Chance · applications

Applications: Chance

10 question types · Primary 5 · each worked step by step with a figure that follows the steps

O-Level · SAT · GCSE Higher

01

A Number Cube at Game Night, and Four Events Placed on the Chance Scale

strategyCount the Faces That Make Each Event Happen / Mark 0 for Impossible and 1 for Certain, and Place the Fractions Between

At game night, Leo rolls a fair number cube with the numbers 1 to 6 on its faces. Here are four events. A: he rolls a 7. B: he rolls a number less than 7. C: he rolls an even number. D: he rolls a 5 or a 6. (a) Which event is impossible, and which event is certain? Give the number each one has on the chance scale. (b) Write the probabilities of C and D as fractions, and put all four events in order from the least likely to the most likely.

12345601/21impossibleeven chancecertainEach face is 1 of 6 equally likely results.
The six faces are equally likely. Each event's probability is a number of faces out of 6.
Write the six faces: 1, 2, 3, 4, 5, 6. Each face is equally likely, so each event's probability is a number of faces out of 6.
step 1 of 4

The chance scale runs from 0 for impossible to 1 for certain. The cube has 6 faces that are equally likely, so the probability of an event is the number of faces that make it happen, out of 6.

  1. Write the six faces: 1, 2, 3, 4, 5, 6. Each face is equally likely, so each event's probability is a number of faces out of 6.
  2. (a) No face shows a 7, so A happens on 0 faces: 06 = 0. A is impossible. Every face is less than 7, so B happens on all 6 faces: 66 = 1. B is certain.
  3. C happens on 2, 4 and 6: 36 = 12. D happens on 5 and 6: 26 = 13.
  4. (b) On the chance scale, 13 lies to the left of 12, because 26 is less than 36. From the least likely to the most likely, the order is A, D, C, B.

answer(a) A is impossible, at 0; B is certain, at 1. (b) C is 12 and D is 13. The order is A, D, C, B.

techniqueThe Probability Scale · Single-Event Probability

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Calling D more likely than C because D names the two biggest numbers, 5 and 6. How big the numbers are does not matter. Count the faces: C has 3 faces and D has only 2.
  • Calling B only likely, because the cube might show a small number. Every one of the 6 faces is less than 7, so B cannot fail to happen. It is certain, at 1.
02

A Bag of Marbles at the School Fair, and Green Marbles Added for an Even Chance

strategyCount the Marbles That Win, Out of All the Marbles / An Even Chance Needs as Many Winners as All the Others

A bag at the school fair holds 4 red, 5 blue and 3 green marbles of the same size. Nina takes one marble without looking. (a) What is the probability that her marble is blue? (b) Nina puts her marble back. The stall keeper then adds some green marbles, and no other marbles, so that the probability of taking a green marble becomes 12. How many green marbles does he add?

redbluegreen4 + 5 + 3 = 12 marbles
There are 4 + 5 + 3 = 12 marbles, and each is equally likely to be taken.
Count all the marbles: 4 + 5 + 3 = 12. Each of the 12 marbles is equally likely to be taken.
step 1 of 5

Each marble is equally likely to be taken. The probability of a color is the number of marbles of that color, out of all the marbles in the bag.

  1. Count all the marbles: 4 + 5 + 3 = 12. Each of the 12 marbles is equally likely to be taken.
  2. (a) 5 of the 12 marbles are blue, so the probability of blue is 512.
  3. A probability of 12 means that half the marbles are green. Then there must be as many green marbles as all the other marbles together.
  4. The other marbles are 4 red and 5 blue, which is 9. Adding green marbles does not change them, so the bag needs 9 green marbles.
  5. (b) There are 3 green marbles already, so he adds 9 − 3 = 6. Check: the bag now has 18 marbles, and 918 = 12.

answer(a) 512; (b) 6 green marbles

techniqueSingle-Event Probability · The Probability Scale

examsO-Level · SAT

Common pitfalls

  • Writing the probability of blue as 57, the blue marbles against the marbles that are not blue. A probability compares the marbles that win with all 12 marbles in the bag.
  • Adding 3 green marbles to make 6, because 6 is half of 12. The new marbles are in the bag too, so the total grows to 15, and 615 is not 12.
03

A Prize Wheel at a Carnival, Where One Section Fills Half the Wheel

strategyCut the Wheel into Equal Sections First / Count the Equal Sections That Win

A prize wheel at a carnival has five sections. The yellow section is half of the wheel. The other half is cut into four equal sections: two red, one blue and one green. The pointer is equally likely to stop anywhere on the wheel. (a) What is the probability that the pointer stops on red? (b) A player wins a prize when the pointer stops on yellow or on green. What is the probability of winning a prize?

yellowredblueredgreen5 sections, not all the same size
Each small section is one quarter of a half: 18 of the wheel.
Each small section is one quarter of a half, so it is 18 of the wheel. The yellow half is the same size as 4 of these small sections.
step 1 of 4

The five sections are not all the same size, so they are not equally likely. Cut the wheel into equal sections first, and then count.

