Probability with Several Events · applications

Applications: Probability with Several Events

12 question types · Secondary 4 · each worked step by step with a figure that follows the steps

O-Level · H2

01

Students Taking Music and Drama in a Venn Diagram, and a Probability Read from One Region

methodFill the Overlap First, Then Each Circle Alone, Then the Outside, and Take the Whole the Student Is Chosen From

In a year group of 60 Secondary 4 students, 27 take music, 21 take drama and 8 take both. (a) Draw a Venn diagram with the number of students in every region, and find the probability that a student chosen at random takes exactly one of the two subjects. (b) A student is chosen at random from those who take music. Find the probability that this student also takes drama.

MusicDrama60 students8both subjects: 8
The 8 students who take both subjects go in the overlap first.
Write the 8 students who take both subjects in the overlap of the two circles.
step 1 of 5

The 27 students who take music include the 8 who take both, and so do the 21 who take drama. Writing the overlap first and subtracting it from each circle gives regions that hold each student exactly once. A probability is then a count over the whole the student is chosen from: all 60 students in (a), and only the 27 music students in (b).

  1. Write the 8 students who take both subjects in the overlap of the two circles.
  2. The music circle holds 27 students, so 27 − 8 = 19 take music only. The drama circle holds 21, so 21 − 8 = 13 take drama only.
  3. The circles hold 19 + 8 + 13 = 40 students, so 60 − 40 = 20 take neither subject. Check: 19 + 8 + 13 + 20 = 60.
  4. (a) Exactly one subject means music only or drama only, which is 19 + 13 = 32 students. The probability is 3260 = 815.
  5. (b) The student is chosen from the music circle, so the whole is 27, not 60. 8 of those 27 students also take drama, so the probability is 827.

answer(a) the regions hold 19 music only, 8 both, 13 drama only and 20 neither, and the probability is 815; (b) 827

techniqueNaming the Regions of a Venn Diagram · Conditional Probability

examsO-Level

Common pitfalls

  • Writing 27 and 21 in the parts of the circles outside the overlap. Those totals already include the 8 students who take both, so the circles would hold 27 + 8 + 21 = 56 students instead of 40, and the 8 would be counted three times.
  • Answering (b) with 860. That is the probability that a student from the whole year group takes both subjects. In (b) the student is already known to take music, so only the 27 students in the music circle can be chosen.
02

Leaving the Jail Square in a Board Game with Two Dice, Read from a Possibility Diagram

methodList All 36 Equally Likely Pairs in a Grid, Shade the Ones That Count, and Add Chances Only for Events That Cannot Happen Together

In a board game, a player in the jail square throws two fair dice and leaves only if the total is 8 or more. (a) Draw a possibility diagram of the totals and find the probability that the player leaves on one throw. (b) The rule is changed so that the player leaves on a total of exactly 7 or exactly 11. Find the new probability, and say which rule makes leaving more likely.

first dice across, second dice down1234561234567234567834567894567891056789101167891011126 × 6 = 36 equally likely pairs
The first dice goes across and the second down. Each of the 36 cells is one equally likely throw, with its total written in it.
Draw a 6 by 6 grid with the first dice across the top and the second dice down the side, and write the total in each of the 36 cells.
step 1 of 5

Each dice has 6 faces, so there are 6 × 6 = 36 pairs of scores, all equally likely. A possibility diagram writes the total of every pair in a grid, so the favorable pairs can be counted without missing any. A total of 7 and a total of 11 cannot happen on the same throw, so their probabilities can be added.

  1. Draw a 6 by 6 grid with the first dice across the top and the second dice down the side, and write the total in each of the 36 cells.
  2. Shade the cells with a total of 8 or more. They lie on diagonals: 5 cells total 8, 4 total 9, 3 total 10, 2 total 11 and 1 totals 12, which is 5 + 4 + 3 + 2 + 1 = 15 cells.
  3. (a) The probability of leaving is 1536 = 512.
  4. Under the new rule, 6 cells total 7 and 2 cells total 11. No cell has both totals, so the two events are mutually exclusive and their probabilities add: 636 + 236 = 836.
  5. (b) The new probability is 836 = 29. Since 836 is less than 1536, the first rule makes leaving more likely.

