Year 11 · Statistical analysis
Quick tips — memory joggers
Theoretical probability
Complementary events
Multi-stage events
Relative & expected frequency
Level 1 · Fluency
A coin is tossed 100 times and lands heads 54 times. Is \(\dfrac{54}{100}\) a theoretical probability or a relative frequency?
A relative frequency (it comes from observed results, not from theory).
A fair die has six faces, so the chance of rolling a 2 is given as \(\dfrac{1}{6}\). Is this a theoretical probability or a relative frequency?
A theoretical probability (it comes from equally likely outcomes, not from observed data).
A spinner is spun 80 times and lands on green 18 times, giving \(\dfrac{18}{80}\). Is this a theoretical probability or a relative frequency?
A relative frequency (it is worked out from observed results).
Level 2 · Application
Explain the difference between the theoretical probability of rolling a 6 on a die and its relative frequency after 30 rolls.
Theoretical probability is \(\dfrac{1}{6}\) (from equally likely outcomes); the relative frequency is the actual proportion of 6s in the 30 rolls, which may differ from \(\dfrac{1}{6}\).
Explain the difference between the theoretical probability of drawing a red card from a standard deck and its relative frequency after 40 draws (with replacement).
Theoretical probability is \(\dfrac{26}{52} = \dfrac{1}{2}\) (from equally likely outcomes); the relative frequency is the actual proportion of red cards in the 40 draws, which may differ from \(\dfrac{1}{2}\).
Explain the difference between the theoretical probability that a coin lands heads and its relative frequency after 50 tosses.
Theoretical probability is \(\dfrac{1}{2}\) (from equally likely outcomes); the relative frequency is the observed fraction of heads in the 50 tosses, which may be more or less than \(\dfrac{1}{2}\).
Level 3 · Further Application
A spinner is spun 200 times; red comes up 88 times. Its theoretical probability of red is 0.5.
(a) \(88 \div 200 = 0.44\).
(b) 0.44 is close to but below the theoretical 0.5; with more spins the relative frequency would be expected to move closer to 0.5.
A die is rolled 120 times and shows a six 15 times. The theoretical probability of a six is about 0.167.
(a) \(15 \div 120 = 0.125\).
(b) 0.125 is a little below the theoretical 0.167; with more rolls the relative frequency would be expected to move closer to 0.167.
A coin is tossed 400 times and lands heads 216 times. Its theoretical probability of heads is 0.5.
(a) \(216 \div 400 = 0.54\).
(b) 0.54 is close to but above the theoretical 0.5; over more tosses the relative frequency would be expected to settle closer to 0.5.
Level 1 · Fluency
How many outcomes are in the sample space when a single die is rolled?
6 outcomes (1, 2, 3, 4, 5, 6).
How many outcomes are in the sample space when a single coin is tossed?
2 outcomes (Heads, Tails).
A letter is chosen at random from the word CAT. How many outcomes are in the sample space?
3 outcomes (C, A, T).
Level 2 · Application
Two coins are tossed. List the sample space and state the number of outcomes.
{HH, HT, TH, TT} — 4 outcomes.
A coin is tossed and a spinner numbered 1–3 is spun. List the sample space and state the number of outcomes.
{H1, H2, H3, T1, T2, T3} — 6 outcomes.
Two spinners, each numbered 1–2, are spun. List the sample space and state the number of outcomes.
{(1,1), (1,2), (2,1), (2,2)} — 4 outcomes.
Level 3 · Further Application
A die is rolled and a coin is tossed.
(a) \(6 \times 2 = 12\) outcomes.
(b) H1, H2, H3, H4, H5, H6.
A spinner numbered 1–4 is spun and a coin is tossed.
(a) \(4 \times 2 = 8\) outcomes.
(b) 2H, 2T, 4H, 4T.
Two dice, one red and one blue, are rolled.
(a) \(6 \times 6 = 36\) outcomes.
(b) (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
Level 1 · Fluency
An event has probability 0. How likely is it?
Impossible — it cannot happen.
An event has probability 1. How likely is it?
Certain — it is sure to happen.
An event has probability 0.5. How likely is it?
An even chance — equally likely to happen or not.
Level 2 · Application
Place these on a probability scale from 0 to 1: “certain”, “even chance”, “unlikely”.
Unlikely (near 0) < even chance (0.5) < certain (1).
Place these on a probability scale from 0 to 1: “impossible”, “likely”, “even chance”.
Impossible (0) < even chance (0.5) < likely (between 0.5 and 1).
