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) = number of favourable outcomes / total number of outcomes $

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

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(red) = 5/8 = 0.625

P(not red) = 1 $ - 5/8 = 3/8 = 0.375 $

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

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)

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(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).

3

Tree Diagrams & Conditional Probability

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

Conditional probability

$ P(A | B) = P(A \cap B) / P(B) $

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

Worked Example

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

P(1st red) = 4/10

P(2nd red | 1st red) = 3/9 (one fewer red, one fewer total)

P(both red) = 4/10 $ \times 3/9 = 12/90 = 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

Expected value of a discrete random variable

$ E(X) = \Sigma 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(win $0) = 1 $ - 0.1 - 0.3 = 0.6 $

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

E(profit) = E(winnings) $ - $ cost = 3.50 $ - 5 = - $$1.50

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

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 ~ 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 $ ~ 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 $ ~ 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($ -10^{99}, a, \mu , \sigma ) $

P($ a $ < $ X $ < $ b) \to $ normalcdf($ a, b, \mu , \sigma ) $

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
For P(X < a): normalcdf(-1E99, a, $ \mu , \sigma ) $
For P(X > a): normalcdf(a, 1E99, $ \mu , \sigma ) $

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: enter Area (left tail), $ \sigma , \mu \to $ gives $ x- $value
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($ -10^{99}, 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.

7

Exam Traps & Key Reminders

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

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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.