answer(a) 1536 = 512; (b) 29, so the first rule makes leaving more likely

techniquePossibility Diagrams · Mutually Exclusive Events

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Counting the totals 2 to 12 as 11 equally likely outcomes, so that a total of 8 or more has probability 511. The totals are not equally likely: a total of 7 comes from 6 pairs and a total of 12 from only one. The 36 pairs are the equally likely outcomes.
  • Counting (3, 5) and (5, 3) as one outcome. They are different throws, because the first dice shows 3 in one and 5 in the other. Merging such pairs leaves 21 outcomes that are not equally likely, and gives the wrong answer 921 for part (a).
03

A Coin Rolled onto a Board of Squares at a Fair, Where the Chance of a Win Is an Area

methodFollow the Center of the Coin: Mark the Region Where It Must Land, and Divide That Area by the Area of One Square

At a school fair, a coin 2 cm across is rolled onto a large board ruled into squares 5 cm by 5 cm. The player wins if the coin comes to rest wholly inside one square, touching no line. The center of the coin is equally likely to land anywhere in a square. (a) Find the probability of a win. (b) The organizers want the chance of a win to be 49. What size of square should they rule?

3 cm5 cmthe center stays 1 cm from every sidewinning square: 5 − 2 = 3 cm
The coin touches no line when its center is at least 1 cm from every side, so the center must land in the shaded square of side 3 cm.
The radius of the coin is 1 cm. The coin touches no line when its center is at least 1 cm from all four sides, which is a smaller square of side 5 − 2 = 3 cm.
step 1 of 5

The coin touches a line exactly when its center is less than one radius, 1 cm, from it. So the coin wins when its center lands in a smaller square, 1 cm in from every side. Because the center is equally likely to land anywhere, the probability is the area of that smaller square divided by the area of the whole square.

  1. The radius of the coin is 1 cm. The coin touches no line when its center is at least 1 cm from all four sides, which is a smaller square of side 5 − 2 = 3 cm.
  2. The smaller square has area 3 × 3 = 9 cm2 and the whole square has area 5 × 5 = 25 cm2.
  3. (a) The probability of a win is 925 = 0.36.
  4. Let the squares be s cm wide. The winning square has side s − 2, so (s − 2)2s2 = 49. Both lengths are positive, so take the positive square root of each side: s − 2s = 23.
  5. (b) 3(s − 2) = 2s, so 3s − 6 = 2s and s = 6. The squares should be 6 cm by 6 cm. Check: the winning square is 4 cm wide, and 4 × 46 × 6 = 1636 = 49.

answer(a) 925 = 0.36; (b) squares 6 cm by 6 cm

techniqueChance from Areas

Common pitfalls

  • Taking the winning square as 5 − 1 = 4 cm wide. The center must keep 1 cm from the left side and 1 cm from the right side, so 2 cm comes off the width, and the same off the height.
  • Using the ratio of the side lengths, 35, as the probability. The center can land anywhere in a flat square, so the chance is a ratio of areas, and the ratio of the sides has to be squared: 35 × 35 = 925.
04

A Raffle That Pays on a Multiple of 3 or a Multiple of 5, and the Tickets Counted Twice

methodAdd the Two Probabilities and Take the Overlap Off Once; Add Without Taking Anything Off Only When the Events Cannot Happen Together

At a fair, one ticket is drawn at random from tickets numbered 1 to 30. A prize is given if the number is a multiple of 3 or a multiple of 5. (a) Find the probability that the ticket drawn wins a prize. (b) A second prize is added for a multiple of 7 that has not already won the first prize. Find the probability that the ticket drawn wins one of the two prizes.

123456789101112131415161718192021222324252627282930multiples of 3 (gold): 10 ticketsmultiples of 5: 6 tickets
10 of the tickets are multiples of 3 and 6 are multiples of 5.
There are 10 multiples of 3 from 3 to 30, so P(multiple of 3) = 1030. There are 6 multiples of 5 from 5 to 30, so P(multiple of 5) = 630.
step 1 of 6

The addition rule says P(A or B) = P(A) + P(B) − P(A and B). Adding the two probabilities counts each ticket that is a multiple of both 3 and 5 twice, so the overlap is taken off once. The second prize goes only to tickets that did not win the first, so the two prizes cannot both happen and their probabilities simply add.