Order these events from least to most likely: rolling a 7 on an ordinary die; tossing a head; the sun rising tomorrow.
Rolling a 7 (impossible, 0) < tossing a head (0.5) < the sun rising (certain, 1).
Level 3 · Further Application
An event has probability 0.85.
(a) Very likely (close to certain).
(b) \(P(\text{not}) = 1 - 0.85 = 0.15\).
An event has probability 0.1.
(a) Unlikely (close to impossible).
(b) \(P(\text{not}) = 1 - 0.1 = 0.9\).
An event has probability \(\dfrac{1}{4}\).
(a) Unlikely (less than an even chance).
(b) \(P(\text{not}) = 1 - \dfrac{1}{4} = \dfrac{3}{4}\).
Level 1 · Fluency
Can a probability be 1.4? Explain.
No — probabilities must lie between 0 and 1 inclusive.
Can a probability be \(-0.2\)? Explain.
No — probabilities cannot be negative; they must lie between 0 and 1 inclusive.
Which of these could be a probability: \(\dfrac{5}{4}\), \(0.75\), \(-0.1\)?
Only \(0.75\) (the other two fall outside \(0 \le P \le 1\)).
Level 2 · Application
The probability of rain is given as \(\dfrac{3}{5}\). Express this as a decimal and a percentage.
\(\dfrac{3}{5} = 0.6 = 60\%\).
A probability is given as \(\dfrac{7}{20}\). Express it as a decimal and a percentage.
\(\dfrac{7}{20} = 0.35 = 35\%\).
A probability is \(0.08\). Express it as a fraction in simplest form and as a percentage.
\(0.08 = \dfrac{8}{100} = \dfrac{2}{25} = 8\%\).
Level 3 · Further Application
Events A, B and C are the only outcomes, with \(P(A) = 0.5\) and \(P(B) = 0.2\).
(a) \(P(C) = 1 - 0.5 - 0.2 = 0.3\).
(b) \(P(B) + P(C) = 0.5 = P(A)\), so A is equally likely as B and C combined.
Events R, S and T are the only outcomes, with \(P(R) = \dfrac{1}{2}\) and \(P(S) = \dfrac{1}{6}\).
(a) \(P(T) = 1 - \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{6}{6} - \dfrac{3}{6} - \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}\).
(b) \(P(T) = \dfrac{1}{3} < \dfrac{1}{2} = P(R)\), so T is less likely than R.
A four-colour spinner gives \(P(\text{red}) = 0.4\), \(P(\text{blue}) = 0.25\) and \(P(\text{green}) = 0.2\); the rest is yellow.
(a) \(P(\text{yellow}) = 1 - 0.4 - 0.25 - 0.2 = 0.15\).
(b) Red, with the highest probability of 0.4.
Level 1 · Fluency
A bag has 4 red and 6 blue balls. Find \(P(\text{red})\).
\(\dfrac{2}{5}\)
\(\dfrac{4}{10} =\) \(\dfrac{2}{5}\).
A bag has 5 green and 3 yellow counters. Find \(P(\text{yellow})\).
\(\dfrac{3}{8}\)
Total \(= 5 + 3 = 8\), so \(P(\text{yellow}) =\) \(\dfrac{3}{8}\).
A standard die is rolled. Find the probability of rolling a 5.
\(\dfrac{1}{6}\)
One favourable face out of 6 → \(\dfrac{1}{6}\).
Level 2 · Application
A card is drawn from a standard 52-card deck. Find the probability it is a heart.
\(\dfrac{1}{4}\)
\(\dfrac{13}{52} =\) \(\dfrac{1}{4}\).
A card is drawn from a standard 52-card deck. Find the probability it is a King.
\(\dfrac{1}{13}\)
4 Kings in the deck → \(\dfrac{4}{52} =\) \(\dfrac{1}{13}\).
A card is drawn from a standard 52-card deck. Find the probability it is a face card (Jack, Queen or King).
\(\dfrac{3}{13}\)
12 face cards (3 in each of 4 suits) → \(\dfrac{12}{52} =\) \(\dfrac{3}{13}\).
Level 3 · Further Application
A spinner has 8 equal sectors numbered 1–8.
\(\dfrac{1}{2}\)
(a) Numbers 6, 7, 8 → \(\dfrac{3}{8}\).
(b) Even numbers 2, 4, 6, 8 → \(\dfrac{4}{8} =\) \(\dfrac{1}{2}\).
A spinner has 10 equal sectors numbered 1–10.
