues in 300 spins.

👀 Show answer
✍️ How your working should look in the exam
x + x + 0.1 + 2x − 0.05 + 0.15 = 1   ✓ M1
4x + 0.2 = 1 → x = 0.2   ✓ A1
P(blue) = 0.3   →   0.3 × 300 = 90   ✓ M1 A1
Why this works
Collect the x terms (4x) and the numbers (0.1 − 0.05 + 0.15 = 0.2) carefully. Then substitute x into the blue expression BEFORE multiplying by 300. Check: 0.2 + 0.3 + 0.35 + 0.15 = 1 ✔

🔢 Calculator[3 marks]

Q5. A drawing pin is dropped 400 times. It lands point up 132 times. (a) Find the relative frequency of landing point up. (b) Estimate how many times it will land point up in 1500 drops.

👀 Show answer
✍️ How your working should look in the exam
(a) 132400 = 0.33   ✓ B1
(b) 0.33 × 1500 = 495   ✓ M1 A1
Why this works
A drawing pin has no “equally likely” outcomes, so experiments are the only way to estimate the probability. Then expected frequency = probability × number of trials.

🔢 Calculator[3 marks]

Q6. A spinner has 5 equal sections numbered 1 to 5. It is spun 500 times and lands on 5 a total of 162 times. Is the spinner fair? Show your working.

👀 Show answer
✍️ How your working should look in the exam
expected if fair: 15 × 500 = 100   ✓ M1
relative frequency = 162500 = 0.324 (compared with 0.2)   ✓ M1
162 is much more than 100 after many spins → not fair (biased towards 5)   ✓ A1
Why this works
One fact on its own is not enough — you must compare with what a fair spinner would give. 500 spins is a large number, so a difference of 62 is far too big to be just chance.

🚫 No calculator[3 marks]

Q7. A bag contains only red and blue beads. There are 12 red beads. The probability of picking a blue bead is \(\dfrac{2}{5}\). How many blue beads are in the bag?

👀 Show answer
✍️ How your working should look in the exam
P(red) = 1 − 25 = 35   ✓ M1
35 of total = 12 → total = 20   ✓ M1
blue = 20 − 12 = 8   ✓ A1
Why this works
The 12 red beads are the red share of the bag. If three-fifths is 12, one-fifth is 4, so the whole bag holds 20. Check: \(\dfrac{8}{20} = \dfrac{2}{5}\) ✔

🚫 No calculator[4 marks]

Q8. A box contains n counters. 4 of them are white. When 6 more white counters are added, the probability of picking a white counter doubles. Find n.

👀 Show answer
✍️ How your working should look in the exam
10n + 6 = 2 × 4n   ✓ M1 M1
10n = 8(n + 6) → 10n = 8n + 48   ✓ M1
n = 24   ✓ A1
Why this works
After adding 6 white counters there are 10 white and n + 6 in total. “Doubles” means new probability = 2 × old probability. Check: before \(\dfrac{4}{24} = \dfrac{1}{6}\), after \(\dfrac{10}{30} = \dfrac{1}{3}\) — doubled ✔

🚫 No calculator[3 marks]

Q9. Twenty cards are numbered 1 to 20. One card is chosen at random. Find (a) P(prime number), (b) P(multiple of 3 or multiple of 5). Give your answers as fractions in their simplest form.

👀 Show answer
✍️ How your working should look in the exam
(a) 2, 3, 5, 7, 11, 13, 17, 19 → 820 = 25   ✓ B1
(b) 3, 5, 6, 9, 10, 12, 15, 18, 20   ✓ M1
920   ✓ A1
Why this works
1 is NOT prime, and 2 IS prime. In (b), 15 is a multiple of both 3 and 5 — list it once only. Six multiples of 3 plus four multiples of 5, minus the double-counted 15, gives 9.

4 Common mistakes — don’t lose easy marks
  • Probabilities written as ratios or words: “1 : 4” or “1 in 4” lose the mark — write \(\dfrac{1}{4}\), 0.25 or 25%.
  • Forgetting the total changes: adding counters to a bag changes the denominator as well as the numerator.
  • Stopping at x: in a table question, substitute x back into the expression the question asks about.
  • Using the theoretical probability for a biased object: once you know it is biased, use the relative frequency.
  • Judging fairness from a few trials: 20 spins prove nothing; conclusions need many trials and a comparison with the expected value.
  • Expected frequency given as a probability: “expected number” is a count (like 90), not a fraction.
5 How the exam asks this
  • Paper 2: quick 1–2 mark items — P(not A), a probability from a bag or a set of cards, expected frequency.
  • Paper 4: the opening parts of a longer probability question — a table with x to find (2–3 marks), followed by tree diagrams or combined events (next unit).
  • Command words: “Find”, “Estimate” (use relative frequency), “Explain” or “Is the dice fair? Give a reason” — you must compare with the expected result.
  • Give probabilities as exact fractions or decimals; if a decimal has to be rounded, give it to 3 significant figures.
✅ Take the Check Yourself quiz below, then continue with the next Probability unit.