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Theoretical probability · 5 marks

Write the sample space and calculate the probability based on the given information.

  1. (i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
  2. (ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
  3. (iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
  4. (iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?
  5. (v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?
Answer: (i) 836 = 29 (ii) 5272 = 1318 (iii) 28 = 14 (iv) 1224 = 12 (v) 964

Step-by-step solution

Idea: Write the sample space so that its outcomes are equally likely (label identical balls, keep the order of draws, list every arrangement), count the favourable outcomes, then divide. For long lists, a tree or a multiplication count (e.g. 6 × 6 = 36) gives n(S).

(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

  1. S = {(1, 1), (1, 2), …, (1, 6), (2, 1), …, (6, 6)}: 6 × 6 = 36 equally likely pairs. Possible sums: 2 to 12; primes greater than 5 among them: 7 and 11.½ mark
  2. Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1): 6 pairs. Sum 11: (5, 6), (6, 5): 2 pairs. Favourable = 8. P = 836 = 29.½ mark
29

(ii) A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

  1. Label the balls R1–R4, G1–G3, B1, B2. S = all ordered pairs (first ball, second ball) of different balls: 9 × 8 = 72 equally likely outcomes. By colour: {RR, RG, RB, GR, GG, GB, BR, BG, BB}.½ mark
  2. Same colour: RR 4 × 3 = 12, GG 3 × 2 = 6, BB 2 × 1 = 2; total 20. Different colours: 72 − 20 = 52. P = 5272 = 1318.
    (Tree check: P(same) = 49 × 38 + 39 × 28 + 29 × 18 = 2072, so P(different) = 1 − 2072 = 5272.)½ mark
1318

(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

  1. S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8.½ mark
  2. First coin H and exactly two heads: {HHT, HTH}, 2 outcomes (THH has a tail first; HHH has three heads). P = 28 = 14.½ mark
14

(iv) A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

  1. S = {1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}: 4 × 3 × 2 × 1 = 24 numbers.½ mark
  2. A number is even when its last digit is even: 2 or 4. Ending in 2: 6 numbers; ending in 4: 6 numbers. Favourable = 12. P = 1224 = 12.½ mark
12

(v) A student takes a multiple-choice test with 3 questions, each having 4 options (A, B, C, D), with only one correct answer. What is the probability that the student guesses and gets exactly 2 answers correct?

  1. S = all possible answer sheets {AAA, AAB, …, DDD}: 4 × 4 × 4 = 64 equally likely outcomes. For each answer: correct (C) with 1 choice, wrong (W) with 3 choices.½ mark
  2. Exactly 2 correct: the pattern is CCW, CWC or WCC (3 ways to choose the wrong one), and the wrong answer can be any of 3 options: 3 × 3 = 9 sheets. P = 964 ≈ 0.14.
    (Tree check: each pattern has probability 14 × 14 × 34 = 364, and 3 × 364 = 964.)½ mark
964
(i) 8/36 = 2/9 (ii) 52/72 = 13/18 (iii) 2/8 = 1/4 (iv) 12/24 = 1/2 (v) 9/64

Check: (v) Number correct 0, 1, 2, 3 occurs on 27, 27, 9, 1 sheets: 27 + 27 + 9 + 1 = 64 ✓. (ii) 20 same + 52 different = 72 ✓.

Answer to write in the exam

(i)

S = {(1, 1), (1, 2), …, (1, 6), (2, 1), …, (6, 6)}; n(S) = 36

Primes greater than 5 (sum ≤ 12): 7, 11

Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1); sum 11: (5, 6), (6, 5); n(E) = 8

∴ P = 836 = 29

(ii)

S = ordered pairs of different balls (balls labelled); n(S) = 9 × 8 = 72

Same colour: RR = 4 × 3 = 12, GG = 3 × 2 = 6, BB = 2 × 1 = 2; total = 20

Different colours = 72 − 20 = 52

∴ P(different colours) = 5272 = 1318

(iii)

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; n(S) = 8

E = {HHT, HTH}; n(E) = 2

∴ P = 28 = 14

(iv)

S = {1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}; n(S) = 24

Even: last digit 2 or 4; n(E) = 6 + 6 = 12

∴ P(even) = 1224 = 12

(v)

S = {AAA, AAB, …, DDD}; n(S) = 4 × 4 × 4 = 64

Exactly 2 correct: patterns CCW, CWC, WCC; each has 1 × 1 × 3 = 3 answer sheets

n(E) = 3 × 3 = 9

∴ P(exactly 2 correct) = 964

Common mistakes that cost marks

  • In (i), counting the sum 5 (5 is prime but not greater than 5) or forgetting 11.
  • In (ii), using the colour pairs {RR, RG, …} as if they were equally likely (giving 69). Count with labelled balls, or multiply along a tree.
  • In (v), answering 38 by treating correct and wrong as equally likely. A guess is correct with probability 14 only.

How this can come in the exam

MCQ (1 mark)

Two dice are thrown. The probability that the sum is 10 is

  1. 112
  2. 118
  3. 19
  4. 536
Show answer

(A) 112
(4, 6), (5, 5), (6, 4): 336 = 112.

Short answer (2 marks)

A three-digit number is formed from the digits 3, 5 and 6 without repetition. List the sample space and find the probability that the number is divisible by 5.

Show answerS = {356, 365, 536, 563, 635, 653}, n(S) = 6. Divisible by 5: ends in 5: {365, 635}. P = 26 = 13.

Try one yourself

Two dice are rolled. Find the probability that the sum is a prime number greater than 7.

Show answer

Only 11: (5, 6), (6, 5). P = 236 = 118.

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