\(\dfrac{1}{2}\)
(a) Multiples of 3: 3, 6, 9 → \(\dfrac{3}{10}\).
(b) Even numbers 2, 4, 6, 8, 10 → \(\dfrac{5}{10} =\) \(\dfrac{1}{2}\).
A bag contains 12 discs numbered 1–12; one is drawn at random.
\(\dfrac{1}{2}\)
(a) Multiples of 4: 4, 8, 12 → \(\dfrac{3}{12} = \dfrac{1}{4}\).
(b) Factors of 12: 1, 2, 3, 4, 6, 12 → \(\dfrac{6}{12} =\) \(\dfrac{1}{2}\).
Level 1 · Fluency
State the formula for the probability of an event with equally likely outcomes.
\(P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\).
In the formula \(P(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}\), what does “favourable outcomes” mean?
The outcomes that make the event \(E\) happen — the ones counted as successes.
In the formula \(P(E) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}\), what condition must the outcomes satisfy for it to apply?
All of the outcomes must be equally likely.
Level 2 · Application
Two dice are rolled and the numbers added. Find the probability the total is 7.
\(\dfrac{1}{6}\)
Favourable: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) = 6 out of 36 = \(\dfrac{1}{6}\).
Two dice are rolled and the numbers added. Find the probability the total is 5.
\(\dfrac{1}{9}\)
Favourable: (1,4),(2,3),(3,2),(4,1) = 4 out of 36 = \(\dfrac{4}{36} =\) \(\dfrac{1}{9}\).
Two dice are rolled. Find the probability that both dice show the same number (a double).
\(\dfrac{1}{6}\)
Doubles: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) = 6 out of 36 = \(\dfrac{6}{36} =\) \(\dfrac{1}{6}\).
Level 3 · Further Application
A spinner numbered 1–4 is spun and a card numbered 1–6 is drawn; the score is the spinner number \(\times\) the card number.
\(\dfrac{1}{8}\)
(a) \(4 \times 6 = 24\) outcomes.
(b) Products equal to 12: \((2\times6),(3\times4),(4\times3)\) = 3 outcomes → \(\dfrac{3}{24} =\) \(\dfrac{1}{8}\).
A spinner numbered 1–3 is spun and a die numbered 1–6 is rolled; the score is the spinner number \(+\) the die number.
\(\dfrac{1}{6}\)
(a) \(3 \times 6 = 18\) outcomes.
(b) Sums equal to 5: \((1,4),(2,3),(3,2)\) = 3 outcomes → \(\dfrac{3}{18} =\) \(\dfrac{1}{6}\).
Two dice are rolled; the score is the product of the two numbers.
\(\dfrac{1}{9}\)
(a) \(6 \times 6 = 36\) outcomes.
(b) Products equal to 6: \((1,6),(6,1),(2,3),(3,2)\) = 4 outcomes → \(\dfrac{4}{36} =\) \(\dfrac{1}{9}\).
Level 1 · Fluency
If \(P(\text{win}) = 0.3\), find \(P(\text{not win})\).
0.7
\(1 - 0.3 =\) 0.7.
If \(P(\text{rain}) = 0.45\), find \(P(\text{no rain})\).
0.55
\(1 - 0.45 =\) 0.55.
If \(P(\text{late}) = \dfrac{1}{5}\), find \(P(\text{not late})\).
\(\dfrac{4}{5}\)
\(1 - \dfrac{1}{5} =\) \(\dfrac{4}{5}\).
Level 2 · Application
The probability that at least one of two components works is 0.96. Find the probability that neither works.
0.04
\(1 - 0.96 =\) 0.04.
The probability a student passes at least one of two exams is 0.88. Find the probability they pass neither.
0.12
\(1 - 0.88 =\) 0.12.
The probability that a randomly chosen light globe is faulty is 0.015. Find the probability it is not faulty.
0.985
\(1 - 0.015 =\) 0.985.
Level 3 · Further Application
A bag has 12 balls, some red. \(P(\text{red}) = \dfrac{1}{3}\).
8 balls
(a) \(1 - \dfrac{1}{3} = \dfrac{2}{3}\).
(b) \(\dfrac{2}{3} \times 12 =\) 8 balls.
A bag has 20 balls, some green. \(P(\text{green}) = \dfrac{2}{5}\).
12 balls
(a) \(1 - \dfrac{2}{5} = \dfrac{3}{5}\).
(b) \(\dfrac{3}{5} \times 20 =\) 12 balls.
A spinner is coloured so that \(P(\text{blue}) = 0.35\). It is spun 60 times.
