Probability.

The Scene at this step

Hats

0fixed

Tosses

2defaulthidden

Books

3defaulthidden

The start / All counted

All countedfixed

Other settings

Mode/nodes/board/props/mode = "tree"

Other settings

Tree/nodes/board/props/tree = "choices"

Other settings

B1/nodes/board/props/b1 = 3

Other settings

B2/nodes/board/props/b2 = 2

Choices

• Correct $12$⁠

• Wrong $7$⁠

  Misconception: Adds the options at each stage instead of multiplying them: 3 shirts and 2 trousers make 5 outfits, two dice make 12 outcomes.

  Feedback from the catalogue

  Every option at the first stage meets every option at the second, so you multiply: each of the $3$⁠ shirts goes with $2$⁠ trousers, which is $3 \times 2 = 6$⁠ outfits. The tree shows it: every branch splits again. Adding only counts the branches of one stage after another, not the paths from the root to the end.

• Wrong $8$⁠

  Misconception: Adds the options at each stage instead of multiplying them: 3 shirts and 2 trousers make 5 outfits, two dice make 12 outcomes.

  Feedback from the catalogue

  Every option at the first stage meets every option at the second, so you multiply: each of the $3$⁠ shirts goes with $2$⁠ trousers, which is $3 \times 2 = 6$⁠ outfits. The tree shows it: every branch splits again. Adding only counts the branches of one stage after another, not the paths from the root to the end.

• Wrong $6$⁠, because a hat does not change the shirt and trousers

  Feedback

  Each of the $6$⁠ outfits can go with either of the two hats, so every outfit becomes two.

Once the prediction is locked in, the Scene changes

Hats

2fixed

The start / All counted

All countedfixed

The Scene then shows: 12 · Grid: cells; tree: leaves (paths)/nodes/board/outputs/outcomes

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Hats

0defaulthidden

Tosses

2defaulthidden

Books

3defaulthidden

The start / All counted

The startfixed

Other settings

Mode/nodes/board/props/mode = "grid"

Other settings

Cell text/nodes/board/props/cellText = "pair"

Accepted within

± 0.01

Computed from the Scene at this step’s setting

Grid: cells; tree: leaves (paths)/nodes/board/outputs/outcomes

Wrong answers it recognises

• 12 ± 0.01

  Misconception: Adds the options at each stage instead of multiplying them: 3 shirts and 2 trousers make 5 outfits, two dice make 12 outcomes.

  Feedback from the catalogue

  Every option at the first stage meets every option at the second, so you multiply: each of the $3$⁠ shirts goes with $2$⁠ trousers, which is $3 \times 2 = 6$⁠ outfits. The tree shows it: every branch splits again. Adding only counts the branches of one stage after another, not the paths from the root to the end.

• 21 ± 0.01

  Misconception: Counts 3 then 4 and 4 then 3 as one outcome of two dice, giving 21 outcomes instead of 36.

  Feedback from the catalogue

  The two dice are different, even when they look alike: paint one of them red. $3$⁠ on the red one and $4$⁠ on the blue one is a different outcome from $4$⁠ on the red and $3$⁠ on the blue, and the table gives them two separate squares. Only then are all $36$⁠ outcomes equally likely.

Hint

Worked solution

The Scene at this step

Hats

0defaulthidden

Tosses

2

Books

3defaulthidden

The start / All counted

All counted

Other settings

Mode/nodes/board/props/mode = "tree"

Other settings

Tree/nodes/board/props/tree = "coins"

Other settings

Heads/nodes/board/props/heads = -1

Read from the Scene at

Grid: cells; tree: leaves (paths)/nodes/board/outputs/outcomes

One setting that reaches it

Tosses

4

Hint

Worked solution

The Scene at this step

Hats

0defaulthidden

Tosses

2defaulthidden

Books

3fixed

The start / All counted

All countedfixed

Other settings

Mode/nodes/board/props/mode = "tree"

Other settings

Tree/nodes/board/props/tree = "arrange"

Choices

• Correct $24$⁠

• Wrong $16$⁠

  Misconception: Counts orders of distinct items as if an item could be used again: 4 books in 4 × 4 × 4 × 4 orders.

