Probability 1 · Outcomes, events & complements

Module 27 of 47Lesson 1 of 7 · Probability 1 · Outcomes, events & complements

Probability · Essential understanding 1

Outcomes, events & complements

IB AI SL 4.5with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Probability is a number from 0 to 1 that tells you how likely something is. You can work it out (when all outcomes are equally likely) or estimate it from experiments. Once you know a probability, you can also say how many times to expect the event in many trials.

By the end of this lesson you can:

use the words trial, outcome, sample space and event correctly
calculate $P(A) = \dfrac{n(A)}{n(U)}$ for equally likely outcomes
estimate a probability from an experiment using relative frequency
use the complement rule $P(A') = 1 - P(A)$
find the expected number of occurrences $n \times P(A)$

🧑🏾 Imagine this…

You have a bag with 10 sweets that feel exactly the same: 3 are orange and 7 are red. You close your eyes and take one. You cannot be sure what you get, but you can say something useful: red is more likely than orange. Probability puts a number on that feeling. Here, 7 out of 10 sweets are red, so the chance of red is 7 out of 10 = 0.7.

Trial

one go at the experiment: one roll of a die, one spin, one customer asked.

Outcome

one possible result of a trial, e.g. rolling a 4.

Sample space U

the list (set) of all possible outcomes. For a die: U = {1, 2, 3, 4, 5, 6}, so n(U) = 6.

Event

one outcome or a group of outcomes we are interested in, e.g. “an even number” = {2, 4, 6}. n(A) means “how many outcomes are in event A”.

Equally likely

every outcome has the same chance (a fair die, a well-mixed bag).

Relative frequency

\dfrac{\text{number of times the event happened}}{\text{number of trials}}

. It is an estimate of the probability.

Complement A′

the event “A does not happen”. Read it as “not A”.

★ The rules of this lesson

Theoretical probability (equally likely outcomes): $P(A) = \dfrac{n(A)}{n(U)}$
Every probability is between 0 (impossible) and 1 (certain). It can be written as a fraction, a decimal or a percentage.
Complement rule: $P(A') = 1 - P(A)$ . Equivalently $P(A) + P(A') = 1$ .
Expected number of times A happens in $n$ trials: $n \times P(A)$
Experimental estimate: $P(A) \approx \dfrac{\text{frequency of } A}{\text{number of trials}}$ . More trials give a better estimate.

⚠️ “Equally likely” must be true

The formula

\dfrac{n(A)}{n(U)}

only works when every outcome in U has the same chance. A school bus is either on time or late, but that does not make it 50/50. Two outcomes is not the same as “equally likely”. If outcomes are not equally likely, use data (relative frequency).

⚠️ Expected does not mean “will happen”

If P(rain) = 0.4, then in 10 days you expect 4 rainy days. You might get 3 or 6. Expected number is the long-run average, so it can even be a decimal such as 12.4 days. Do not round it unless the question asks you to.

💡 Command terms

Write down: a probability straight from the information, no working needed. Estimate: use relative frequency from data. Find the expected number: show n × P(A). Always give probabilities as exact fractions or to 3 significant figures.

🧪 Try it yourselfCheck your feel for probability

Use the numbers from this lesson: a bag of 10 sweets (3 orange, 7 red).

Worked examples

Example 1.1Estimating from data (relative frequency)

A bus company in Gaborone recorded the arrival of the 07:15 bus on 120 school days. It was late on 34 of those days.

(a) Estimate the probability that the bus is late tomorrow.
(b) Estimate how many of the next 90 school days the bus will be late.

🧠 Think it through

We cannot list equally likely outcomes here, so we use the relative frequency.
Relative frequency $= \dfrac{34}{120} = \dfrac{17}{60}$ .
Expected number in 90 days $= 90 \times \dfrac{17}{60}$ .