39 times
(a) \(1 - 0.35 = 0.65\).
(b) \(0.65 \times 60 =\) 39 times.
Level 1 · Fluency
On a probability tree, what do you do with the probabilities along a single path (branch to branch)?
Multiply them.
On a probability tree, what do you do with the probabilities of the separate paths that each satisfy the event?
Add them.
In a two-way table (array) of equally likely outcomes, how do you find a probability from it?
Count the favourable cells and divide by the total number of cells (outcomes).
Level 2 · Application
For the tree shown, calculate the probability that both marbles chosen are blue (B then B).
\(\dfrac{1}{3}\)
\(P(BB) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} =\) \(\dfrac{1}{3}\).
A coin is tossed twice. Using a tree diagram, find the probability of getting two heads.
\(\dfrac{1}{4}\)
\(P(HH) = \dfrac{1}{2} \times \dfrac{1}{2} =\) \(\dfrac{1}{4}\).
A bag has 3 red and 2 green counters. Two are drawn without replacement. Find the probability both are red.
\(\dfrac{3}{10}\)
\(P(RR) = \dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} =\) \(\dfrac{3}{10}\).
Level 3 · Further Application
Using the marble tree above:
\(\dfrac{8}{15}\)
(a) \(P(RB) + P(BR) = \left(\dfrac{4}{10} \times \dfrac{6}{9}\right) + \left(\dfrac{6}{10} \times \dfrac{4}{9}\right) = \dfrac{24}{90} + \dfrac{24}{90} = \dfrac{48}{90} =\) \(\dfrac{8}{15}\).
(b) Multiplied along each path, then added the two paths that give different colours.
A bag has 3 red and 2 green counters; two are drawn without replacement.
\(\dfrac{3}{5}\)
(a) \(P(RG) + P(GR) = \left(\dfrac{3}{5} \times \dfrac{2}{4}\right) + \left(\dfrac{2}{5} \times \dfrac{3}{4}\right) = \dfrac{6}{20} + \dfrac{6}{20} = \dfrac{12}{20} =\) \(\dfrac{3}{5}\).
(b) Multiplied along each path, then added the two paths that give different colours.
Two cards are drawn without replacement from four cards numbered 1, 2, 3, 4.
\(\dfrac{1}{6}\)
(a) The even cards are 2 and 4, so \(P(\text{both even}) = \dfrac{2}{4} \times \dfrac{1}{3} = \dfrac{2}{12} =\) \(\dfrac{1}{6}\).
(b) Multiplied the probabilities along the single path (even, then even).
Level 1 · Fluency
For two tosses of a coin, how many branches are on the second stage of the tree from each first-stage outcome?
2 branches (Heads and Tails) from each.
For rolling a die twice, how many branches come from each first-stage outcome on the tree?
6 branches (one for each face 1–6).
A tree is drawn for tossing a coin then rolling a die. How many complete paths (final outcomes) does the tree have?
\(2 \times 6 = 12\) paths.
Level 2 · Application
A bag has 4 red (R) and 6 blue (B) marbles. Two are drawn without replacement. State the probabilities on the first two branches (first draw).
First draw: \(P(R) = \dfrac{4}{10}\) and \(P(B) = \dfrac{6}{10}\).
A bag has 5 red (R) and 5 white (W) marbles. Two are drawn without replacement. State the probabilities on the first two branches (first draw).
First draw: \(P(R) = \dfrac{5}{10} = \dfrac{1}{2}\) and \(P(W) = \dfrac{5}{10} = \dfrac{1}{2}\).
A box has 3 apples and 7 oranges. Two pieces of fruit are taken without replacement. State the probabilities on the first two branches (first draw).
First draw: \(P(\text{apple}) = \dfrac{3}{10}\) and \(P(\text{orange}) = \dfrac{7}{10}\).
Level 3 · Further Application
For the same bag (4 R, 6 B, without replacement):
(a) After an R is drawn, 9 remain (3 R, 6 B): \(P(R) = \dfrac{3}{9}\), \(P(B) = \dfrac{6}{9}\).
(b) One marble has been removed and not replaced, so the total falls from 10 to 9.
A bag has 5 red and 5 white marbles, drawn without replacement:
(a) After a white is drawn, 9 remain (5 red, 4 white): \(P(R) = \dfrac{5}{9}\), \(P(W) = \dfrac{4}{9}\).
(b) One marble has been removed and not replaced, so the total falls from 10 to 9.