  1. Each small section is one quarter of a half, so it is 18 of the wheel. The yellow half is the same size as 4 of these small sections.
  2. Now the wheel is 8 equal sections: 4 yellow, 2 red, 1 blue and 1 green. Each of the 8 is equally likely.
  3. (a) Red is 2 of the 8 equal sections, so the probability of red is 28 = 14.
  4. (b) Yellow or green is 4 + 1 = 5 of the 8 equal sections, so the probability of winning a prize is 58.

answer(a) 14; (b) 58

techniqueSingle-Event Probability

examsO-Level · SAT

Common pitfalls

  • Saying the probability of red is 25, because 2 of the 5 sections are red. The sections are different sizes, so they are not equally likely: the yellow section is as big as four small ones.
  • Giving yellow a probability of 15 in part (b). Yellow is half the wheel, 48, so yellow or green is 48 + 18 = 58.
04

A Coin and a Three-Color Spinner That Decide Who Moves First

strategyList Every Pair of Results in a Table / Count the Pairs That Win, Each One Only Once

To start a board game, Ana tosses a fair coin and spins a spinner with three equal sections colored red, blue and yellow. (a) List all the possible results, such as heads and red. How many results are there? (b) Ana moves first if the coin shows heads, or the spinner shows red, or both. What is the probability that Ana moves first?

redblueyellowheadstails2 rows and 3 columns
The rows are the two sides of the coin. The columns are the three colors.
Draw a table. The rows are heads (H) and tails (T). The columns are red, blue and yellow.
step 1 of 4

Each side of the coin can go with each color of the spinner. A table with the coin down the side and the colors across the top lists every pair once.

  1. Draw a table. The rows are heads (H) and tails (T). The columns are red, blue and yellow.
  2. (a) Each cell is one result: H and red, H and blue, H and yellow, T and red, T and blue, T and yellow. There are 2 × 3 = 6 results, and they are equally likely.
  3. Shade the cells where Ana moves first: all 3 cells in the heads row, and the tails cell under red. The cell H and red is shaded once, not twice.
  4. (b) 4 of the 6 cells are shaded, so the probability that Ana moves first is 46 = 23.

answer(a) 6 results; (b) 23

techniqueSample Space · Single-Event Probability

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Adding 12 for heads and 13 for red to get 56. The result H and red is in both groups, so it is counted twice. Count the shaded cells instead.
  • Listing only 5 results: heads, tails, red, blue and yellow. Each result is a pair, one side of the coin with one color, so there are 2 × 3 = 6 results.
05

A City Bus That Can Be Early, On Time or Late

strategyNot Happening Is 1 Minus Happening / Take Away Every Case That Does Not Count

The Route 9 bus takes Mia to school. On any school day, the probability that it is late is 0.25 and the probability that it is early is 0.1. Otherwise it is on time. (a) What is the probability that the bus is not late? (b) What is the probability that the bus is on time?

All daysearly?latetotal 10.10.25
The whole bar is all the school days, probability 1: early 0.1 and late 0.25.
Draw one bar for all the school days, with probability 1. Mark 0.1 at one end for early and 0.25 at the other end for late.
step 1 of 4

On each day the bus is early, on time or late, and only one of these can happen. So the three probabilities add up to 1.

  1. Draw one bar for all the school days, with probability 1. Mark 0.1 at one end for early and 0.25 at the other end for late.
  2. (a) Not late is the complement of late, so its probability is 1 − 0.25 = 0.75.
  3. The days that are not late are the early days and the on-time days together.
  4. (b) Take away the early days: 0.75 − 0.1 = 0.65. Check: 0.1 + 0.65 + 0.25 = 1.

answer(a) 0.75; (b) 0.65

techniqueComplementary Events

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Answering part (b) with 0.75. The days that are not late include the early days, and an early bus is not on time, so take away the 0.1 as well.
  • Giving 0.25 + 0.1 = 0.35 as the probability that the bus is on time. That is the probability that it is early or late, the opposite of on time, so on time is 1 − 0.35 = 0.65.
06

A Tombola at the School Fair, and the Chance That a Ticket Wins Nothing

strategyCount the Tickets That Do Not Win, Out of All the Tickets / Change Both Counts When Tickets Are Taken Out

A tombola drum at the school fair holds 200 folded tickets. 30 of them win a prize and the rest win nothing. Each person takes one ticket without looking, and no ticket goes back. (a) What is the probability that the first ticket taken wins nothing? (b) By noon, 50 tickets have been taken, and 10 of them won prizes. What is the probability that the next ticket taken wins nothing?