  Feedback from the catalogue

  A book already on the shelf cannot appear again. The first place has $4$⁠ books to choose from, the second only $3$⁠ left, the third $2$⁠ and the last one book: $4 \times 3 \times 2 \times 1 = 24$$4 \times 3 \times 2 \times 1 =$$24$⁠. The tree has one branch fewer at each stage.

• Wrong $256$⁠

  Misconception: Counts orders of distinct items as if an item could be used again: 4 books in 4 × 4 × 4 × 4 orders.

  Feedback from the catalogue

  A book already on the shelf cannot appear again. The first place has $4$⁠ books to choose from, the second only $3$⁠ left, the third $2$⁠ and the last one book: $4 \times 3 \times 2 \times 1 = 24$$4 \times 3 \times 2 \times 1 =$$24$⁠. The tree has one branch fewer at each stage.

• Wrong $10$⁠

  Misconception: Adds the options at each stage instead of multiplying them: 3 shirts and 2 trousers make 5 outfits, two dice make 12 outcomes.

  Feedback from the catalogue

  Every option at the first stage meets every option at the second, so you multiply: each of the $3$⁠ shirts goes with $2$⁠ trousers, which is $3 \times 2 = 6$⁠ outfits. The tree shows it: every branch splits again. Adding only counts the branches of one stage after another, not the paths from the root to the end.

Once the prediction is locked in, the Scene changes

Books

4fixed

The start / All counted

All countedfixed

The Scene then shows: 24 · Grid: cells; tree: leaves (paths)/nodes/board/outputs/outcomes

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Total

5hidden

All outcomes / The event counted

All outcomesfixed

Other settings

Second/nodes/board/props/second = "none"

Other settings

Event/nodes/board/props/event = "sumAtLeast"

Other settings

Cell text/nodes/board/props/cellText = "pair"

Choices

• Correct $\frac{2}{6} = \frac{1}{3}$$\frac{2}{6} =$$\frac{1}{3}$⁠

• Wrong $\frac{2}{4} = \frac{1}{2}$$\frac{2}{4} =$$\frac{1}{2}$⁠

  Misconception: Divides the outcomes that give the event by the outcomes that do not (2/4), instead of by all the outcomes (2/6).

  Feedback from the catalogue

  A probability is a part of all the outcomes: the outcomes that fit on top, all the outcomes underneath. On a die $2$⁠ faces out of $6$⁠ give $5$⁠ or more, so the chance is $\frac{2}{6} = \frac{1}{3}$$\frac{2}{6} =$$\frac{1}{3}$⁠. The ratio $2$⁠ to $4$⁠ compares what fits with what does not, which is something else.

• Wrong $\frac{1}{2}$⁠, because it either happens or it doesn’t

  Misconception: Thinks any event has chance 1/2 because it either happens or it does not.

  Feedback from the catalogue

  “It will happen or it won’t” is two possibilities, but not equally likely ones. What counts is how many of the equally likely outcomes fit: on a die only $2$⁠ of the $6$⁠ faces are $5$⁠ or more, so the chance is $\frac{1}{3}$⁠, not a half.

• Wrong $\frac{5}{6}$⁠

  Feedback

  $\frac{5}{6}$⁠ is the chance of a $2$⁠ or more: five faces. Here only the faces $5$⁠ and $6$⁠ fit.

Once the prediction is locked in, the Scene changes

All outcomes / The event counted

The event countedfixed

The Scene then shows: 2 · Grid: cells in either event; tree (coins): paths with exactly `heads` heads/nodes/board/outputs/favourable

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Total

2fixed

All outcomes / The event counted

The event countedfixed

Choices

• Correct A total of $7$⁠

• Wrong Both are equally likely, because each total is one outcome

  Misconception: Treats every total of two dice (or every count of heads, or every bin) as equally likely: a total of 2 as likely as 7, 1/11 each.