  1. There are 10 multiples of 3 from 3 to 30, so P(multiple of 3) = 1030. There are 6 multiples of 5 from 5 to 30, so P(multiple of 5) = 630.
  2. The tickets 15 and 30 are multiples of both 3 and 5, so they are in both lists, and P(both) = 230.
  3. By the addition rule, P(prize) = 1030 + 630 − 230 = 1430.
  4. (a) The probability of a prize is 1430 = 715.
  5. The multiples of 7 are 7, 14, 21 and 28. The ticket 21 is a multiple of 3 and has already won, so the second prize goes to 7, 14 and 28. No ticket wins both prizes, so the events are mutually exclusive: 1430 + 330 = 1730.
  6. (b) The probability of winning one of the two prizes is 1730. Check: 14 + 3 = 17 of the 30 tickets win a prize.

answer(a) 1430 = 715; (b) 1730

techniqueThe Addition Rule for Overlapping Events · Mutually Exclusive Events · Naming the Regions of a Venn Diagram

examsO-Level · H2

Common pitfalls

  • Adding 1030 + 630 = 1630 for part (a). The tickets 15 and 30 are in both lists, so they are counted twice, and only 14 different tickets win.
  • Adding 430 for all four multiples of 7 in part (b). The ticket 21 is already one of the 14 winners of the first prize, so adding it again counts it twice.
05

Two Lifts in an Office Block That Break Down Independently, on a Tree Diagram

methodDraw the Same Second Branches After Every First Branch, Multiply Along Each Path, and Add the Paths That Answer the Question

An office block has two lifts, which break down independently. On any day, lift A is out of order with probability 15 and lift B is out of order with probability 14. (a) Find the probability that both lifts are out of order on a given day. (b) Find the probability that exactly one of the lifts is working.

lift A1/5out4/5workslift B1/4out3/4works1/4out3/4worksthe same branches for B after either A
The lifts are independent, so the branches for lift B are the same after either branch for lift A.
Draw the first pair of branches for lift A: out of order 15 and working 45. After each of them draw the branches for lift B: out of order 14 and working 34, the same both times because the lifts are independent.
step 1 of 5

Independent means that whether lift A is working does not change the chances for lift B. On a tree diagram the branches for lift B are therefore the same after either branch for lift A. The probability of a path is the product of the probabilities along it, and the four paths together have probability 1.

  1. Draw the first pair of branches for lift A: out of order 15 and working 45. After each of them draw the branches for lift B: out of order 14 and working 34, the same both times because the lifts are independent.
  2. Multiply along each path: 15 × 14 = 120, 15 × 34 = 320, 45 × 14 = 420 and 45 × 34 = 1220. Check: 1 + 3 + 4 + 12 = 20, so the four paths add up to 1.
  3. (a) Both lifts are out of order on the first path, with probability 120.
  4. Exactly one lift is working on two paths: A out and B working, 320, and A working and B out, 420. These cannot both happen on the same day, so add them.
  5. (b) The probability that exactly one lift is working is 320 + 420 = 720.

answer(a) 120; (b) 720

techniqueTree Diagrams · Independent Events · Multiplying Along a Tree Without Replacement

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Adding 15 + 14 = 920 for the chance that both lifts are out. Both out means A out and B out, and for independent events that is a product. Adding gives more than either chance on its own, but both lifts failing must be less likely than lift A failing.
  • Answering (b) with one path only, such as 320. Exactly one lift working can happen in two ways, lift A working with lift B out or lift B working with lift A out, and both paths must be added.
06

A Fairground Game That Pays on at Least One Six in Three Rolls, Worked by the Complement

methodFind the Chance That No Six Comes Up at All, Then Take It Away from 1

A fairground stall charges a player to roll a fair dice three times, and pays a prize if at least one six comes up. The stall-holder says the chance of a prize is 36, because each roll has a 16 chance of a six. (a) Find the true probability of at least one six in three rolls. (b) What is the smallest number of rolls for which the chance of at least one six is more than 12?

