10 questions that get harder as you go. No calculator. Write your working on paper, pick A–E, then open the solution. Q1–3 are warm-ups, Q4–6 a step up, Q7–10 ESAT standard.
Q1 · Warm-up
A fair six-sided dice is rolled. What is the probability of an even number?
A 1/6 · B 1/3 · C 1/2 · D 2/3 · E 1
Show solution
C. Even numbers: 2, 4, 6. That’s 3 out of 6 = 1/2.
Q2 · Warm-up
A bag has 3 red and 5 blue counters. One is picked at random. What is the probability it is blue?
A 3/8 · B 5/8 · C 3/5 · D 5/3 · E 1/2
Show solution
B. 5 blue out of 8 counters in total = 5/8. (D can’t be right: a probability is never more than 1.)
Q3 · Warm-up
The Venn diagram shows which of 30 students play football and which play tennis. One student is chosen at random. What is the probability they play exactly one of the two sports?
A 17/30 · B 5/6 · C 1/6 · D 2/3 · E 13/30
Show solution
D. “Exactly one” means the parts of the circles that don’t overlap: football only (12) + tennis only (8) = 20. So 20/30 = 2/3.
💡 B is the trap: 25/30 = 5/6 includes the 5 who play both, which is “at least one”. A (17/30) is everyone in the football circle. Check the numbers add to the total: 12 + 5 + 8 + 5 = 30 ✓
Q4 · Step up
Two fair coins are flipped. What is the probability of getting two heads?
A 1/4 · B 1/3 · C 1/2 · D 1/8 · E 3/4
Show solution
A. Independent events, so multiply: ½ × ½ = ¼. Or list them: HH, HT, TH, TT, and only 1 of the 4 is HH. (B is the trap: thinking there are only 3 outcomes.)
Q5 · Step up
The probability a spinner lands on red is 0.15. It is spun 200 times. How many times would you expect it to land on red?
A 15 · B 20 · C 85 · D 170 · E 30
Show solution
E. Expected frequency = probability × number of trials = 0.15 × 200 = 30.
Q6 · Step up
Two fair dice are rolled and the scores are added. What is the probability the total is 7?
A 1/12 · B 1/6 · C 7/36 · D 1/36 · E 1/2
Show solution
B. There are 6 × 6 = 36 equally likely outcomes. Totals of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). That’s 6 out of 36 = 1/6.
Q7 · ESAT standard
A bag has 4 red and 2 blue balls. Two balls are taken out, one after the other, without putting the first back. What is the probability both are red?
A 4/9 · B 1/3 · C 2/3 · D 2/5 · E 3/5
Show solution
D. First red: 4/6. Then 3 red are left out of 5: 3/5. Multiply: 4/6 × 3/5 = 12/30 = 2/5.
💡 A is the trap: 4/6 × 4/6 = 4/9 is the answer with replacement.
Q8 · ESAT standard
A fair dice is rolled twice. What is the probability of getting at least one six?
A 1/36 · B 1/3 · C 11/36 · D 25/36 · E 1/6
Show solution
C. “At least one” is easiest the back way round. P(no six either time) = 5/6 × 5/6 = 25/36. So P(at least one six) = 1 − 25/36 = 11/36.
💡 B is the trap: 1/6 + 1/6 counts double six twice.
Q9 · ESAT standard
Events A and B are independent. P(A) = 0.3 and P(B) = 0.5. What is the probability that A or B (or both) happens?
A 0.15 · B 0.65 · C 0.80 · D 0.35 · E 0.85
Show solution
B. Neither happens: 0.7 × 0.5 = 0.35. So at least one happens: 1 − 0.35 = 0.65.
Or: P(A) + P(B) − P(both) = 0.3 + 0.5 − 0.15 = 0.65. (C is the trap: adding without taking off the overlap.)
Q10 · ESAT standard
A bag contains n balls, and 3 of them are red. Two balls are taken without replacement. The probability that both are red is 1/15. What is n?
A 10 · B 6 · C 9 · D 12 · E 15
Show solution
A. P(both red) = 3/n × 2/(n − 1) = 6 / (n(n − 1)). Set this equal to 1/15: n(n − 1) = 90.
Two whole numbers next to each other that multiply to 90: 10 × 9. So n = 10.
Check: 3/10 × 2/9 = 6/90 = 1/15 ✓. In multiple choice you could also just test the options.
🔥 Challenge: full ESAT level
Five harder questions with up to 8 options, answers in symbols and “which statements are correct?” questions, just like the real test. Aim for 90 seconds each.
C1
A bag has r red and b blue balls. Two balls are taken out without replacement. What is the probability that one is red and one is blue?
A rb/(r + b)² · B 2rb/(r + b)² · C 2rb/((r + b)(r + b − 1)) · D rb/((r + b)(r + b − 1)) · E 2rb/(r + b − 1)² · F r(r − 1)/((r + b)(r + b − 1)) · G (r² + b²)/(r + b)² · H 2/(r + b)
Show solution
C. Let N = r + b. P(red then blue) = (r/N) × (b/(N − 1)). P(blue then red) = (b/N) × (r/(N − 1)), which is the same. Add them: 2rb/(N(N − 1)).
💡 D forgets the second order; B is the answer with replacement. Check with r = b = 1: the answer must be 1, and only C gives 2/(2 × 1) = 1.
C2
Two fair six-sided dice are rolled. Which of these statements are correct?
I The probability that the total is even is 1/2.
II The probability of at least one six is 11/36.
III The probability that the product of the scores is odd is 1/2.
A none · B I only · C II only · D III only · E I and II only · F I and III only · G II and III only · H I, II and III
Show solution
E. I: the total is even when both are odd or both are even: 1/4 + 1/4 = 1/2. True. II: 1 − (5/6)² = 1 − 25/36 = 11/36. True. III: the product is odd only if both are odd: 1/2 × 1/2 = 1/4. False.
C3
1% of people have a condition. A test gives a positive result for 90% of people who have it, and also for 10% of people who don’t. A person tests positive. What is the probability they have the condition?
A 1/12 · B 1/10 · C 9/10 · D 1/100 · E 1/2 · F 1/9 · G 10/11 · H 9/100
Show solution
A. Imagine 1000 people. 10 have it, and 9 of them test positive. 990 don’t, and 99 of them test positive. Positives = 9 + 99 = 108, of which 9 really have it. 9/108 = 1/12.
💡 C is the trap: 9/10 is P(positive | has it), not P(has it | positive). Using a frequency tree with 1000 people makes these questions quick.
C4
A fair coin is flipped n times. What is the smallest n for which the probability of getting at least one head is greater than 0.9?
A 2 · B 3 · C 4 · D 5 · E 9 · F 10 · G 6 · H 1
Show solution
C. P(at least one head) = 1 − (1/2)ⁿ. We need (1/2)ⁿ < 0.1. (1/2)³ = 1/8 = 0.125, too big. (1/2)⁴ = 1/16 = 0.0625, which works. So n = 4.
💡 “At least one” → always do 1 − P(none).
C5
Two whole numbers are each chosen at random from 1, 2, 3, 4, 5 (the same number can be picked twice). What is the probability that the first number is larger than the second?
A 1/5 · B 1/2 · C 9/25 · D 2/5 · E 4/5 · F 12/25 · G 1/4 · H 3/5
Show solution
D. There are 25 equally likely pairs. 5 of them are equal, so P(same) = 1/5. By symmetry, “first larger” and “second larger” are equally likely, so each is (1 − 1/5) ÷ 2 = 2/5.
💡 B is the trap: it forgets that the two numbers can be the same.
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