  Feedback from the catalogue

  The totals are not the equally likely outcomes: the $36$⁠ squares are, and each total has a different number of squares. A total of $7$⁠ has six ways ($1+6, 2+5, \dots, 6+1$⁠), a total of $2$⁠ only one ($1+1$⁠). In the same way, the number of heads, or the bin a ball lands in, are not equally likely: many more paths lead to the middle than to the edges.

• Wrong A total of $2$⁠

  Feedback

  A total of $2$⁠ has only one way, $1+1$⁠: one square out of $36$⁠. Try finding how many squares in the table show $7$⁠.

Once the prediction is locked in, the Scene changes

Total

7fixed

All outcomes / The event counted

The event countedfixed

The Scene then shows: 6 · Grid: cells in either event; tree (coins): paths with exactly `heads` heads/nodes/board/outputs/favourable

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Total

8fixed

All outcomes / The event counted

All outcomesfixed

Accepted within

± 0.5

Computed from the Scene at this step’s setting

Favourable ÷ outcomes, 0 to 1/nodes/board/outputs/probability

Wrong answers it recognises

• 9.09 ± 0.5

  Misconception: Treats every total of two dice (or every count of heads, or every bin) as equally likely: a total of 2 as likely as 7, 1/11 each.

  Feedback from the catalogue

  The totals are not the equally likely outcomes: the $36$⁠ squares are, and each total has a different number of squares. A total of $7$⁠ has six ways ($1+6, 2+5, \dots, 6+1$⁠), a total of $2$⁠ only one ($1+1$⁠). In the same way, the number of heads, or the bin a ball lands in, are not equally likely: many more paths lead to the middle than to the edges.

Hint

Worked solution

The Scene at this step

Total

7

All outcomes / The event counted

The event counted

Read from the Scene at

Grid: cells in either event; tree (coins): paths with exactly `heads` heads/nodes/board/outputs/favourable

One setting that reaches it

Total

4

Hint

Worked solution

The Scene at this step

Total

7fixed

All outcomes / The event counted

The event counted

Accepted within

± 0.5

Computed from the Scene at this step’s setting

1 − probability/nodes/board/outputs/complement

Hint

Worked solution

The Scene at this step

First total

6hidden

Second total

2defaulthidden

All outcomes / The events counted

The events counted

Other settings

Event/nodes/board/props/event = "noneIs"

Other settings

Event2/nodes/board/props/event2 = "none"

Other settings

Cell text/nodes/board/props/cellText = "pair"

Accepted within

± 0.01

Computed from the Scene at this step’s setting

Outcomes − favourable/nodes/board/outputs/unfavourable

Wrong answers it recognises

• 12 ± 0.01

  Misconception: Adds the chances of two events that can happen together, counting the outcomes in both twice: 6/36 + 6/36 for a six on either die.

  Feedback from the catalogue

  Adding works only for events that cannot happen together. A six on the first die and a six on the second can happen together, in the square $(6,6)$⁠, and adding $6 + 6$⁠ counts it twice. So take it away once: $6 + 6 - 1 = 11$⁠ squares, which is $\frac{11}{36}$⁠.

Hint

Worked solution

The Scene at this step

First total

6hidden

Second total

6hidden

All outcomes / The events counted

All outcomesfixed

Other settings

Event/nodes/board/props/event = "firstIs"

Other settings

Event2/nodes/board/props/event2 = "secondIs"

Other settings

Cell text/nodes/board/props/cellText = "pair"

Choices

• Correct $\frac{11}{36}$⁠

• Wrong $\frac{12}{36} = \frac{1}{3}$$\frac{12}{36} =$$\frac{1}{3}$⁠

  Misconception: Adds the chances of two events that can happen together, counting the outcomes in both twice: 6/36 + 6/36 for a six on either die.

