Chapter 7 · VI · Statistics and Probability

Introduction to Probability Worksheet

12 questions Probability scale Empirical and theoretical probability Sample spaces, trees and tables Simulation

Try each question before opening its solution.

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Foundation

Randomness and probability scale · 1 practice mark

Question 1

Which number cannot be a probability?

  1. 0
  2. 0.4
  3. 1
  4. 1.2
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(d) 1.2

  1. Probabilities lie on the scale from 0 to 1 inclusive.
  2. 1.2 is larger than 1 and therefore cannot be a probability.

Sample spaces and theoretical probability · 1 practice mark

Question 2

For one throw of a fair six-sided die, the probability of a prime number is:

  1. 1/6
  2. 1/3
  3. 1/2
  4. 2/3
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(c) 1/2

  1. The equally likely sample space is {1,2,3,4,5,6}.
  2. Prime outcomes are 2,3,5, so probability = 3/6 = 1/2.

Sample spaces and theoretical probability · Randomness and probability scale · 1 practice mark

Question 3

An impossible event has probability:

  1. −1
  2. 0
  3. 1/2
  4. 1
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(b) 0

  1. An impossible event contains no possible outcomes.
  2. Its probability is 0. A certain event has probability 1.

Equally likely outcomes · 1 practice mark

Question 4

A fair die is thrown. The probability of a number greater than 4 is:

  1. 1/6
  2. 1/3
  3. 1/2
  4. 2/3
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(b) 1/3

  1. The favourable outcomes are 5 and 6.
  2. There are six equally likely outcomes, giving 2/6 = 1/3.

Practice

Experimental/empirical probability · Randomness and probability scale · 3 practice marks

Question 5

A spinner lands on red 47 times in 100 trials. Estimate its probability of red. Does this prove its theoretical probability is exactly 0.47?

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Empirical estimate 0.47; it does not prove an exact theoretical probability.

  1. Relative frequency = red outcomes/total trials = 47/100.
  2. Finite experiments fluctuate randomly, so the observed fraction need not equal an exact theoretical value.
  3. A theoretical model requires information about the spinner and outcome probabilities; more trials can help assess the estimate.

Experimental estimate · 3 practice marks

Question 6

A coin gives 84 heads in 150 tosses. Estimate P(head). Explain why this does not prove the coin is unfair.

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Empirical estimate 84/150 = 14/25 = 0.56.

  1. Divide the head count by the total trials.
  2. Reduce 84/150 by 6 to get 14/25.
  3. A finite sample can fluctuate even for a fair coin. Further evidence is needed to judge fairness; one observed frequency is not an exact theoretical probability.

Challenge

Probability trees and tables · Sample spaces and theoretical probability · 4 practice marks

Question 7

Two fair coins are tossed independently. Use a tree or table to list the sample space and find probabilities of exactly one head and at least one head.

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Sample space {HH, HT, TH, TT}; probabilities 1/2 and 3/4.

  1. Branch H or T for the first coin, then H or T for the second.
  2. Independence gives each ordered outcome probability (1/2)(1/2) = 1/4.
  3. Exactly one head is {HT,TH}, so its probability is 2/4 = 1/2.
  4. At least one head is {HH,HT,TH}, so its probability is 3/4.
Probability tree for two independent fair coin tossesH: ½T: ½HH: ¼HT: ¼TH: ¼TT: ¼

Simulation and computational thinking · Experimental/empirical probability · Sample spaces and theoretical probability · 4 practice marks

Question 8

Write a step-by-step simulation of 1000 fair die throws to estimate the probability of an even outcome. State the theoretical target and why a simulation may differ from it.

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Choose uniform random integers 1–6; estimated probability = even count/1000; theoretical target 1/2.

  1. Initialise an even-outcome counter to zero.
  2. Repeat 1000 times: sample one of 1,2,3,4,5,6 with equal probability and increment the counter for 2,4 or 6.
  3. Divide the final counter by 1000.
  4. Three of six equally likely outcomes are even, so theoretical probability is 1/2. Random fluctuations mean a finite estimate need not be exactly 0.5.

Probability trees and tables · Sample spaces and theoretical probability · 4 practice marks

Question 9

A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Draw a probability tree and find the probability that the colours differ.

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3/5.

  1. First branches have probabilities R = 3/5 and B = 2/5.
  2. After R, the next blue probability is 2/4. After B, the next red probability is 3/4.
  3. P(RB) = (3/5)(2/4) = 3/10 and P(BR) = (2/5)(3/4) = 3/10.
  4. The paths are disjoint, so add them: 3/10 + 3/10 = 3/5. Without replacement the second probabilities change.

Experimental/empirical probability · Randomness and probability scale · Sample spaces and theoretical probability · 4 practice marks

Question 10

A player wins 18 of 30 games and then 12 of 20 games. Find the combined empirical win probability. Can we conclude the next 10 games must contain exactly 6 wins?

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Estimate 30/50 = 0.6; no exact future count is guaranteed.

  1. Add successes: 18 + 12 = 30. Add trials: 30 + 20 = 50.
  2. Combined relative frequency is 30/50 = 0.6.
  3. This is an empirical estimate under comparable playing conditions, not a guaranteed theoretical probability.
  4. Even if a suitable model had probability 0.6, future random outcomes need not match 6 out of 10 exactly.

Simulation and computational thinking · Probability trees and tables · Sample spaces and theoretical probability · 4 practice marks

Question 11

A learner lists sums 2,3,…,12 for two fair dice and assumes all sums are equally likely. Correct the model using a table and describe a simulation for estimating P(sum = 7).

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Use 36 equally likely ordered pairs; P(sum = 7) = 6/36 = 1/6.

  1. Make a 6-by-6 table: rows identify the first die and columns the second, giving 36 ordered pairs.
  2. Sum 7 occurs at (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so its probability is 1/6.
  3. Sums are not equally likely: sum 2 has only one ordered pair, while sum 7 has six.
  4. Simulate independent uniform throws of each die repeatedly, count sums of 7, and divide by total trials. Compare this relative frequency with 1/6.

With and without replacement · 4 practice marks

Question 12

A bag has 4 red and 3 blue balls. Find the probability of two red draws (i) with replacement, (ii) without replacement. Explain the changed branch probabilities.

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With replacement 16/49; without replacement 2/7.

  1. With replacement both red probabilities are 4/7; multiply to obtain 16/49.
  2. Without replacement, after a red ball is removed, 3 red balls remain among 6 balls.
  3. Multiply (4/7)(3/6) = 12/42 = 2/7.
  4. Replacement restores the initial composition; without it the second branch depends on the first result.