IB Math — Probability Workbook

1

Q1. ⭐ Very Easy

A fair six-sided die is rolled once.
What is the probability of rolling a number greater than 4?

The sample space is \(S = \{1, 2, 3, 4, 5, 6\}\), so \(n(S) = 6\).
Numbers greater than 4 are \(\{5, 6\}\), so the event has 2 outcomes.
\(P = \dfrac{2}{6} = \dfrac{1}{3}\)

Favorable outcomes: \(\{5, 6\}\) → count = 2
Total outcomes: \(n(S) = 6\)
\(P(\text{greater than 4}) = \dfrac{2}{6} = \dfrac{1}{3}\) ✓

2

Q2. ⭐ Very Easy

A bag contains 3 red, 5 blue, and 2 green marbles.
A marble is picked at random. What is \(P(\text{blue})\)?

Total marbles = 3 + 5 + 2 = 10
Blue marbles = 5
\(P(\text{blue}) = \dfrac{5}{10} = \dfrac{1}{2}\) ✓

3

Q3. ⭐ Very Easy

The probability that it rains tomorrow is 0.35.
What is the probability that it does NOT rain?

Key formula: \(P(A') = 1 - P(A)\)
The complement rule — all probabilities must add up to 1.

\(P(\text{no rain}) = 1 - P(\text{rain}) = 1 - 0.35 = 0.65\) ✓
⚠️ Common mistake: forgetting to subtract from 1, or adding instead.

4

Q4. ⭐⭐ Easy

A card is drawn at random from a standard deck of 52 cards.
What is the probability of drawing a King OR a Heart?

Addition Rule: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Don't forget to subtract the overlap! King of Hearts is counted in BOTH.

\(P(\text{King}) = \dfrac{4}{52}\), \(P(\text{Heart}) = \dfrac{13}{52}\), \(P(\text{King of Hearts}) = \dfrac{1}{52}\)

\(P(\text{King} \cup \text{Heart}) = \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}\) ✓

⚠️ Most common mistake: forgetting to subtract \(P(A \cap B)\) = King of Hearts!

5

Q5. ⭐⭐ Easy

In a class of 30 students: 18 play Football, 12 play Basketball, and 5 play both.
How many students play NEITHER sport?

Using the formula: \(n(F \cup B) = n(F) + n(B) - n(F \cap B)\)
\(= 18 + 12 - 5 = 25\)

Neither = Total − those playing at least one sport
\(= 30 - 25 = 5\) ✓

⚠️ Draw the Venn diagram first — it prevents double-counting errors!

6

Q6. ⭐⭐ Easy

Given \(P(A) = 0.4\), \(P(B) = 0.5\), and \(P(A \cap B) = 0.2\).
Find \(P(A \cup B)\).

\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.4 + 0.5 - 0.2 = 0.7\) ✓

⚠️ Choice D (1.1) is a trap — forgetting to subtract the intersection!

7

Q7. ⭐⭐⭐ Medium

The table shows survey results about pets:

	Has Dog	No Dog	Total
Has Cat	8	12	20
No Cat	15	5	20
Total	23	17	40

A person is chosen at random. Find \(P(\text{has cat} \mid \text{has dog})\).

Conditional probability: \(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}\)
Or simply: restrict your sample space to ONLY the given condition (dog owners = 23).

Given "has dog" → restrict to dog owners only: total = 23
Among dog owners, has cat = 8

\(P(\text{cat} \mid \text{dog}) = \dfrac{8}{23}\) ✓

⚠️ Mistake: using 40 (whole group) instead of 23 (restricted group). With conditional probability, your denominator ALWAYS shrinks!

8

Q8. ⭐⭐ Easy

A coin is flipped and a die is rolled.
What is the probability of getting Heads AND a 6?

These are independent events — the coin flip does NOT affect the die roll.
\(P(\text{H} \cap 6) = P(\text{H}) \times P(6) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12}\)

Coin and die are independent events.
\(P(\text{Heads}) = \dfrac{1}{2}\), \(P(6) = \dfrac{1}{6}\)
\(P(\text{Heads AND 6}) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12}\) ✓

9

Q9. ⭐⭐⭐ Medium

A bag has 4 red and 6 blue balls. Two balls are drawn WITHOUT replacement.
What is \(P(\text{both red})\)?