  Feedback from the catalogue

  Adding works only for events that cannot happen together. A six on the first die and a six on the second can happen together, in the square $(6,6)$⁠, and adding $6 + 6$⁠ counts it twice. So take it away once: $6 + 6 - 1 = 11$⁠ squares, which is $\frac{11}{36}$⁠.

• Wrong $\frac{1}{36}$⁠

  Feedback

  $\frac{1}{36}$⁠ is the chance that both dice show a six, that is “and”. “Or” also takes in the squares where only one of them is a six.

• Wrong $\frac{6}{36}$⁠

  Feedback

  That is only one of the events. “Or” takes in the squares of both.

Once the prediction is locked in, the Scene changes

All outcomes / The events counted

The events countedfixed

The Scene then shows: 11 · Grid: cells in either event; tree (coins): paths with exactly `heads` heads/nodes/board/outputs/favourable

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

First total

7fixed

Second total

2

All outcomes / The events counted

The events counted

Other settings

Event/nodes/board/props/event = "sum"

Other settings

Event2/nodes/board/props/event2 = "sum"

Other settings

Cell text/nodes/board/props/cellText = "sum"

Read from the Scene at

Grid: cells in either event; tree (coins): paths with exactly `heads` heads/nodes/board/outputs/favourable

One setting that reaches it

Second total

4

Hint

Worked solution

The Scene at this step

First total

7defaulthidden

Second total

2defaulthidden

All outcomes / The events counted

All outcomesfixed

Other settings

Mode/nodes/board/props/mode = "tree"

Other settings

Tree/nodes/board/props/tree = "coins"

Other settings

Stages/nodes/board/props/stages = 3

Other settings

Heads/nodes/board/props/heads = 3

Choices

• Correct $\frac{1}{8}$⁠

• Wrong $\frac{1}{2}$⁠, because each toss is a half

  Misconception: Gives a run of independent events the chance of one of them: three heads in a row as 1/2 because each toss is 1/2.

  Feedback from the catalogue

  Each toss alone is a half, but three tosses and means all of them must succeed. Half the time the first is heads, of those half the time the second is too, and of those half the time the third: $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} =$$\frac{1}{8}$⁠. On the tree it is one path out of $8$⁠.

• Wrong $\frac{1}{4}$⁠, because you can get $0, 1, 2$⁠ or $3$⁠ heads

  Misconception: Treats every total of two dice (or every count of heads, or every bin) as equally likely: a total of 2 as likely as 7, 1/11 each.

  Feedback from the catalogue

  The totals are not the equally likely outcomes: the $36$⁠ squares are, and each total has a different number of squares. A total of $7$⁠ has six ways ($1+6, 2+5, \dots, 6+1$⁠), a total of $2$⁠ only one ($1+1$⁠). In the same way, the number of heads, or the bin a ball lands in, are not equally likely: many more paths lead to the middle than to the edges.

• Wrong $\frac{3}{8}$⁠

  Feedback

  $\frac{3}{8}$⁠ is the chance of exactly two heads: three paths. Only one path leads to three heads.

Once the prediction is locked in, the Scene changes

All outcomes / The events counted

The events countedfixed

The Scene then shows: 1 · Grid: cells in either event; tree (coins): paths with exactly `heads` heads/nodes/board/outputs/favourable

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

First total

7defaulthidden

Second total

2defaulthidden

All outcomes / The events counted

The events counted

Other settings

Mode/nodes/board/props/mode = "tree"

Other settings

Tree/nodes/board/props/tree = "coins"

Other settings

Stages/nodes/board/props/stages = 4

Other settings

Heads/nodes/board/props/heads = 4

Choices

Shown in a random order.

• Correct $\frac{1}{2}$⁠

• Wrong Less than a half, because tails is due

  Misconception: Thinks that after a run of heads, tails is due: the next toss of a fair coin is less likely to be heads.

  Feedback from the catalogue

  A coin has no memory. The earlier tosses have already happened, and the next toss is heads or tails, half and half, once again. A run of four heads is indeed rare ($\frac{1}{16}$⁠), but that is the chance of the whole run before it starts, not of the toss after it. This is the “gambler’s fallacy”.