✍️ By hand

P(\text{late}) \approx \dfrac{34}{120} = \dfrac{17}{60} \approx 0.283

\text{Expected} = 90 \times \dfrac{17}{60} = 25.5

🔢 On your TI-Nspire CX II

On a Calculator page type 34/120 and press enter ↵💻 Enter. The TI shows the answer as a fraction.
Press ctrl💻 Ctrl enter ↵💻 Enter to get a decimal. Then type 90*34/120 enter ↵💻 Enter.

✅ Answer (a) P(late) ≈

\dfrac{17}{60}

≈ 0.283. (b) About 25.5 days (so roughly 25 or 26).

This is an estimate: a different set of 120 days would give a slightly different answer.

Example 1.2Equally likely outcomes

Twelve cards are numbered 1 to 12 and shuffled. One card is picked at random.

(a) Write down the sample space and n(U).
(b) Find the probability that the card is a multiple of 3.
(c) Find the probability that the card is a prime number.
(d) Find the probability that the card is odd and greater than 6.

🧠 Think it through

U = {1, 2, 3, …, 12}, so n(U) = 12. Each card is equally likely.
Multiples of 3: {3, 6, 9, 12} → 4 cards.
Primes: {2, 3, 5, 7, 11} → 5 cards. (1 is not prime.)
Odd AND greater than 6: {7, 9, 11} → 3 cards.

✍️ By hand

P(\text{multiple of }3) = \dfrac{4}{12} = \dfrac{1}{3}

P(\text{prime}) = \dfrac{5}{12}

P(\text{odd and} > 6) = \dfrac{3}{12} = \dfrac{1}{4}

🔢 On your TI-Nspire CX II

Check the fraction with the calculator: type 4/12 enter ↵💻 Enter and 3/12 enter ↵💻 Enter.

✅ Answer (a) U = {1, …, 12}, n(U) = 12. (b)

\dfrac{1}{3}

. (c)

\dfrac{5}{12}

. (d)

\dfrac{1}{4}

Always list the outcomes in the event first (like {3, 6, 9, 12}). Counting by eye often misses one.

Example 1.3Complementary events

A raffle sells 250 tickets and 15 of them win a prize. Kagiso buys one ticket.

(a) Find the probability that Kagiso wins a prize.
(b) Find the probability that Kagiso does not win a prize.
(c) A fair 20-sided die is rolled. Find the probability that the result is not a multiple of 4.

🧠 Think it through

(a) $P(\text{win}) = \dfrac{15}{250}$ .
(b) “Not win” is the complement, so subtract from 1. (Or count the 235 losing tickets: both methods agree.)
(c) Multiples of 4 on the die: 4, 8, 12, 16, 20 → 5 of 20. Complement: 1 − $\dfrac{5}{20}$ .

✍️ By hand

P(W) = \dfrac{15}{250} = \dfrac{3}{50} = 0.06

P(W') = 1 - \dfrac{3}{50} = \dfrac{47}{50} = 0.94

P(\text{not a multiple of }4) = 1 - \dfrac{5}{20} = \dfrac{3}{4}

🔢 On your TI-Nspire CX II

Type 1-15/250 enter ↵💻 Enter and 1-5/20 enter ↵💻 Enter.

✅ Answer (a)

\dfrac{3}{50}

= 0.06. (b)

\dfrac{47}{50}

= 0.94. (c)

\dfrac{3}{4}

The complement is a big time-saver for “not” and, later, for “at least one” questions.

Example 1.4Expected number of occurrences

In Kasane in December, the probability that a given day has rain is 0.4.

(a) Find the expected number of rainy days in December (31 days).
(b) A factory makes phone chargers. The probability that a charger is faulty is 0.03. Find the expected number of faulty chargers in a batch of 800.

🧠 Think it through

Use $\text{expected number} = n \times P(A)$ .
(a) n = 31 days, P = 0.4.
(b) n = 800 chargers, P = 0.03.

✍️ By hand

31 \times 0.4 = 12.4

800 \times 0.03 = 24

🔢 On your TI-Nspire CX II

Type 31*0.4 enter ↵💻 Enter and 800*0.03 enter ↵💻 Enter.

