Conditional probability.

The Scene at this step

Total

10fixed

First die

6fixed

All outcomes / The events counted

The events counted

Other settings

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

Other settings

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

Accepted within

± 0.5

Computed from the Scene at this step’s setting

Grid: P(A | B), cells in both ÷ cells in the second event (NaN when it has none)/nodes/board/outputs/aGivenB

Wrong answers it recognises

• 8.333 ± 0.5

  Misconception: Confuses the chance of A and B with the chance of A given B: divides by all the outcomes instead of by the outcomes in B, or the reverse.

  Feedback from the catalogue

  “$A$⁠ and $B$⁠” and “$A$⁠ given $B$⁠” count the same outcomes, those in both events, but divide by different things. For “and” you divide by all the outcomes. For “given $B$⁠” you already know $B$⁠ happened, so you divide only by the outcomes in $B$⁠: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$P(A \mid B) =$$\frac{P(A \cap B)}{P(B)}$⁠. The condition shrinks the denominator, so the conditional chance is the larger one.

• 16.67 ± 0.5

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback from the catalogue

  The condition is new information, and you have to use it. “Given $B$⁠” says: cross out every outcome not in $B$⁠, and count again inside what is left. Sometimes the chance comes out the same, and then the events are independent, but you find that out only after counting.

Hint

Worked solution

The Scene at this step

Total

11fixed

First die

6fixed

All outcomes / The events counted

The events counted

Other settings

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

Other settings

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

Accepted within

± 0.5

Computed from the Scene at this step’s setting

Grid: P(B | A), cells in both ÷ cells in the first event (NaN when it has none)/nodes/board/outputs/bGivenA

Wrong answers it recognises

• 33.33 ± 0.5

  Misconception: Treats the chance of A given B as the chance of B given A.

  Feedback from the catalogue

  $P(A \mid B)$⁠ and $P(B \mid A)$⁠ share a numerator, the outcomes in both events, but their denominators differ: once all the outcomes in $B$⁠, once all the outcomes in $A$⁠. They are equal only when $A$⁠ and $B$⁠ are the same size. “Most ill people test positive” is not “most people who test positive are ill”, and mixing them up is called the prosecutor’s fallacy.

• 5.556 ± 0.5

  Misconception: Confuses the chance of A and B with the chance of A given B: divides by all the outcomes instead of by the outcomes in B, or the reverse.

  Feedback from the catalogue

  “$A$⁠ and $B$⁠” and “$A$⁠ given $B$⁠” count the same outcomes, those in both events, but divide by different things. For “and” you divide by all the outcomes. For “given $B$⁠” you already know $B$⁠ happened, so you divide only by the outcomes in $B$⁠: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$P(A \mid B) =$$\frac{P(A \cap B)}{P(B)}$⁠. The condition shrinks the denominator, so the conditional chance is the larger one.

• 16.67 ± 0.5

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback from the catalogue

  The condition is new information, and you have to use it. “Given $B$⁠” says: cross out every outcome not in $B$⁠, and count again inside what is left. Sometimes the chance comes out the same, and then the events are independent, but you find that out only after counting.

Hint

Worked solution

The Scene at this step

Total

7fixed

First die

6fixed

All outcomes / The events counted

All outcomesfixed

Other settings

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

Other settings

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

Choices

• Correct Still $\frac{1}{6}$⁠

• Wrong More than $\frac{1}{6}$⁠, since $6$⁠ is most of the way to $7$⁠

  Misconception: Assumes that learning one event happened must change the chance of the other.

  Feedback from the catalogue

  Sometimes the information does not help. When the first die shows $6$⁠, the second still has six possibilities, and exactly one of them, $1$⁠, makes a total of $7$⁠: the chance stays $\frac{1}{6}$⁠, just as without the information. Events like these are called independent: $P(A \mid B) = P(A)$⁠, or equivalently $P(A \cap B) = P(A) \cdot P(B)$$P(A \cap B) =$$P(A) \cdot P(B)$⁠.

• Wrong Less than $\frac{1}{6}$⁠, since only one way is left

  Misconception: Assumes that learning one event happened must change the chance of the other.

  Feedback from the catalogue

  Sometimes the information does not help. When the first die shows $6$⁠, the second still has six possibilities, and exactly one of them, $1$⁠, makes a total of $7$⁠: the chance stays $\frac{1}{6}$⁠, just as without the information. Events like these are called independent: $P(A \mid B) = P(A)$⁠, or equivalently $P(A \cap B) = P(A) \cdot P(B)$$P(A \cap B) =$$P(A) \cdot P(B)$⁠.

