Probability SL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

1

Basic Probability & Sample Space

Probability of an event

$ P(A) = \dfrac{n(A)}{n(U)} = \dfrac{\text{favourable outcomes}}{\text{total outcomes}} $

$ 0 \leq P(A) \leq 1 $ $ P(A') = 1 - P(A) $ (complement)

Expected number of occurrences

$ \text{expected occurrences} = n \times P(\text{event}) $

Over $ n $ trials, an event of probability $ p $ is expected to occur $ np $ times.

The sample space is the set of all possible outcomes. List it systematically (table, tree, or list) to avoid missing outcomes.

Worked Example

A bag contains 5 red and 3 blue balls. One ball is drawn at random. Find P(red) and P(not red).

Total outcomes = 5 + 3 = 8

$ P(\text{red}) = 5/8 = 0.625 $

$ P(\text{not red}) = 1 - 5/8 = 3/8 = 0.375 $

Answer: P(red) = 5/8, P(not red) = 3/8

Worked Example — expected occurrences

A fair die is rolled 600 times. How many sixes are expected?

$ P(\text{six}) = \tfrac{1}{6} $

expected sixes $ = n \times P = 600 \times \tfrac{1}{6} = 100 $

Answer: 100 sixes

2

Venn Diagrams

Venn diagrams show relationships between events. Fill in from the intersection first, then work outward.

Union (A or B)

$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $

Key notation

$ A \cap B $ = intersection (A and B)
$ A' $ = complement (not A)

Mutually exclusive events

$ P(A \cap B) = 0 \quad\Rightarrow\quad P(A \cup B) = P(A) + P(B) $

A and B cannot both happen, so there is no overlap to subtract.

Worked Example

In a class of 30: 18 play football, 12 play tennis, 5 play both. Find P(football or tennis) and P(neither).

$ P(F \cup T) = 18/30 + 12/30 - 5/30 = 25/30 = 5/6 $

$ P(\text{neither}) = 1 - 25/30 = 5/30 = 1/6 $

Answer: P(football or tennis) = 5/6, P(neither) = 1/6

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Filling a Venn diagram: Always start with the intersection. Then subtract to find "only A" and "only B". Then find "neither" = total $ - $ (all regions inside the circles).

Sample-space diagrams & tables of outcomes

For a two-stage equally-likely experiment, draw a grid of outcomes and count. Sum of two fair dice (36 equally-likely cells):

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Count favourable cells: e.g. $ P(\text{sum}=7) = 6/36 = 1/6 $.

Probability SL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

3

Tree Diagrams & Conditional Probability

Tree diagrams show sequential events. Multiply along branches, add between branches for combined probabilities.

Conditional probability

$ P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} $

"The probability of A given that B has occurred."

Independent events

$ P(A \cap B) = P(A)\,P(B) $

Equivalently $ P(A \mid B) = P(A) $. Use this to test independence.

Worked Example

A bag has 4 red and 6 blue balls. Two are drawn without replacement. Find P(both red).

$ P(\text{1st red}) = 4/10 $

$ P(\text{2nd red} \mid \text{1st red}) = 3/9 $ (one fewer red, one fewer total)

$ P(\text{both red}) = \tfrac{4}{10} \times \tfrac{3}{9} = \tfrac{12}{90} = \tfrac{2}{15} $

Answer: P(both red) = 2/15 = 0.133

✗

Common error: Forgetting "without replacement" changes the denominator. After removing one ball, the total drops by 1. With replacement, probabilities stay the same on each draw.

Probability SL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

4

Expected Value

Probability distribution table

A discrete random variable $ X $ is described by a table of its values and probabilities. The probabilities must sum to 1:

$ x $	0	1	2
$ P(X=x) $	0.2	$ k $	0.3

$ \sum P(X = x) = 1 $

Use this to find an unknown: here $ 0.2 + k + 0.3 = 1 \Rightarrow k = 0.5 $.

Expected value of a discrete random variable

$ E(X) = \sum x \cdot P(X = x) $

The long-run average. Does not have to be a possible value of $ X $.

Worked Example

A game costs \$5 to play. You win \$20 with probability 0.1, \$5 with probability 0.3, and \$0 otherwise. Find the expected profit.

$ P(\text{win } \$0) = 1 - 0.1 - 0.3 = 0.6 $

$ E(\text{winnings}) = 20(0.1) + 5(0.3) + 0(0.6) = 2 + 1.5 + 0 = \$3.50 $

$ E(\text{profit}) = E(\text{winnings}) - \text{cost} = 3.50 - 5 = -\$1.50 $

Answer: Expected profit = −\$1.50 (you lose \$1.50 on average per game)

Probability SL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

5

Binomial Distribution

Use the binomial distribution when there are a fixed number of independent trials, each with the same probability of success.

Conditions for binomial $ X \sim B(n, p) $

1. Fixed number of trials $ n $ 2. Two outcomes (success/fail) 3. Constant probability $ p $ 4. Independent trials

Probability (use GDC)

$ P(X = k) \to $ binompdf$ (n, p, k) $
$ P(X \leq k) \to $ binomcdf$ (n, p, k) $

Expected value & std dev

$ E(X) = np $
$ \sigma = \sqrt{np(1 - p)} $

TI-84 Plus CE — Binomial

[2nd][DISTR]
• binompdf(n, p, k) → P(X = k) [exact]
• binomcdf(n, p, k) → P(X $ \leq $ k) [cumulative]
Tip: $ P(X \geq 3) = 1 - $ binomcdf(n, p, 2)

TI-Nspire CX II — Binomial

[Menu] → Statistics → Distributions →
• Binomial Pdf(n, p, k) → P(X = k)
• Binomial Cdf(n, p, lower, upper) → P(lower $ \leq $ X $ \leq $ upper)
Or type directly: binomPdf(n, p, k)

Casio fx-CG50 — Binomial

[MENU] → Statistics → DIST (F5) → BINM
• Bpd: P(X = k) — enter x = k, Numtrial = n, p
• Bcd: P(X $ \leq $ k) — enter x = k, Numtrial = n, p

Worked Example

A fair die is rolled 10 times. Find P(exactly 3 sixes) and P(at least 2 sixes).