Tree diagram approach (WITHOUT replacement = dependent!):
1st draw: \(P(R) = \dfrac{4}{10}\)
2nd draw (given 1st was red): \(P(R) = \dfrac{3}{9}\) ← one less red AND one less total!
\(P(\text{both red}) = \dfrac{4}{10} \times \dfrac{3}{9}\)

Without replacement → dependent events (denominator decreases!)

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

⚠️ Classic mistake: using \(\dfrac{4}{10} \times \dfrac{4}{10} = \dfrac{16}{100}\) — this is WITH replacement!
When WITHOUT replacement: both numerator AND denominator decrease.

10

Q10. ⭐⭐⭐ Medium

\(P(A) = 0.3\), \(P(B) = 0.4\). Events A and B are independent.
Find \(P(A \cup B)\).

Since independent: \(P(A \cap B) = P(A) \times P(B) = 0.3 \times 0.4 = 0.12\)

Then: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.3 + 0.4 - 0.12 = 0.58\) ✓

⚠️ Trap: Answer A (0.70) = forgetting to subtract the intersection. Two steps needed!

11

Q11. ⭐⭐⭐ Medium

\(P(A) = 0.5\), \(P(B) = 0.4\), \(P(A \cap B) = 0.2\).
Find \(P(A \mid B)\) and determine if A and B are independent.

Independence test: Events are independent if \(P(A \mid B) = P(A)\)
Equivalently: \(P(A \cap B) = P(A) \times P(B)\)

\(P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.2}{0.4} = 0.5\)

Since \(P(A \mid B) = 0.5 = P(A)\), the events ARE independent ✓

Cross-check: \(P(A) \times P(B) = 0.5 \times 0.4 = 0.2 = P(A \cap B)\) ✓

Independence means knowing B gives NO extra information about A.

12

Q12. ⭐⭐⭐ Medium

In a school, 60% of students study French (F) and 40% study Spanish (S).
20% study both. A student is chosen at random.
Given they study French, what is the probability they also study Spanish?

We need \(P(S \mid F)\)

\(P(S \mid F) = \dfrac{P(S \cap F)}{P(F)} = \dfrac{0.20}{0.60} = \dfrac{1}{3}\) ✓

Think of it as: out of the 60 French students (per 100), 20 also do Spanish. So \(20/60 = 1/3\).

13

Q13. ⭐⭐⭐ Medium

A fair coin is tossed 3 times.
What is the probability of getting at least one Head?

🧠 KEY TRICK: "At least one" → use complement!
\(P(\text{at least 1 Head}) = 1 - P(\text{no Heads at all})\)
Much easier than listing all cases!

\(P(\text{no heads}) = P(\text{TTT}) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}\)

\(P(\text{at least 1 Head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}\) ✓

⚠️ Without complement, you'd need to list: H, HH, HHH for 1, 2, or 3 heads — very slow! Always use the complement for "at least one."

14

Q14. ⭐⭐⭐⭐ Hard

Box A has 3 red, 2 white balls. Box B has 1 red, 4 white balls.
A box is chosen at random, then one ball is drawn.
What is the probability the ball is red?

This requires the Total Probability Rule (tree diagram)!
\(P(\text{Red}) = P(\text{Box A}) \cdot P(\text{R}|\text{A}) + P(\text{Box B}) \cdot P(\text{R}|\text{B})\)

\(P(\text{Box A}) = \dfrac{1}{2}\), \(P(\text{Box B}) = \dfrac{1}{2}\)
\(P(\text{Red}|\text{A}) = \dfrac{3}{5}\), \(P(\text{Red}|\text{B}) = \dfrac{1}{5}\)

\(P(\text{Red}) = \dfrac{1}{2} \cdot \dfrac{3}{5} + \dfrac{1}{2} \cdot \dfrac{1}{5} = \dfrac{3}{10} + \dfrac{1}{10} = \dfrac{4}{10} = \dfrac{2}{5}\)

⚠️ This is the Law of Total Probability — multiply down each branch, then add across final branches.

15

Q15. ⭐⭐⭐⭐ Hard

Events A and B are mutually exclusive.
\(P(A) = 0.3\), \(P(B) = 0.4\). Find \(P(A \cup B)\).

Mutually Exclusive means \(P(A \cap B) = 0\) — they CANNOT happen at the same time!
Example: rolling a 2 AND rolling a 5 on one die (impossible).

Mutually exclusive → \(P(A \cap B) = 0\)

\(P(A \cup B) = P(A) + P(B) - 0 = 0.3 + 0.4 = 0.7\) ✓

⚠️ Don't confuse mutually exclusive (can't both happen) with independent (one doesn't affect the other). They're very different concepts!