1 roll5/61/6left: no six, right: at least one sixone roll: no six 5/6, a six 1/6
On one roll the chance of no six is 56. The complement of at least one six is no six on any roll.
The complement of at least one six is no six in any of the three rolls. Each roll misses a six with probability 56.
step 1 of 5

At least one six can happen in many ways: a six on the first roll only, sixes on the first and third rolls, and so on. Its complement, no six at all, happens in one way only: every roll misses. The rolls are independent, so the chance that every roll misses is 56 multiplied together once for each roll.

  1. The complement of at least one six is no six in any of the three rolls. Each roll misses a six with probability 56.
  2. The rolls are independent, so P(no six) = 56 × 56 × 56 = 125216.
  3. (a) P(at least one six) = 1 − 125216 = 91216, which is about 0.421 — the game is worse than even, not the stall-holder's even chance. Adding 16 three times the chances of events that can happen together; with seven rolls the same reasoning would give 76, which is more than 1.
  4. Three rolls give less than a half, since 91216 is less than 108216. Try a fourth roll: P(no six) = 6251296, so P(at least one six) = 6711296.
  5. (b) The smallest number is 4 rolls: 6711296 is more than 12 = 6481296, and three rolls give less.

answer(a) 91216, about 0.421; (b) 4 rolls

techniqueThe Chance of At Least One · Independent Events

examsO-Level · SAT

Common pitfalls

  • Adding 16 three times to get 36. A six on the first roll and a six on the second roll can both happen, so adding counts those throws more than once. The complement has no overlap to worry about.
  • Working out 16 × 16 × 16. That is the chance of a six on every one of the three rolls, which is far smaller than the chance of at least one six.
07

Two Socks Taken from a Drawer in the Dark, Without Replacement

methodChange the Second Branches to Match What Is Left in the Drawer, Then Multiply Along Each Path

A drawer holds 5 black socks and 3 gray socks. In the dark, Wei Ming takes out one sock and then a second, without putting the first back. (a) Find the probability that the two socks are the same color. (b) Find the probability that he takes one sock of each color.

first sock5/8black3/8gray8 socks: 5/8 black, 3/8 gray
The first sock is black with probability 58 and gray with probability 38.
The first sock is black with probability 58 and gray with probability 38.
step 1 of 5

After the first sock is taken, only 7 socks are left, and which colors are left depends on the first sock. So the second branches of the tree are different after a black sock and after a gray one. Each path's probability is the product along it, and the paths that answer the question are added.

  1. The first sock is black with probability 58 and gray with probability 38.
  2. After a black sock, 4 black and 3 gray socks are left, so the second sock is black with probability 47 and gray with probability 37. After a gray sock, 5 black and 2 gray socks are left, so the probabilities are 57 and 27.
  3. Multiply along each path: black then black 58 × 47 = 2056, black then gray 1556, gray then black 1556, and gray then gray 38 × 27 = 656. Check: 20 + 15 + 15 + 6 = 56.
  4. (a) The same color is black then black or gray then gray: 2056 + 656 = 2656 = 1328.
  5. (b) One of each color is the two middle paths: 1556 + 1556 = 3056 = 1528. Check: 1328 + 1528 = 1, because the two socks are either the same color or not.

answer(a) 1328; (b) 1528

techniqueWithout Replacement · Multiplying Along a Tree Without Replacement

examsO-Level · GCSE Higher

Common pitfalls

  • Using 58 and 38 again on the second branches. That describes putting the first sock back. Without replacement there are only 7 socks for the second draw, and the color of the first sock has one sock fewer.
  • Counting only black then gray for part (b), which gives 1556. Gray then black is a different path that also gives one sock of each color, so both paths must be added.
08

A Survey of Cyclists and Helmets, Where the Given Condition Chooses the Row or the Column

methodFind the Group That the Word Given Names, Use Its Total as the Denominator, and Count Only Inside It

A council counted 160 cyclists on a cycle path and recorded their age group and whether they wore a helmet. Of the 48 cyclists under 18, 36 wore a helmet. Of the 112 cyclists aged 18 or over, 70 wore a helmet. (a) A cyclist under 18 is chosen at random. Find the probability that this cyclist wore a helmet, and compare it with the same probability for the older cyclists. (b) A cyclist who wore no helmet is chosen at random. Find the probability that this cyclist is under 18.