Once the prediction is locked in, the Scene changes

All outcomes / The events counted

The events countedfixed

The Scene then shows: 0.1667 · Grid: P(A | B), cells in both ÷ cells in the second event (NaN when it has none)/nodes/board/outputs/aGivenB

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

8

First die

6fixed

All outcomes / The events counted

The events counted

Other settings

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

Other settings

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

Read from the Scene at

Grid: 1 when P(A and B) = P(A) × P(B), else 0 (events that are certain or impossible count as independent)/nodes/board/outputs/independent

One setting that reaches it

Total

7

Hint

Worked solution

The Scene at this step

Total

2fixed

First die

6fixed

All outcomes / The events counted

The events counted

Other settings

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

Other settings

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

Choices

Shown in a random order.

• Correct No: if one happened, the other certainly did not

• Wrong Yes: they have nothing to do with each other

  Misconception: Thinks events that cannot happen together are independent.

  Feedback from the catalogue

  Mutually exclusive events, which cannot happen together, are the very opposite of independent: if one happened, you know for certain the other did not. Independent means knowing one does not change the chance of the other, so they must be able to happen together: $P(A \cap B) = P(A) \cdot P(B) > 0$$P(A \cap B) =$$P(A) \cdot P(B) >$$0$⁠.

• Wrong You cannot tell without rolling

  Feedback

  You can tell from counting: $P(A \cap B) = 0$⁠, but $P(A) \cdot P(B) = \frac{1}{36} \cdot \frac{1}{6} > 0$$P(A) \cdot P(B) =$$\frac{1}{36} \cdot \frac{1}{6} >$$0$⁠.

Hint

Worked solution

The Scene at this step

How common the disease is

10 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Who is ill

Accepted within

± 0.5

Computed from the Scene at this step’s setting

Positives who are ill/nodes/crowd/outputs/truePositives

Wrong answers it recognises

• 900 ± 0.5

  Misconception: Applies a branch's rate to the wrong group: the test's accuracy to the whole crowd, or the ill people's rate to the healthy.

  Feedback from the catalogue

  On a tree, every rate belongs to its own branch. $90\%$⁠ of the ill test positive: that is $90\%$⁠ of the ill, not of everyone. And $10\%$⁠ of the healthy test positive: that is $10\%$⁠ of the healthy. First split the population by the first branch, and only then apply each group’s own rate.

• 100 ± 0.5

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback

  $100$⁠ is all the ill. Not all of them test positive: the test misses $10\%$⁠ of them.

Hint

Worked solution

The Scene at this step

How common the disease is

10 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Who is ill

Accepted within

± 0.5

Computed from the Scene at this step’s setting

Positives who are healthy/nodes/crowd/outputs/falsePositives

Wrong answers it recognises

• 10 ± 0.5

  Misconception: Applies a branch's rate to the wrong group: the test's accuracy to the whole crowd, or the ill people's rate to the healthy.

  Feedback from the catalogue

  On a tree, every rate belongs to its own branch. $90\%$⁠ of the ill test positive: that is $90\%$⁠ of the ill, not of everyone. And $10\%$⁠ of the healthy test positive: that is $10\%$⁠ of the healthy. First split the population by the first branch, and only then apply each group’s own rate.

• 100 ± 0.5

  Misconception: Applies a branch's rate to the wrong group: the test's accuracy to the whole crowd, or the ill people's rate to the healthy.

  Feedback from the catalogue

  On a tree, every rate belongs to its own branch. $90\%$⁠ of the ill test positive: that is $90\%$⁠ of the ill, not of everyone. And $10\%$⁠ of the healthy test positive: that is $10\%$⁠ of the healthy. First split the population by the first branch, and only then apply each group’s own rate.

Hint

Worked solution

The Scene at this step

How common the disease is

10 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Positives sorted

Accepted within

± 0.5

Computed from the Scene at this step’s setting

People who test positive/nodes/crowd/outputs/positives

Wrong answers it recognises

• 9 ± 0.5

  Misconception: Counts only the true positives as the positives, forgetting the healthy people the test flags.

  Feedback from the catalogue

  A positive result comes from two branches: ill people the test caught, and healthy people it got wrong. When the disease is rare there are many more healthy people, so even a small error rate gives many false positives. Count all the positives from both branches together.

• 90 ± 0.5

  Misconception: Applies a branch's rate to the wrong group: the test's accuracy to the whole crowd, or the ill people's rate to the healthy.

  Feedback from the catalogue

  On a tree, every rate belongs to its own branch. $90\%$⁠ of the ill test positive: that is $90\%$⁠ of the ill, not of everyone. And $10\%$⁠ of the healthy test positive: that is $10\%$⁠ of the healthy. First split the population by the first branch, and only then apply each group’s own rate.

Hint

Worked solution

The Scene at this step

How common the disease is

10 %

Test accuracy

90 %

Retest the positives

off

Sort the results

Positives sorted

Accepted within

± 0.5

A fixed value in the question

Wrong answers it recognises

• 16 ± 0.5

  Misconception: Uses the same chance for the second draw as for the first, as if what was drawn were put back.