16

Q16. ⭐⭐⭐⭐ Hard

Two dice are rolled. What is the probability that the sum equals 7?

Total outcomes: \(6 \times 6 = 36\)
Pairs that sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → list them ALL!

All favorable pairs for sum = 7:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 pairs

\(P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}\) ✓

💡 Sum of 7 is actually the most common outcome when rolling two dice!

17

Q17. ⭐⭐⭐⭐ Hard

\(P(A') = 0.3\), \(P(B) = 0.6\), \(P(A \cap B) = 0.5\).
Are A and B mutually exclusive? Are they independent?

First: \(P(A) = 1 - P(A') = 1 - 0.3 = 0.7\)

Mutually exclusive? Check if \(P(A \cap B) = 0\).
\(P(A \cap B) = 0.5 \neq 0\) → NOT mutually exclusive ✓

Independent? Check if \(P(A) \times P(B) = P(A \cap B)\).
\(0.7 \times 0.6 = 0.42 \neq 0.5\) → NOT independent ✓

⚠️ Remember: two events CAN'T be both mutually exclusive AND independent (unless one has probability 0).

18

Q18. ⭐⭐⭐⭐⭐ Exam

A medical test for a disease has:
• Probability the test is Positive given disease: \(P(+|D) = 0.95\)
• Probability the test is Positive given no disease: \(P(+|D') = 0.08\)
• Probability of having the disease: \(P(D) = 0.01\)

A person tests positive. What is \(P(D \mid +)\)?

This is Bayes' Theorem!
\(P(D|+) = \dfrac{P(+|D) \cdot P(D)}{P(+|D) \cdot P(D) + P(+|D') \cdot P(D')}\)

Numerator: \(P(+|D) \cdot P(D) = 0.95 \times 0.01 = 0.0095\)
Denominator: \(0.0095 + P(+|D') \cdot P(D') = 0.0095 + 0.08 \times 0.99\)
\(= 0.0095 + 0.0792 = 0.0887\)

\(P(D|+) = \dfrac{0.0095}{0.0887} \approx 0.107\) ✓

💡 Surprising result! Even with a 95% accurate test, if the disease is rare (1%), only ~10.7% of positive tests are true positives. This is why doctors test again!

19

Q19. ⭐⭐⭐⭐⭐ Exam

Three students A, B, C independently solve a problem.
\(P(\text{A solves}) = \dfrac{1}{2}\), \(P(\text{B solves}) = \dfrac{1}{3}\), \(P(\text{C solves}) = \dfrac{1}{4}\).

What is the probability that the problem is solved by at least one student?

Use complement: \(P(\text{at least one}) = 1 - P(\text{none solve it})\)

\(P(\text{A fails}) = \dfrac{1}{2}\), \(P(\text{B fails}) = \dfrac{2}{3}\), \(P(\text{C fails}) = \dfrac{3}{4}\)

Since independent:
\(P(\text{none}) = \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} = \dfrac{6}{24} = \dfrac{1}{4}\)

\(P(\text{at least one}) = 1 - \dfrac{1}{4} = \dfrac{3}{4}\) ✓

⚠️ Always use complement for "at least one" — trying to add all cases directly is a nightmare!

20

Q20. ⭐⭐⭐⭐⭐ IB Exam Level

A biased coin has \(P(\text{H}) = 0.6\). It is tossed twice.
Given that at least one Head occurred, what is the probability that both tosses were Heads?

This is a CONDITIONAL probability on a restricted sample space!
\(P(\text{HH} \mid \text{at least 1 H}) = \dfrac{P(\text{HH} \cap \text{at least 1 H})}{P(\text{at least 1 H})}\)
Note: HH ⊂ "at least 1 H" so the intersection is just HH itself.

\(P(\text{HH}) = 0.6 \times 0.6 = 0.36\)
\(P(\text{at least 1 H}) = 1 - P(\text{TT}) = 1 - (0.4)^2 = 1 - 0.16 = 0.84\)

Since HH ⊆ "at least 1 H":
\(P(\text{HH} \mid \text{at least 1 H}) = \dfrac{0.36}{0.84} = \dfrac{36}{84} = \dfrac{9}{21} = \dfrac{3}{7}\) ✓

⚠️ This is a very common IB exam trap. The conditional probability changes the denominator from 1 to 0.84 — don't use 0.36 alone!

📐 IB Mathematics