At the startwin 30nothing 170200 in all170 of 200 win nothing: 170/200 = 17/20
(a) 200 − 30 = 170 tickets win nothing: 170200 = 1720.
(a) 200 − 30 = 170 tickets win nothing. The probability is 170200 = 1720. Check with the complement: 1 − 30200 = 170200.
step 1 of 4

A ticket either wins a prize or it does not, so the tickets that win nothing are all the others. When tickets are taken out, the total in the drum changes, and so does the number of winning tickets left.

  1. (a) 200 − 30 = 170 tickets win nothing. The probability is 170200 = 1720. Check with the complement: 1 − 30200 = 170200.
  2. At noon, 200 − 50 = 150 tickets are left in the drum.
  3. 10 prizes have gone, so 30 − 10 = 20 winning tickets are left. The other 150 − 20 = 130 tickets win nothing.
  4. (b) The probability that the next ticket wins nothing is 130150 = 1315.

answer(a) 1720; (b) 1315

techniqueComplementary Events · Single-Event Probability

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Keeping 200 as the total in part (b). The 50 tickets taken are not in the drum any more, so the next ticket comes from 150.
  • Taking all 50 tickets away from the 170 that win nothing. Only 40 of the 50 won nothing, so 170 − 40 = 130 are left, the same as 150 − 20.
07

A Thumbtack Tossed 200 Times, and an Estimate for the Next 50 Tosses

strategyDivide the Times It Happened by the Number of Tosses / Multiply That Chance by the Tosses to Come

Ravi tosses a thumbtack 200 times, in four rounds of 50. It lands point up or point down. In the four rounds it lands point up 30, 34, 31 and 33 times, and point down 20, 16, 19 and 17 times. (a) What is the experimental probability that the thumbtack lands point up? Give it as a decimal. (b) Ravi will toss the thumbtack 50 more times. Estimate how many of those tosses will land point down.

round 1round 2round 3round 4totalup30343133128down2016191772128 + 72 = 200 tosses
Point up: 128 in all. Point down: 72 in all.
Add each row of the table. Point up: 30 + 34 + 31 + 33 = 128. Point down: 20 + 16 + 19 + 17 = 72. Check: 128 + 72 = 200.
step 1 of 4

A thumbtack is not the same shape on both sides, so point up and point down need not be equally likely. The chance has to be measured: the experimental probability is the number of times something happened, out of the number of tosses.

  1. Add each row of the table. Point up: 30 + 34 + 31 + 33 = 128. Point down: 20 + 16 + 19 + 17 = 72. Check: 128 + 72 = 200.
  2. (a) The experimental probability of point up is 128200 = 64100 = 0.64.
  3. The experimental probability of point down is 72200 = 0.36. It is also 1 − 0.64 = 0.36.
  4. (b) In 50 tosses, expect about 50 × 0.36 = 18 to land point down. The real number may be a little more or a little less.

answer(a) 0.64; (b) about 18 tosses

techniqueExperimental Probability · Expected Counts from a Chance · Complementary Events

examsSAT · GCSE Higher

Common pitfalls

  • Using only one round, such as 3450 = 0.68. All 200 tosses together give a better estimate than any one round of 50.
  • Answering part (b) with 50 × 0.64 = 32. That is the estimate for point up, and the question asks about point down.
08

Free Throws in a Week of Practice, Checked Against the Coach's Claim

strategyTotal the Shots and the Baskets Before Dividing / Turn the Claim into a Count for the Same Number of Shots

Jada practices free throws after school. She takes 20 free throws each day. She scores 13 baskets on Monday, 15 on Tuesday, 14 on Wednesday, 12 on Thursday and 16 on Friday. (a) What is the experimental probability that Jada scores a free throw? Give it as a decimal. (b) Her coach says Jada scores 3 out of every 4 free throws. How many baskets would the coach's claim predict for the week's free throws, and how many fewer did Jada really score?

MonTueWedThuFritotalshots2020202020100baskets1315141216705 × 20 = 100 shots, 70 baskets
She took 5 × 20 = 100 free throws and scored 70.
Jada took 5 × 20 = 100 free throws. She scored 13 + 15 + 14 + 12 + 16 = 70 baskets.
step 1 of 4

The experimental probability uses every shot of the week. The coach's claim is a probability too, so it predicts a number of baskets for the same number of shots.