• Wrong $\frac{1}{32}$⁠

  Misconception: Gives the next toss the chance of the whole run: after four heads, a fifth head as 1/32.

  Feedback from the catalogue

  $\frac{1}{32}$⁠ is the chance of five heads in a row before you start tossing. Here the first four are already in, and the only question left is the one next toss: heads or tails, half and half.

• Wrong More than a half, because the coin is on a run of heads

  Feedback

  A “hot streak” also assumes the coin has a memory. A fair coin has none: every toss is half and half.

Hint

Worked solution

The Scene at this step

Rows of pegs

4fixed

Chance of bouncing right

50 %fixed

Balls / Bell curve

Balls

Accepted within

± 0.01

Computed from the Scene at this step’s setting

Left-or-right routes through the pegs, 2^rows/nodes/board/outputs/paths

Wrong answers it recognises

• 8 ± 0.01

  Misconception: Multiplies 2 by the number of stages instead of multiplying 2 by itself: 4 rows of pegs give 2 × 4 = 8 paths.

  Feedback from the catalogue

  At every row each path splits in two, so the number of paths doubles with every row: $2, 4, 8, 16$⁠. That is $2 \times 2 \times 2 \times 2 = 2^4 = 16$$2 \times 2 \times 2 \times 2 =$$2^4 =$$16$⁠, not $2 \times 4$⁠. The chance of one particular path goes the same way: $\left(\frac{1}{2}\right)^{n}$⁠, which shrinks very fast.

• 5 ± 0.01

  Misconception: Counts the bins at the bottom of a Galton board as its paths: 5 paths for 4 rows.

  Feedback from the catalogue

  There are only $5$⁠ bins, but several different paths lead to each bin. Six ways lead to the middle one: right twice and left twice, in any order. Count every way from the top to the bottom: $2 \times 2 \times 2 \times 2 = 16$$2 \times 2 \times 2 \times 2 =$$16$⁠.

Hint

Worked solution

The Scene at this step

Rows of pegs

4fixed

Chance of bouncing right

50 %fixed

Balls / Bell curve

Bell curve

Other settings

Probe/nodes/board/props/probe = 2

Accepted within

± 0.5

Computed from the Scene at this step’s setting

The probe bin's chance, 0 to 1/nodes/board/outputs/binChance

Wrong answers it recognises

• 20 ± 0.5

  Misconception: Treats every total of two dice (or every count of heads, or every bin) as equally likely: a total of 2 as likely as 7, 1/11 each.

  Feedback from the catalogue

  The totals are not the equally likely outcomes: the $36$⁠ squares are, and each total has a different number of squares. A total of $7$⁠ has six ways ($1+6, 2+5, \dots, 6+1$⁠), a total of $2$⁠ only one ($1+1$⁠). In the same way, the number of heads, or the bin a ball lands in, are not equally likely: many more paths lead to the middle than to the edges.

• 50 ± 0.5

  Misconception: Expects a fair process to land on its average almost surely: exactly 5 heads in 10 tosses most of the time.

  Feedback from the catalogue

  Five heads out of ten is the most likely result, but not the result you get almost every time: exactly $5$⁠ comes up only about a quarter of the time. Most of the time you get a little more or a little less, $4$⁠ or $6$⁠ and sometimes $3$⁠ or $7$⁠. Over many tosses the proportion gets close to a half, but the exact count spreads around it in the shape of a bell.

Hint

Worked solution

The Scene at this step

Rows of pegs

12fixed

Chance of bouncing right

50 %fixed

Balls / Bell curve

Ballsfixed

Other settings

Probe/nodes/board/props/probe = 12

Choices

• Correct $\frac{1}{4096}$⁠, less than one ball in $4000$⁠

• Wrong $\frac{1}{13}$⁠, because there are $13$⁠ bins

  Misconception: Treats every total of two dice (or every count of heads, or every bin) as equally likely: a total of 2 as likely as 7, 1/11 each.