A box has 3 apples and 7 oranges, taken without replacement:
(a) After an apple is taken, 9 remain (2 apples, 7 oranges): \(P(\text{apple}) = \dfrac{2}{9}\), \(P(\text{orange}) = \dfrac{7}{9}\).
(b) One apple has been removed and not replaced, so both the apple count (3 → 2) and the total (10 → 9) drop.
Level 1 · Fluency
On a tree, the path for “Heads then Tails” has probabilities \(\dfrac{1}{2}\) and \(\dfrac{1}{2}\). Find \(P(HT)\).
\(\dfrac{1}{4}\)
\(\dfrac{1}{2} \times \dfrac{1}{2} =\) \(\dfrac{1}{4}\).
On a tree, the path “red then red” has probabilities \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\). Find \(P(RR)\).
\(\dfrac{1}{6}\)
\(\dfrac{1}{2} \times \dfrac{1}{3} =\) \(\dfrac{1}{6}\).
On a tree, the path “six then six” has probabilities \(\dfrac{1}{6}\) and \(\dfrac{1}{6}\). Find \(P(\text{two sixes})\).
\(\dfrac{1}{36}\)
\(\dfrac{1}{6} \times \dfrac{1}{6} =\) \(\dfrac{1}{36}\).
Level 2 · Application
Using the marble tree above, find the probability that both marbles are red.
\(\dfrac{2}{15}\)
\(P(RR) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} =\) \(\dfrac{2}{15}\).
A bag has 5 red and 3 blue marbles; two are drawn without replacement. Find the probability both are blue.
\(\dfrac{3}{28}\)
\(P(BB) = \dfrac{3}{8} \times \dfrac{2}{7} = \dfrac{6}{56} =\) \(\dfrac{3}{28}\).
A drawer has 4 black and 6 white socks; two are taken without replacement. Find the probability both are black.
\(\dfrac{2}{15}\)
\(P(BB) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} =\) \(\dfrac{2}{15}\).
Level 3 · Further Application
A student takes two buses to school. \(P(\text{bus 1 late}) = 0.2\) and \(P(\text{bus 2 late}) = 0.3\), independently.
0.44
(a) \(0.2 \times 0.3 = 0.06\).
(b) \(P(\text{neither late}) = 0.8 \times 0.7 = 0.56\), so \(P(\text{at least one late}) = 1 - 0.56 =\) 0.44.
A student sets two alarms. \(P(\text{alarm 1 fails}) = 0.1\) and \(P(\text{alarm 2 fails}) = 0.05\), independently.
0.995
(a) \(0.1 \times 0.05 = 0.005\).
(b) \(P(\text{both fail}) = 0.005\), so \(P(\text{at least one works}) = 1 - 0.005 =\) 0.995.
Two machines run independently. \(P(\text{machine A breaks}) = 0.15\) and \(P(\text{machine B breaks}) = 0.2\).
0.32
(a) \(0.15 \times 0.2 = 0.03\).
(b) \(P(\text{neither breaks}) = 0.85 \times 0.8 = 0.68\), so \(P(\text{at least one breaks}) = 1 - 0.68 =\) 0.32.
Level 1 · Fluency
To simulate a fair coin, how many equally likely outcomes must the random device give?
Two (e.g. 0 and 1, or odd and even).
To simulate rolling a fair die, how many equally likely outcomes must the random device give?
Six (one for each face 1–6).
You want to simulate an event with probability \(\dfrac{1}{2}\) using a die. How many of the six faces should represent the event?
Three, since \(\dfrac{3}{6} = \dfrac{1}{2}\).
Level 2 · Application
Explain how a single die could be used to simulate an event with probability \(\dfrac{1}{3}\).
Assign two of the six faces (e.g. 1 and 2) to the event; since \(\dfrac{2}{6} = \dfrac{1}{3}\), rolling a 1 or 2 simulates the event.
Explain how a single die could be used to simulate an event with probability \(\dfrac{1}{2}\).
Assign three of the six faces (e.g. 1, 2, 3) to the event; since \(\dfrac{3}{6} = \dfrac{1}{2}\), those faces simulate the event.
Explain how a spinner could be used to simulate an event with probability 0.25.
Divide the spinner into four equal sectors and let one sector be the event; since \(\dfrac{1}{4} = 0.25\), landing on that sector simulates the event.
Level 3 · Further Application
A weather model gives a 40% chance of rain each day. You use a spinner to simulate 10 days.
(a) Divide the spinner so 40% of the area = “rain” and 60% = “no rain”; spin once per day.