HelmetNo helmetTotalUnder 1836124818 and over7042112Total1065416048 − 36 = 12 and 112 − 70 = 4236 + 70 = 106 and 12 + 42 = 54
Complete the table: 12 and 42 cyclists wore no helmet, and the columns total 106 and 54.
Complete the two-way table. Under 18, 48 − 36 = 12 wore no helmet. Aged 18 or over, 112 − 70 = 42 wore no helmet. The column totals are 36 + 70 = 106 with a helmet and 12 + 42 = 54 without.
step 1 of 5

P(A | B) means the probability of A given that B is known to have happened. The condition B picks out one row or one column of the two-way table, and that total is the whole. P(helmet | under 18) uses the Under 18 row, while P(under 18 | no helmet) uses the No helmet column, so the two probabilities have different denominators.

  1. Complete the two-way table. Under 18, 48 − 36 = 12 wore no helmet. Aged 18 or over, 112 − 70 = 42 wore no helmet. The column totals are 36 + 70 = 106 with a helmet and 12 + 42 = 54 without.
  2. (a) Given under 18, the whole is the Under 18 row, 48 cyclists. P(helmet | under 18) = 3648 = 34. For the older cyclists it is 70112 = 58, so a younger cyclist was more likely to wear a helmet.
  3. Given no helmet, the whole is the No helmet column, 54 cyclists, and 12 of them are under 18.
  4. (b) P(under 18 | no helmet) = 1254 = 29.
  5. This is not the same as P(no helmet | under 18) = 1248 = 14. Both have the same 12 cyclists on top, but the condition changes the whole underneath: 54 in one and 48 in the other.

answer(a) 34, compared with 58 for the cyclists aged 18 or over; (b) 29

techniqueConditional Probability Notation · Conditional Probability

examsSAT · H2

Common pitfalls

  • Dividing 36 by 160 in part (a). That is the probability that a cyclist from the whole count was under 18 and wore a helmet. Given under 18, only the 48 younger cyclists can be chosen.
  • Answering (b) with 1248. That is the probability that a cyclist under 18 wore no helmet. The question gives no helmet and asks about the age, so the denominator is the 54 cyclists without a helmet.
09

The Game Show with Three Doors, and Whether to Switch

methodList Every Place the Car Can Be, Follow What the Host Must Do in Each Case, and Count the Cases Where Switching Wins

On a game show a car is hidden behind one of three closed doors, each equally likely and a goat behind each of the other two. You choose door 1. The host, who knows where the car is, always opens one of the other two doors to show a goat, and then offers you the chance to switch to the last closed door. (a) By listing every case, find the probability that you win the car if you switch. (b) In a version with 10 doors, you choose one, and the host opens 8 of the other 9 doors to show 8 goats. Find the probability that you win if you switch.

door 1door 2door 3car behind door 1cargoatgoatcar behind door 2goatcargoatcar behind door 3goatgoatcaryour choice: door 1 (gold)three cases, each 1/3
You choose door 1. The car is behind door 1, 2 or 3, each with probability 13.
The car is behind door 1, door 2 or door 3, each with probability 13. You have chosen door 1 in every case.
step 1 of 5

The car is equally likely to be behind any of the three doors, so there are three equally likely cases. In each case the host's move is either forced by where the car is or makes no difference to the result. Listing the cases shows that switching loses only when the first choice was right, which happens with probability 13.