✅ Answer (a) 12.4 days. (b) 24 faulty chargers.

12.4 is a perfectly good expected value. You cannot have 12.4 rainy days in one December, but it is the long-run average.

Example 1.5An unknown number of items (exam style)

A bag contains 5 red balls, 3 blue balls and n green balls. The probability of picking a green ball at random is

\dfrac{1}{3}

(a) Find n.
(b) Find the probability that a ball picked at random is not red.
(c) A ball is picked, its colour noted, and it is put back. This is done 90 times. Find the expected number of green balls.

🧠 Think it through

Total balls = 5 + 3 + n = 8 + n.
Form the equation $\dfrac{n}{8 + n} = \dfrac{1}{3}$ , then solve: $3n = 8 + n$ .
Now the bag has 5 red, 3 blue, 4 green = 12 balls.
Not red = 12 − 5 = 7 balls (or use 1 − $\dfrac{5}{12}$ ).
Expected green = 90 × P(green).

✍️ By hand

\dfrac{n}{8 + n} = \dfrac{1}{3} \Rightarrow 3n = 8 + n \Rightarrow 2n = 8 \Rightarrow n = 4

P(\text{not red}) = 1 - \dfrac{5}{12} = \dfrac{7}{12}

90 \times \dfrac{1}{3} = 30

🔢 On your TI-Nspire CX II

Open menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve and type nSolve(n/(8+n)=1/3,n) enter ↵💻 Enter.

✅ Answer (a) n = 4. (b)

\dfrac{7}{12}

. (c) 30 green balls.

In an “unknown n” question, write the probability as a fraction first, then make an equation.

Challenge questions

⭐ Challenge 1.6Not whitePaper 1 style

A bag has 7 yellow, 5 green and 8 white counters. One counter is picked at random.

(a) Write down the probability that it is white.
(b) Find the probability that it is not white.
(c) The counter is replaced and the process is repeated 150 times. Find the expected number of times a counter that is not white is picked.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Total = 7 + 5 + 8 = 20.
P(white) = $\dfrac{8}{20}$ = $\dfrac{2}{5}$ .
P(not white) = 1 − $\dfrac{2}{5}$ = $\dfrac{3}{5}$ .
Expected = 150 × $\dfrac{3}{5}$ .

🔢 On your TI-Nspire CX II

Type 150*(1-8/20) enter ↵💻 Enter.

✅ Answer (a)

\dfrac{2}{5}

. (b)

\dfrac{3}{5}

. (c) 90.

⭐ Challenge 1.7How do learners get to school?Paper 2 style

A random sample of 400 learners at a school in Maun was asked how they travel to school.

Travel	Walk	Bus	Car	Bicycle	Other
Number	132	118	64	36	50

(a) Estimate the probability that a learner chosen at random walks to school.
(b) Estimate the probability that a learner does not come by car.
(c) The school has 1500 learners. Estimate how many come by bus.
(d) Mpho says: “Another sample of 400 learners would give exactly the same numbers.” Comment on this statement.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Check the total: 132 + 118 + 64 + 36 + 50 = 400.
Relative frequency of walking = $\dfrac{132}{400}$ .
Not car = 1 − $\dfrac{64}{400}$ (or $\dfrac{336}{400}$ ).
Bus: $\dfrac{118}{400}$ × 1500.
Samples vary, so estimates vary.

🔢 On your TI-Nspire CX II

Type 132/400, 1-64/400 and 118/400*1500.

✅ Answer (a)

\dfrac{33}{100}

= 0.33. (b)

\dfrac{21}{25}

= 0.84. (c) 442.5, so about 443 learners. (d) Not true: a different random sample would give different frequencies, so the estimates would be similar but not identical. A larger sample usually gives a more reliable estimate.

Next: Lesson 2 · Sample space diagrams & tables of outcomes

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL Probability 1 · Outcomes, events & complements

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

Outcomes, events & complements

Worked examples

Challenge questions