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ESAT Maths 1 Probability questions

12 original ESAT Maths 1 probability questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: List the sample space of equally likely outcomesProbability without replacement: update after each pickConditional probability: restrict to the given group

Question 1ESAT Maths 1Warm-up

Two fair 4-sided dice, each numbered 1 to 4, are rolled. What is the probability that the total is 4?
  1. A332\dfrac{3}{32}
  2. B18\dfrac{1}{8}
  3. C316\dfrac{3}{16}
  4. D14\dfrac{1}{4}
  5. E516\dfrac{5}{16}
  6. F1316\dfrac{13}{16}
Show the answer and worked solution

Answer: C · Method: List the sample space of equally likely outcomes

  1. There are 4×4=164 \times 4 = 16 equally likely outcomes.
  2. 33 of them work, so the probability is 316\dfrac{3}{16}.

Why the tempting wrong answers are wrong

  • F: That's the probability of the opposite event.

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Question 2ESAT Maths 1Warm-up

A fair 4-sided dice numbered 1 to 4 and a fair 6-sided dice numbered 1 to 6 are rolled. What is the probability that the total is at least 9?
  1. A116\dfrac{1}{16}
  2. B112\dfrac{1}{12}
  3. C18\dfrac{1}{8}
  4. D16\dfrac{1}{6}
  5. E78\dfrac{7}{8}
Show the answer and worked solution

Answer: C · Method: List the sample space of equally likely outcomes

  1. There are 4×6=244 \times 6 = 24 equally likely outcomes.
  2. 33 of them work, so the probability is 18\dfrac{1}{8}.

Why the tempting wrong answers are wrong

  • E: That's the probability of the opposite event.

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Question 3ESAT Maths 1Warm-up

Two fair 4-sided dice, each numbered 1 to 4, are rolled. What is the probability that the product of the two scores is even?
  1. A14\dfrac{1}{4}
  2. B38\dfrac{3}{8}
  3. C58\dfrac{5}{8}
  4. D1116\dfrac{11}{16}
  5. E34\dfrac{3}{4}
  6. F1316\dfrac{13}{16}
  7. G78\dfrac{7}{8}
Show the answer and worked solution

Answer: E · Method: List the sample space of equally likely outcomes

  1. There are 4×4=164 \times 4 = 16 equally likely outcomes.
  2. 1212 of them work, so the probability is 34\dfrac{3}{4}.

Why the tempting wrong answers are wrong

  • A: That's the probability of the opposite event.

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Question 4ESAT Maths 1Warm-up

A bag contains 7 red counters and 6 blue counters. Two counters are taken out at random, one after the other, without replacement. What is the probability that they are different colours?
  1. A726\dfrac{7}{26}
  2. B84169\dfrac{84}{169}
  3. C2039\dfrac{20}{39}
  4. D4178\dfrac{41}{78}
  5. E713\dfrac{7}{13}
  6. F4378\dfrac{43}{78}
  7. G2239\dfrac{22}{39}
  8. H1526\dfrac{15}{26}
Show the answer and worked solution

Answer: E · Method: Probability without replacement: update after each pick

  1. P(red then blue) =713×612= \dfrac{7}{13} \times \dfrac{6}{12}, and P(blue then red) is the same.
  2. That gives 713\dfrac{7}{13}.

Why the tempting wrong answers are wrong

  • A: Red then blue and blue then red are both possible: add both orders.
  • B: That's with replacement. After the first counter is taken, there's one fewer in the bag.

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Question 5ESAT Maths 1Warm-up

A bag contains 5 red counters and 2 blue counters. Two counters are taken out at random, one after the other, without replacement. What is the probability that they are both red?
  1. A521\dfrac{5}{21}
  2. B821\dfrac{8}{21}
  3. C37\dfrac{3}{7}
  4. D1021\dfrac{10}{21}
  5. E2549\dfrac{25}{49}
  6. F1121\dfrac{11}{21}
  7. G47\dfrac{4}{7}
  8. H1321\dfrac{13}{21}
Show the answer and worked solution

Answer: D · Method: Probability without replacement: update after each pick

  1. P(both red) =57×46= \dfrac{5}{7} \times \dfrac{4}{6}.
  2. That gives 1021\dfrac{10}{21}.

Why the tempting wrong answers are wrong

  • E: That's with replacement. After the first counter is taken, there's one fewer in the bag.
  • F: That's the probability of the opposite event.

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Question 6ESAT Maths 1Warm-up

The table summarises some students.
Year 12: 19 study physics, 9 do not study physics
Year 13: 10 study physics, 20 do not study physics
What is the probability that a student chosen at random is in Year 12 and studies physics?
  1. A928\dfrac{9}{28}
  2. B1958\dfrac{19}{58}
  3. C1029\dfrac{10}{29}
  4. D1429\dfrac{14}{29}
  5. E12\dfrac{1}{2}
  6. F1929\dfrac{19}{29}
  7. G1928\dfrac{19}{28}
Show the answer and worked solution

Answer: B · Method: Conditional probability: restrict to the given group

  1. 1919 of the 5858 students are in Year 12 and study, so the probability is 1958\dfrac{19}{58}.

Why the tempting wrong answers are wrong

  • G: Check which group you're restricting to.