  1. The car is behind door 1, door 2 or door 3, each with probability 13. You have chosen door 1 in every case.
  2. If the car is behind door 1, the host opens door 2 or door 3, and switching loses. If it is behind door 2, the host must open door 3, and switching to door 2 wins. If it is behind door 3, the host must open door 2, and switching to door 3 wins.
  3. (a) Switching wins in 2 of the 3 equally likely cases, so P(win by switching) = 23. Sticking wins only when the car is behind door 1, with probability 13.
  4. With 10 doors, your first choice is right with probability 110, and then switching loses. With probability 910 the car is behind one of the other 9 doors, and the host must leave that door closed when he opens 8 doors with goats.
  5. (b) Switching wins exactly when the first choice was wrong, so P(win by switching) = 910.

answer(a) 23; (b) 910

techniqueThe Monty Hall Problem · Conditional Probability

examsSAT · GCSE Higher · H2

Common pitfalls

  • Saying that once a goat is shown, two doors are left and each has a 12 chance. The two doors are not alike: the host chose which door to open knowing where the car is, and he could never open yours. Your door keeps the 13 it had at the start.
  • Counting the case where the car is behind door 1 as two cases, because the host can open door 2 or door 3. Those are two halves of one case with probability 13, each 16, and switching loses in both of them.
10

Three Friends Born on the Same Day of the Week, and 23 People Sharing a Birthday

methodFind the Chance That Everyone Is Different, One Person at a Time, Then Take It Away from 1

(a) Three friends are each equally likely to have been born on any of the 7 days of the week, independently of one another. Find the probability that at least two of them were born on the same day of the week. (b) There are 23 people in a room. Take every birthday to be equally likely to fall on any of 365 days. The chance that all 23 birthdays are different is 365365 × 364365 × ⋯ × 343365, which is 0.493 to 3 decimal places. Find the probability that at least two of the people share a birthday.

first friend7/7the first friend: any day, 7/7
The complement of at least two sharing is all three different. The first friend can have any day.
The complement of at least two sharing a day is all three born on different days. The first friend can be born on any day, with probability 77.
step 1 of 5

At least two sharing can happen in many ways, but its complement, everyone different, can be built up one person at a time. Each new person must avoid every day already taken, so the fractions shrink by one day each time. The chance that at least two share is 1 minus the chance that all are different.

  1. The complement of at least two sharing a day is all three born on different days. The first friend can be born on any day, with probability 77.
  2. The second friend must avoid the first friend's day, 67, and the third must avoid both, 57. So P(all different) = 77 × 67 × 57 = 210343 = 3049.
  3. (a) P(at least two share a day) = 1 − 3049 = 1949.
  4. For 23 people the same chain has 23 fractions, one for each person, from 365365 down to 343365. Its product is given as 0.493.
  5. (b) P(at least two share a birthday) = 1 − 0.493 = 0.507, so a shared birthday is slightly more likely than not. There are 23 × 222 = 253 pairs of people, and every pair is a chance for a match.

answer(a) 1949; (b) 0.507

techniqueThe Birthday Paradox · The Chance of At Least One

examsO-Level · SAT

Common pitfalls

  • Comparing 23 with 365 and expecting a chance of about 23365. That is close to the chance that someone shares one particular person's birthday. Any of the 253 pairs can match, so the chance is far larger.
  • Giving 77 × 67 × 57 = 3049 as the answer to (a). That product is the chance that all three friends were born on different days; the chance that at least two share is 1 minus it.
11

A Bank's Fraud Flag, and the Chance That a Flagged Payment Is Really Fraud

methodPicture 1000 Payments, Count the True Flags and the False Flags Separately, and Divide the True Ones by All the Flags

A bank's fraud flag checks card payments. 1 in every 100 payments is fraudulent. The flag catches 9 out of every 10 fraudulent payments, but it also wrongly flags 1 in every 30 genuine payments. (a) Out of 1000 payments, how many are flagged? Find the probability that a flagged payment is really fraudulent. (b) At an online shop, 1 in every 10 payments is fraudulent, and the same flag is used. Find the probability that a flagged payment there is really fraudulent.

1000 payments: 10 fraudulent (gold)990 genuine
Each dot is one of 1000 payments. 1100 of them, 10 payments, are fraudulent.
Of 1000 payments, 1100 of 1000, which is 10, are fraudulent, and 990 are genuine.
step 1 of 6

When the thing a check looks for is rare, a small rate of false flags among the many genuine payments can outnumber the true flags among the few fraudulent ones. Counting a round number of payments turns every rate into a whole number of payments that can be counted. The probability that a flagged payment is fraudulent is the number of true flags divided by the number of all flags.