  1. Jada took 5 × 20 = 100 free throws. She scored 13 + 15 + 14 + 12 + 16 = 70 baskets.
  2. (a) The experimental probability is 70100 = 0.7.
  3. The coach's claim is 34. For 100 free throws it predicts 100 × 34 = 75 baskets.
  4. (b) Jada scored 75 − 70 = 5 fewer than the claim predicts. A gap of 5 in 100 shots is small, and chance alone often makes a gap that size, so this week does not show that the coach is wrong.

answer(a) 0.7; (b) the claim predicts 75, and she scored 5 fewer

techniqueExperimental Probability · Expected Counts from a Chance

examsSAT · GCSE Higher

Common pitfalls

  • Dividing 70 by 5 days to get 14. That is the mean number of baskets in a day, not a probability. A probability is baskets out of shots: 70100.
  • Saying the coach must be wrong because 70 is not 75. Results that depend on chance are rarely exactly what a probability predicts. A small gap over 100 shots is to be expected.
09

A Class Spinner, and Its Greens After 10, 100 and 1000 Spins

strategyFind the Theoretical Probability from the Equal Sections / Read Each Running Total as a Fraction and Compare the Gaps

A spinner has 4 equal sections colored green, red, blue and yellow. A class spins it many times and keeps a running total of the greens. After 10 spins there are 5 greens, after 100 spins there are 29 greens, and after 1000 spins there are 256 greens. (a) What is the theoretical probability of green, as a decimal? What is the experimental probability of green after the first 10 spins? (b) What is the experimental probability of green after all 1000 spins, and how far is it from the theoretical probability?

theory 0.251 of 4 equal sections: 1/4 = 0.25
One of the 4 equal sections is green: 14 = 0.25 in theory.
1 of the 4 equal sections is green, so the theoretical probability of green is 14 = 0.25.
step 1 of 5

The theoretical probability comes from the equal sections. The experimental probability is the greens so far, out of the spins so far. The more spins there are, the closer the experimental probability usually comes to the theoretical one. This is the law of large numbers.

  1. 1 of the 4 equal sections is green, so the theoretical probability of green is 14 = 0.25.
  2. (a) After 10 spins, the experimental probability is 510 = 0.5. That is twice the theoretical probability. In only 10 spins, a gap this big can happen.
  3. After 100 spins, it is 29100 = 0.29, which is 0.29 − 0.25 = 0.04 away from 0.25.
  4. (b) After 1000 spins, it is 2561000 = 0.256, which is only 0.256 − 0.25 = 0.006 away from the theoretical probability.
  5. The gaps shrink from 0.25 to 0.04 to 0.006. With more spins, the experimental probability settles close to 0.25.

answer(a) 0.25 in theory, and 0.5 after 10 spins; (b) 0.256, which is 0.006 away from 0.25

techniqueThe Law of Large Numbers · Experimental Probability · Single-Event Probability

examsSAT · GCSE Higher

Common pitfalls

  • Deciding after 10 spins that the spinner is unfair because green came up half the time. Ten spins are too few: a fair spinner often gives results far from 14 in a short run.
  • Expecting exactly 250 greens in 1000 spins. The law of large numbers says the fraction of greens comes close to 14, not that the count lands exactly on 250.
10

Visitors to a Science Museum Shop, Expected on Saturday and on Sunday

strategyMultiply the Number of Visitors by the Probability / Work Backward from the Expected Count to the Visitors

At a science museum, the probability that a visitor buys something in the shop is 310, on every day of the week. (a) On Saturday, 300 people visit the museum. How many of them would you expect to buy something? (b) On Sunday, the shop expects 72 buyers. How many visitors does the museum expect on Sunday?

10 visitors3 buy
A probability of 310 means about 3 buyers in every 10 visitors.
A probability of 310 means about 3 buyers in every 10 visitors.
step 1 of 4

An expected count is the probability times the number of tries, which here is the number of visitors. It tells you about how many, not exactly how many.

  1. A probability of 310 means about 3 buyers in every 10 visitors.
  2. (a) 300 visitors make 300 ÷ 10 = 30 groups of 10. Expect about 30 × 3 = 90 buyers. This is the same as 300 × 310 = 90.
  3. On Sunday, 72 buyers make 72 ÷ 3 = 24 groups of 3, and each group of 3 buyers comes from a group of 10 visitors.
  4. (b) Expect about 24 × 10 = 240 visitors. Check: 240 × 310 = 72.

answer(a) about 90 buyers; (b) about 240 visitors

techniqueExpected Counts from a Chance

examsSAT · GCSE Higher

Common pitfalls

  • Multiplying 72 by 310 in part (b) to get 21.6. The 72 buyers are only part of the visitors, so there must be more visitors than 72: divide by 3, then multiply by 10.
  • Saying that exactly 90 people will buy something on Saturday. An expected count is about how many. On a real Saturday it might be 84 or 97.
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