(b) Repeat the 10-day simulation many times and record the proportion of runs with 5 or more rainy days.
A basketball player scores 70% of free throws. You use a spinner to simulate 15 throws.
(a) Divide the spinner so 70% of the area = “score” and 30% = “miss”; spin once per throw.
(b) Repeat the 15-throw simulation many times and record the proportion of runs with 10 or more scores.
A game is won 30% of the time. You simulate playing 8 games.
(a) Divide the spinner so 30% = “win” and 70% = “lose”; spin once per game.
(b) Repeat the 8-game simulation many times and record the proportion of runs with 4 or more wins.
Level 1 · Fluency
A spreadsheet function returns a random number between 0 and 1. To simulate an event with probability 0.3, which outputs count as the event?
Random values less than 0.3 (that happens 30% of the time).
A spreadsheet returns a random number between 0 and 1. To simulate an event with probability 0.7, which outputs count as the event?
Random values less than 0.7 (which happens 70% of the time).
A random number between 0 and 1 is generated. To simulate an event with probability \(\dfrac{1}{4}\), which outputs count as the event?
Values less than 0.25, since \(\dfrac{1}{4} = 0.25\).
Level 2 · Application
Describe how to use =RAND() in a spreadsheet to simulate one toss of a fair coin.
Generate =RAND(); if the value is less than 0.5 call it Heads, otherwise Tails (each has probability 0.5).
Describe how to use =RANDBETWEEN(1,6) in a spreadsheet to simulate one roll of a fair die.
Each call returns a whole number from 1 to 6, all equally likely, so the result is the simulated face rolled.
Describe how to use =RAND() in a spreadsheet to simulate one roll of a fair die (six equally likely outcomes).
Generate =RAND(); split 0–1 into six equal bands (\(0\)–\(\tfrac{1}{6}\), \(\tfrac{1}{6}\)–\(\tfrac{2}{6}\), …) and read off which band the value falls in as the face 1–6.
Level 3 · Further Application
You simulate rolling a die 600 times using technology.
(a) \(600 \times \dfrac{1}{6} = 100\).
(b) Simulations involve random variation, so results fluctuate around the expected value; only over very many trials does the relative frequency settle near \(\dfrac{1}{6}\).
You simulate tossing a coin 500 times using technology.
(a) \(500 \times \dfrac{1}{2} = 250\).
(b) Simulations involve random variation, so the count fluctuates around the expected value; only over very many trials does the relative frequency settle near 0.5.
You simulate a spinner with \(P(\text{win}) = 0.2\) for 400 spins.
(a) \(400 \times 0.2 = 80\).
(b) Each run is random, so the number of wins varies from run to run; the more spins simulated, the closer the relative frequency tends to 0.2.
Level 1 · Fluency
A drawing pin lands “point up” 63 times in 100 drops. Estimate \(P(\text{point up})\).
0.63
\(\dfrac{63}{100} =\) 0.63.
A basketball player scores 34 of 50 free throws. Estimate \(P(\text{score})\).
0.68
\(\dfrac{34}{50} =\) 0.68.
A bus is late on 9 of 30 days. Estimate \(P(\text{late})\).
0.3
\(\dfrac{9}{30} =\) 0.3.
Level 2 · Application
In 250 trials an event occurs 90 times. Estimate its probability and predict the number of occurrences in 400 trials.
144
Estimated probability = \(\dfrac{90}{250} = 0.36\); in 400 trials, expect \(0.36 \times 400 =\) 144.
In 200 trials an event occurs 46 times. Estimate its probability and predict the number of occurrences in 500 trials.
115
Estimated probability = \(\dfrac{46}{200} = 0.23\); in 500 trials, expect \(0.23 \times 500 =\) 115.
A spinner lands on blue 24 times in 80 spins. Estimate \(P(\text{blue})\) and predict the number of blues in 300 spins.
90
Estimated probability = \(\dfrac{24}{80} = 0.3\); in 300 spins, expect \(0.3 \times 300 =\) 90.
Level 3 · Further Application
A machine produces items, and 15 of a sample of 500 are defective.
240 items
(a) \(\dfrac{15}{500} = 0.03\).
(b) \(0.03 \times 8000 =\) 240 items.
A quality check finds 8 faulty phones in a sample of 400.
120 phones
(a) \(\dfrac{8}{400} = 0.02\).
(b) \(0.02 \times 6000 =\) 120 phones.
A survey of 250 households finds 40 own a dog.
640 households
(a) \(\dfrac{40}{250} = 0.16\).
(b) \(0.16 \times 4000 =\) 640 households.