  Feedback from the catalogue

  The totals are not the equally likely outcomes: the $36$⁠ squares are, and each total has a different number of squares. A total of $7$⁠ has six ways ($1+6, 2+5, \dots, 6+1$⁠), a total of $2$⁠ only one ($1+1$⁠). In the same way, the number of heads, or the bin a ball lands in, are not equally likely: many more paths lead to the middle than to the edges.

• Wrong $\frac{1}{24}$⁠

  Misconception: Multiplies 2 by the number of stages instead of multiplying 2 by itself: 4 rows of pegs give 2 × 4 = 8 paths.

  Feedback from the catalogue

  At every row each path splits in two, so the number of paths doubles with every row: $2, 4, 8, 16$⁠. That is $2 \times 2 \times 2 \times 2 = 2^4 = 16$$2 \times 2 \times 2 \times 2 =$$2^4 =$$16$⁠, not $2 \times 4$⁠. The chance of one particular path goes the same way: $\left(\frac{1}{2}\right)^{n}$⁠, which shrinks very fast.

• Wrong $\frac{1}{2}$⁠, because at every peg the chance is a half

  Misconception: Gives a run of independent events the chance of one of them: three heads in a row as 1/2 because each toss is 1/2.

  Feedback from the catalogue

  Each toss alone is a half, but three tosses and means all of them must succeed. Half the time the first is heads, of those half the time the second is too, and of those half the time the third: $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} =$$\frac{1}{8}$⁠. On the tree it is one path out of $8$⁠.

Once the prediction is locked in, the Scene changes

Balls / Bell curve

Bell curvefixed

A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.

Hint

Worked solution

The Scene at this step

Rows of pegs

12fixed

Chance of bouncing right

50 %

Balls / Bell curve

Bell curve

Read from the Scene at

The expected bin (count of rights), rows × chance right/nodes/board/outputs/mean

One setting that reaches it

Chance of bouncing right

75 %

Hint

Worked solution

The Scene at this step

Rows of pegs

10fixed

Chance of bouncing right

50 %fixed

Balls / Bell curve

Bell curve

Other settings

Probe/nodes/board/props/probe = 5

Accepted within

± 0.5

Computed from the Scene at this step’s setting

The probe bin's chance, 0 to 1/nodes/board/outputs/binChance

Wrong answers it recognises

• 50 ± 1

  Misconception: Expects a fair process to land on its average almost surely: exactly 5 heads in 10 tosses most of the time.

  Feedback from the catalogue

  Five heads out of ten is the most likely result, but not the result you get almost every time: exactly $5$⁠ comes up only about a quarter of the time. Most of the time you get a little more or a little less, $4$⁠ or $6$⁠ and sometimes $3$⁠ or $7$⁠. Over many tosses the proportion gets close to a half, but the exact count spreads around it in the shape of a bell.

• 100 ± 1

  Misconception: Expects a fair process to land on its average almost surely: exactly 5 heads in 10 tosses most of the time.

  Feedback from the catalogue

  Five heads out of ten is the most likely result, but not the result you get almost every time: exactly $5$⁠ comes up only about a quarter of the time. Most of the time you get a little more or a little less, $4$⁠ or $6$⁠ and sometimes $3$⁠ or $7$⁠. Over many tosses the proportion gets close to a half, but the exact count spreads around it in the shape of a bell.

• 9.09 ± 0.5

  Misconception: Treats every total of two dice (or every count of heads, or every bin) as equally likely: a total of 2 as likely as 7, 1/11 each.

  Feedback from the catalogue

  The totals are not the equally likely outcomes: the $36$⁠ squares are, and each total has a different number of squares. A total of $7$⁠ has six ways ($1+6, 2+5, \dots, 6+1$⁠), a total of $2$⁠ only one ($1+1$⁠). In the same way, the number of heads, or the bin a ball lands in, are not equally likely: many more paths lead to the middle than to the edges.

Hint

Worked solution