  1. Of 1000 payments, 1100 of 1000, which is 10, are fraudulent, and 990 are genuine.
  2. The flag catches 910 of the 10 fraudulent payments, which is 9. It wrongly flags 130 of the 990 genuine payments, which is 33.
  3. Altogether 9 + 33 = 42 payments are flagged. The false flags outnumber the true ones, because there are 99 genuine payments for every fraudulent one.
  4. (a) 42 payments are flagged, and P(fraudulent | flagged) = 942 = 314, so fewer than 1 flagged payment in 4 is really fraud.
  5. At the online shop, 1000 payments hold 100 fraudulent and 900 genuine ones. The flag catches 910 × 100 = 90 and wrongly flags 130 × 900 = 30, so 120 are flagged.
  6. (b) P(fraudulent | flagged) = 90120 = 34. The flag works exactly as before; only the base rate of fraud has changed.

answer(a) 42 payments are flagged, and the probability is 314; (b) 34

techniqueWhy a Positive Test Can Still Mean Healthy · Conditional Probability Notation

examsSAT · H2

Common pitfalls

  • Taking 910 as the probability that a flagged payment is fraudulent. 910 is the chance that a fraudulent payment is flagged, which is a different question with a different whole: the 10 fraudulent payments, not the 42 flagged ones.
  • Leaving out the genuine payments that are flagged. They are a small fraction, 130, but of a large number, 990, so they give 33 flags, far more than the 9 from fraud.
12

A Faulty Bolt Traced Back to the Machine That Made It, by Reversing a Tree

methodDraw the Tree in the Order Things Happen, Then Divide the Path You Want by All the Paths That Match What Was Seen

A factory makes bolts on two machines. Machine A makes 35 of the bolts and machine B makes the rest. 120 of the bolts from machine A are faulty, and 110 of the bolts from machine B. An inspector picks a bolt at random. (a) Find the probability that it is faulty. (b) The bolt turns out to be faulty. Find the probability that it was made by machine B.

machine3/5A2/5Bbolt1/20faulty19/20good1/10faulty9/10goodfirst the machine, then the bolt
The tree follows the order of events: first the machine, then whether the bolt is faulty.
The first branches are machine A, 35, and machine B, 25. After A the bolt is faulty with probability 120 and good with 1920. After B it is faulty with probability 110 and good with 910.
step 1 of 5

The tree follows the order of events: first the machine, then whether the bolt is faulty. The question in (b) runs the other way, from the result back to the machine. Of all the paths that end in a faulty bolt, the share that passes through machine B is P(B | faulty) = P(B and faulty)P(faulty).

  1. The first branches are machine A, 35, and machine B, 25. After A the bolt is faulty with probability 120 and good with 1920. After B it is faulty with probability 110 and good with 910.
  2. Multiply along each path: A and faulty 35 × 120 = 3100, A and good 57100, B and faulty 25 × 110 = 4100, and B and good 36100. Check: 3 + 57 + 4 + 36 = 100.
  3. (a) Two paths end in a faulty bolt, so P(faulty) = 3100 + 4100 = 7100.
  4. Reverse the tree. Given that the bolt is faulty, only those two paths are possible, and the path through B is 4100 of the 7100: P(B | faulty) = 4100 ÷ 7100.
  5. (b) P(B | faulty) = 47. Machine B makes fewer of the bolts but more than half of the faulty ones. Check with 1000 bolts: A makes 600 with 30 faulty, B makes 400 with 40 faulty, and 4070 = 47.

answer(a) 7100; (b) 47

techniqueReversing a Tree: the Base-Rate Answer · Multiplying Along a Tree Without Replacement · Conditional Probability Notation

examsGCSE Higher

Common pitfalls

  • Answering (b) with 110, the chance that a bolt from machine B is faulty. The question asks the reverse: given a faulty bolt, how likely it is to have come from B. The whole is the faulty bolts, 7100 of all the bolts.
  • Answering (b) with 25, the share of the bolts that machine B makes, as if knowing the bolt is faulty changed nothing. Machine B makes faulty bolts twice as often as machine A, so a faulty bolt is more likely than a random bolt to have come from B.
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