Level 1 · Fluency
As the number of trials increases, the relative frequency tends to get closer to what?
The theoretical probability.
The law of large numbers says the relative frequency approaches which value as the number of trials grows?
The theoretical probability.
After many trials, an experimental (relative frequency) estimate becomes what kind of estimate of the true probability?
A more reliable (more accurate) estimate.
Level 2 · Application
Explain why tossing a coin 1000 times gives a more reliable estimate of P(heads) than tossing it 10 times.
With more tosses, random fluctuations average out, so the relative frequency settles closer to the true probability (0.5) — the law of large numbers.
Explain why rolling a die 600 times gives a more reliable estimate of \(P(\text{six})\) than rolling it 12 times.
With more rolls, random fluctuations average out, so the relative frequency settles closer to the true value \(\dfrac{1}{6}\) — the law of large numbers.
A survey of 1000 people estimates a proportion more reliably than a survey of 20. Explain why.
A larger sample reduces the effect of random variation, so the observed proportion is closer to the true population proportion.
Level 3 · Further Application
A die is rolled; after 20 rolls the relative frequency of 6 is 0.30, but the theoretical probability is about 0.167.
(a) 20 rolls is a small sample, so random variation can make the observed proportion much higher (or lower) than the true probability.
(b) With many more rolls it would move closer to 0.167.
A coin is tossed 15 times and lands heads 10 times, a relative frequency of about 0.67, but the theoretical probability is 0.5.
(a) 15 tosses is a small sample, so random variation can push the observed proportion well above (or below) 0.5.
(b) With many more tosses it would move closer to 0.5.
A spinner with \(P(\text{red}) = 0.25\) is spun 16 times and lands red 6 times (relative frequency 0.375).
(a) 16 spins is too few for the fluctuations to average out, so the observed proportion can be well above 0.25.
(b) As the number of spins increases it settles closer to 0.25.
Level 1 · Fluency
A model assumes each day’s rain is independent. Why might this be unrealistic?
Weather is often correlated day to day — rain one day can make rain the next day more likely, so days are not truly independent.
A model assumes each customer at a shop spends the same average amount. Why might this be unrealistic?
Spending varies a lot between customers (some buy one item, others fill a trolley), so a single fixed average may not reflect reality.
A simulation assumes a spinner is perfectly balanced. Why might a real spinner not match this?
A real spinner may be slightly biased through wear, uneven sectors or friction, so the outcomes are not exactly equally likely.
Level 2 · Application
A simulation assumes a basketball player’s free throws are independent, each with probability 0.8. State one reason this might not hold in a real game.
Factors like fatigue, pressure or a “hot streak” can change the success rate, so throws may not be independent or constant.
A simulation assumes each traffic light a driver meets is independently red with probability 0.5. State one reason this may not hold.
Lights are often synchronised (“green waves”), so meeting one green can make the next green more likely — they are not independent.
A model assumes a call centre receives calls at a constant rate all day, each independent. State one reason this might not hold.
Call volume changes with the time of day (busy in the morning, quiet at night), so the rate is not constant.
Level 3 · Further Application
A queue at a shop is simulated assuming customers arrive at a constant rate.
(a) Arrival rates vary (e.g. lunchtime rushes, weekends), so arrivals are not at a constant rate.
(b) The simulation could badly under-estimate waiting times during peak periods.
A game is simulated assuming a player’s chance of winning stays fixed at 0.4 each round.
(a) A player may improve with practice (or tire), so the win probability changes over rounds.
(b) The simulation could under- or over-estimate the long-run number of wins.
A model simulates daily sales assuming every day is identical and independent.
(a) Sales vary with weekends, holidays and seasons, so days are not identical or independent.
(b) The simulation could badly mis-estimate totals during peak or quiet periods.
Level 1 · Fluency
A raffle sells 500 tickets and you buy 10. Find your probability of winning the single prize.
\(\dfrac{1}{50}\)
\(\dfrac{10}{500} =\) \(\dfrac{1}{50}\).
A raffle sells 800 tickets and you buy 20. Find your probability of winning the single prize.
\(\dfrac{1}{40}\)
\(\dfrac{20}{800} =\) \(\dfrac{1}{40}\).
A prize draw has 250 entries and you submit 5. Find your probability of winning.
\(\dfrac{1}{50}\)
\(\dfrac{5}{250} =\) \(\dfrac{1}{50}\).
Level 2 · Application
A survey finds 0.18 of shoppers use a loyalty card. Predict how many of 1500 shoppers use one.