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Question 7ESAT Maths 1Warm-up

The table summarises some students.
boys: 15 study physics, 7 do not study physics
girls: 15 study physics, 14 do not study physics
What is the probability that a student chosen at random is in boys and studies physics?
  1. A517\dfrac{5}{17}
  2. B722\dfrac{7}{22}
  3. C2251\dfrac{22}{51}
  4. D12\dfrac{1}{2}
  5. E1017\dfrac{10}{17}
  6. F1522\dfrac{15}{22}
Show the answer and worked solution

Answer: A · Method: Conditional probability: restrict to the given group

  1. 1515 of the 5151 students are in boys and study, so the probability is 517\dfrac{5}{17}.

Why the tempting wrong answers are wrong

  • F: Check which group you're restricting to.

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Question 8ESAT Maths 1Warm-up

A fair 4-sided dice numbered 1 to 4 and a fair 6-sided dice numbered 1 to 6 are rolled. What is the probability that the product of the two scores is even?
  1. A14\dfrac{1}{4}
  2. B38\dfrac{3}{8}
  3. C1724\dfrac{17}{24}
  4. D34\dfrac{3}{4}
  5. E1924\dfrac{19}{24}
Show the answer and worked solution

Answer: D · Method: List the sample space of equally likely outcomes

  1. There are 4×6=244 \times 6 = 24 equally likely outcomes.
  2. 1818 of them work, so the probability is 34\dfrac{3}{4}.

Why the tempting wrong answers are wrong

  • A: That's the probability of the opposite event.

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Question 9ESAT Maths 1Warm-up

A bag contains 8 red counters and 2 blue counters. Two counters are taken out at random, one after the other, without replacement. What is the probability that they are both the same colour?
  1. A2990\dfrac{29}{90}
  2. B1645\dfrac{16}{45}
  3. C35\dfrac{3}{5}
  4. D2845\dfrac{28}{45}
  5. E2945\dfrac{29}{45}
  6. F23\dfrac{2}{3}
  7. G1725\dfrac{17}{25}
  8. H3145\dfrac{31}{45}
Show the answer and worked solution

Answer: E · Method: Probability without replacement: update after each pick

  1. P(both red) =810×79= \dfrac{8}{10} \times \dfrac{7}{9} and P(both blue) =210×19= \dfrac{2}{10} \times \dfrac{1}{9}.
  2. That gives 2945\dfrac{29}{45}.

Why the tempting wrong answers are wrong

  • B: That's the probability of the opposite event.
  • G: That's with replacement. After the first counter is taken, there's one fewer in the bag.

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Question 10ESAT Maths 1Standard

The table summarises some students.
boys: 11 play a sport, 6 do not play a sport
girls: 5 play a sport, 20 do not play a sport
What is the probability that a student who plays a sport is in boys?
  1. A1142\dfrac{11}{42}
  2. B821\dfrac{8}{21}
  3. C1742\dfrac{17}{42}
  4. D1117\dfrac{11}{17}
  5. E1116\dfrac{11}{16}
Show the answer and worked solution

Answer: E · Method: Conditional probability: restrict to the given group

  1. Restrict to students who play a sport: 11+5=1611 + 5 = 16, of whom 1111 are in boys.
  2. So the probability is 1116\dfrac{11}{16}.

Why the tempting wrong answers are wrong

  • A: That's out of all students. 'Given' means only count the group you're told about.
  • D: Check which group you're restricting to.

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Question 11ESAT Maths 1Standard

Two fair 4-sided dice, each numbered 1 to 4, are rolled. (The first score is from the first dice.) What is the probability that the total is even, given that the first score is odd?
  1. A14\dfrac{1}{4}
  2. B516\dfrac{5}{16}
  3. C38\dfrac{3}{8}
  4. D716\dfrac{7}{16}
  5. E12\dfrac{1}{2}
  6. F916\dfrac{9}{16}
  7. G58\dfrac{5}{8}
  8. H1116\dfrac{11}{16}
Show the answer and worked solution

Answer: E · Method: List the sample space of equally likely outcomes

  1. There are 4×4=164 \times 4 = 16 equally likely outcomes.
  2. 88 of them have an odd first score, and 44 of those have an even total. So the probability is 12\dfrac{1}{2}.

Why the tempting wrong answers are wrong

  • A: Given the first score is odd, only those outcomes count: divide by them, not by all outcomes.

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Question 12ESAT Maths 1Standard

The table summarises some students.
Year 12: 15 study physics, 9 do not study physics
Year 13: 7 study physics, 6 do not study physics
What is the probability that a student who studies physics is in Year 12?
  1. A722\dfrac{7}{22}
  2. B38\dfrac{3}{8}
  3. C1537\dfrac{15}{37}
  4. D2237\dfrac{22}{37}
  5. E58\dfrac{5}{8}
  6. F2437\dfrac{24}{37}
  7. G1522\dfrac{15}{22}
Show the answer and worked solution

Answer: G · Method: Conditional probability: restrict to the given group

  1. Restrict to students who study physics: 15+7=2215 + 7 = 22, of whom 1515 are in Year 12.
  2. So the probability is 1522\dfrac{15}{22}.

Why the tempting wrong answers are wrong

  • C: That's out of all students. 'Given' means only count the group you're told about.
  • E: Check which group you're restricting to.

Try it interactively →

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