$ X \sim B(10,\ 1/6) $

$ P(X = 3) = $ binompdf(10, 1/6, 3) $ = 0.155 $

$ P(X \geq 2) = 1 - P(X \leq 1) = 1 - $ binomcdf(10, 1/6, 1) $ = 1 - 0.4845 = 0.516 $

Answer: P(exactly 3) = 0.155; P(at least 2) = 0.516

✗

Common error: "At least 2" means $ P(X \geq 2) = 1 - P(X \leq 1) $, NOT $ 1 - P(X \leq 2) $. Be precise with the boundary value.

Probability SL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

6

Normal Distribution

The normal distribution is a continuous bell-shaped distribution defined by the mean $ \mu $ and standard deviation $ \sigma $. Written $ X \sim N(\mu, \sigma^2) $.

Key properties

Symmetric about $ \mu $ 68% within $ \mu \pm \sigma $ 95% within $ \mu \pm 2\sigma $ 99.7% within $ \mu \pm 3\sigma $

Finding probability (use GDC)

$ P(X < a) \to $ normalcdf$ (-1\text{E}99,\ a,\ \mu,\ \sigma) $

$ P(a < X < b) \to $ normalcdf$ (a,\ b,\ \mu,\ \sigma) $ (TI: type the lower −∞ as −1E99)

Inverse normal (finding $ a $)

Given $ P(X < a) = p \to $ invNorm$ (p,\ \mu,\ \sigma) $

Returns the value $ a $ such that the area to the left is $ p $.

TI-84 Plus CE — Normal

[2nd][DISTR]
• normalcdf(lower, upper, $ \mu, \sigma) \to $ P(lower < X < upper)
• invNorm(area, $ \mu, \sigma) \to $ finds $ x $-value (area = left tail)
For P(X < a): normalcdf(−1E99, a, $ \mu, \sigma) $
For P(X > a): normalcdf(a, 1E99, $ \mu, \sigma) $
Tip: type −1E99 as (−) 1 [2nd][,] 99

TI-Nspire CX II — Normal

[Menu] → Statistics → Distributions →
• Normal Cdf: lower, upper, $ \mu, \sigma \to $ probability
• Inverse Normal: area (left tail), $ \mu, \sigma \to x $-value
Or type: normCdf(lower, upper, $ \mu, \sigma) $

Casio fx-CG50 — Normal

[MENU] → Statistics → DIST (F5) → NORM
• Ncd: P(lower < X < upper) — enter lower, upper, $ \sigma, \mu $
• InvN: set Tail: Left, then enter Area, $ \sigma, \mu \to $ gives $ x $-value
(Tail: Right gives the "top p%" value directly.)
Note: Casio asks for $ \sigma $ before $ \mu $ (different order from TI)

Worked Example

Heights of students are normally distributed with $ \mu = 170 $ cm, $ \sigma = 8 $ cm. Find P(height < 165) and the height exceeded by 10% of students.

$ P(X < 165) = $ normalcdf(−1E99, 165, 170, 8) $ = 0.266 $

Top 10% means $ P(X > a) = 0.10 $, so $ P(X < a) = 0.90 $

$ a = $ invNorm(0.90, 170, 8) $ = 180.3 $ cm

Answer: P(height < 165) = 0.266; top 10% threshold = 180 cm

✗

Common error: Confusing "more than" with "less than" in inverse normal. If $ P(X > a) = 0.10 $, the left tail area is 0.90. Always draw a sketch and shade the region.

Probability SL

IB Mathematics: Applications & Interpretation · Topic 4: Statistics & Probability

JMaths

www.jmaths.xyz

7

Exam Traps & Key Reminders

✗

"At least one": $ P(\text{at least 1}) = 1 - P(\text{none}) $. This is easier than adding P(1) + P(2) + ... and avoids errors.

✗

Independent vs dependent events. "With replacement" = independent (probabilities stay the same). "Without replacement" = dependent (probabilities change). Tree diagrams make this clear.

✗

$ N(\mu, \sigma^2) $ notation. The IB writes $ N(100,\ 15^2) $ meaning $ \mu = 100,\ \sigma = 15 $. The second parameter is $ \sigma^2 $ (variance), NOT $ \sigma $. In the GDC, enter $ \sigma = 15 $, not 225.

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Draw a diagram! For every probability question, draw a Venn diagram, tree diagram, or normal curve and shade the required region. This organises your thinking and earns method marks.

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Formula booklet: The combination formula, binomial PDF, and normal PDF are given. At SL you never calculate these by hand — always use the GDC. Know which function to use (pdf vs cdf, normal vs inverse).

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3 sf rule: Give probabilities to 3 significant figures unless told otherwise. Never write P = 0 or P = 1 unless it is exactly 0 or 1.

\( x \)	0	1	2
\( P(X=x) \)	0.2	\( k \)	0.3