270 shoppers
\(0.18 \times 1500 =\) 270 shoppers.
A survey finds 0.24 of commuters cycle to work. Predict how many of 2000 commuters cycle.
480 commuters
\(0.24 \times 2000 =\) 480 commuters.
Past data shows 0.35 of website visitors make a purchase. Predict the number of purchasers among 1200 visitors.
420 purchasers
\(0.35 \times 1200 =\) 420 purchasers.
Level 3 · Further Application
In a factory, past data shows 4% of items are faulty.
2400
(a) \(1 - 0.04 = 0.96\).
(b) Faulty: \(0.04 \times 2500 = 100\); acceptable: \(0.96 \times 2500 =\) 2400.
Records show 6% of eggs from a farm are cracked.
4700 intact
(a) \(1 - 0.06 = 0.94\).
(b) Cracked: \(0.06 \times 5000 = 300\); intact: \(0.94 \times 5000 =\) 4700.
A test correctly detects a fault 92% of the time.
20 missed
(a) \(1 - 0.92 = 0.08\).
(b) Detected: \(0.92 \times 250 = 230\); missed: \(0.08 \times 250 =\) 20.
Level 1 · Fluency
If \(P(\text{event}) = 0.25\) and there are 40 trials, how many occurrences are expected?
10
\(0.25 \times 40 =\) 10.
If \(P(\text{event}) = 0.2\) and there are 60 trials, how many occurrences are expected?
12
\(0.2 \times 60 =\) 12.
If \(P(\text{event}) = \dfrac{1}{5}\) and there are 45 trials, how many occurrences are expected?
9
\(\dfrac{1}{5} \times 45 =\) 9.
Level 2 · Application
A biased coin has \(P(\text{heads}) = 0.6\). In 150 tosses, how many heads are expected?
90 heads
\(0.6 \times 150 =\) 90 heads.
A biased spinner has \(P(\text{red}) = 0.35\). In 200 spins, how many reds are expected?
70 reds
\(0.35 \times 200 =\) 70 reds.
A biased die has \(P(\text{six}) = 0.28\). In 250 rolls, how many sixes are expected?
70 sixes
\(0.28 \times 250 =\) 70 sixes.
Level 3 · Further Application
A quiz has 20 multiple-choice questions, each with 4 options, answered by guessing.
5 correct
(a) \(\dfrac{1}{4} = 0.25\).
(b) Expected = \(0.25 \times 20 =\) 5 correct.
A multiple-choice quiz has 30 questions, each with 5 options, answered by guessing.
6 correct
(a) \(\dfrac{1}{5} = 0.2\).
(b) Expected = \(0.2 \times 30 =\) 6 correct.
A true/false test has 24 questions, answered by guessing.
12 correct
(a) \(\dfrac{1}{2} = 0.5\).
(b) Expected = \(0.5 \times 24 =\) 12 correct.
Level 1 · Fluency
Write the formula for expected frequency in \(n\) trials with probability \(p\).
Expected frequency = \(n \times p\).
In the expected frequency formula \(np\), what does \(n\) stand for?
The number of trials.
State the expected frequency of heads in \(n\) tosses of a fair coin, in terms of \(n\).
\(np = n \times \dfrac{1}{2} = \dfrac{n}{2}\).
Level 2 · Application
A die is rolled 300 times. Using \(np\), find the expected number of 4s.
50
\(n \times p = 300 \times \dfrac{1}{6} =\) 50.
A coin is tossed 240 times. Using \(np\), find the expected number of tails.
120
\(n \times p = 240 \times \dfrac{1}{2} =\) 120.
A spinner has 5 equal sectors. It is spun 200 times. Using \(np\), find the expected number of times it lands on a given sector.
40
\(n \times p = 200 \times \dfrac{1}{5} =\) 40.
Level 3 · Further Application
In a town, 55% of adults own a pet. A sample of 640 adults is surveyed.
288
(a) \(0.55 \times 640 = 352\).
(b) \(0.45 \times 640 =\) 288 (or \(640 - 352 = 288\)).
In a region, 65% of households have internet. A sample of 480 households is taken.
168
(a) \(0.65 \times 480 = 312\).
(b) \(0.35 \times 480 =\) 168 (or \(480 - 312 = 168\)).
A factory finds 8% of bolts are undersized. A batch of 1500 bolts is checked.
1380
(a) \(0.08 \times 1500 = 120\).
(b) \(0.92 \times 1500 =\) 1380 (or \(1500 - 120 = 1380\)).