  Feedback from the catalogue

  Without putting back, the second stage depends on the first. After you draw one red marble out of $5$⁠, there are $4$⁠ left and only one red. So the second branch is $\frac{1}{4}$⁠, not $\frac{2}{5}$⁠: $P(R_1 \cap R_2) = \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$$P(R_1 \cap R_2) =$$\frac{2}{5} \cdot \frac{1}{4} =$$\frac{1}{10}$⁠. A branch of a tree always carries the chance given everything that happened before it.

• 40 ± 0.5

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback

  $40\%$⁠ is the chance the first marble is red. Both are red only if the second is red too.

Hint

Worked solution

The Scene at this step

How common the disease is

5 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Who is ill

Accepted within

± 0.3

Computed from the Scene at this step’s setting

Positives who are ill/nodes/crowd/outputs/truePositives

Wrong answers it recognises

• 90 ± 0.3

  Misconception: Confuses the chance of A and B with the chance of A given B: divides by all the outcomes instead of by the outcomes in B, or the reverse.

  Feedback from the catalogue

  “$A$⁠ and $B$⁠” and “$A$⁠ given $B$⁠” count the same outcomes, those in both events, but divide by different things. For “and” you divide by all the outcomes. For “given $B$⁠” you already know $B$⁠ happened, so you divide only by the outcomes in $B$⁠: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$P(A \mid B) =$$\frac{P(A \cap B)}{P(B)}$⁠. The condition shrinks the denominator, so the conditional chance is the larger one.

• 5 ± 0.3

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback

  $5\%$⁠ is all the ill. Only $90\%$⁠ of them also test positive.

Hint

Worked solution

The Scene at this step

How common the disease is

1 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Test resultsfixed

Choices

• Correct About $8\%$⁠

• Wrong $90\%$⁠, the same as the accuracy

  Misconception: Takes a test's accuracy as the chance that a positive result is right: 90% accurate, so a positive means 90% ill.

  Feedback from the catalogue

  Write $D$⁠ for “ill” and $+$⁠ for “positive”. Accuracy says what the test does to the ill and to the healthy: $P(+ \mid D)$⁠. The question of someone who tested positive is the reverse: $P(D \mid +)$⁠. The answer also depends on how common the disease is. When it is rare, most positives come from the large group of the healthy, so the chance of being ill is far below the accuracy. This is base-rate neglect.

• Wrong About $50\%$⁠

  No feedback of its own; the learner sees the standard message

  Not yet. Look again and try once more.

• Wrong $1\%$⁠, like anyone else

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback from the catalogue

  The condition is new information, and you have to use it. “Given $B$⁠” says: cross out every outcome not in $B$⁠, and count again inside what is left. Sometimes the chance comes out the same, and then the events are independent, but you find that out only after counting.

Once the prediction is locked in, the Scene changes

Sort the results

Positives sortedfixed

The Scene then shows: 8.333 · Per cent chance a positive is ill, by Bayes' rule/nodes/crowd/outputs/posterior

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

How common the disease is

1 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Positives sorted

Accepted within

± 0.5

Computed from the Scene at this step’s setting

Positives who are healthy/nodes/crowd/outputs/falsePositives

Wrong answers it recognises

• 1 ± 0.5

  Misconception: Applies a branch's rate to the wrong group: the test's accuracy to the whole crowd, or the ill people's rate to the healthy.

  Feedback from the catalogue

  On a tree, every rate belongs to its own branch. $90\%$⁠ of the ill test positive: that is $90\%$⁠ of the ill, not of everyone. And $10\%$⁠ of the healthy test positive: that is $10\%$⁠ of the healthy. First split the population by the first branch, and only then apply each group’s own rate.

• 10 ± 0.5

  Misconception: Applies a branch's rate to the wrong group: the test's accuracy to the whole crowd, or the ill people's rate to the healthy.

  Feedback from the catalogue

  On a tree, every rate belongs to its own branch. $90\%$⁠ of the ill test positive: that is $90\%$⁠ of the ill, not of everyone. And $10\%$⁠ of the healthy test positive: that is $10\%$⁠ of the healthy. First split the population by the first branch, and only then apply each group’s own rate.

Hint

Worked solution

The Scene at this step

How common the disease is

1 %

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Positives sorted

Read from the Scene at

Per cent chance a positive is ill, by Bayes' rule/nodes/crowd/outputs/posterior

One setting that reaches it

How common the disease is

9.7 %

Hint

Worked solution

The Scene at this step

How common the disease is

1 %fixed

Test accuracy

90 %fixed

Retest the positives

offfixed

Sort the results

Test resultsfixed

Choices

• Correct About $50\%$⁠

• Wrong $99\%$⁠

  Misconception: Takes a test's accuracy as the chance that a positive result is right: 90% accurate, so a positive means 90% ill.

  Feedback from the catalogue

  Write $D$⁠ for “ill” and $+$⁠ for “positive”. Accuracy says what the test does to the ill and to the healthy: $P(+ \mid D)$⁠. The question of someone who tested positive is the reverse: $P(D \mid +)$⁠. The answer also depends on how common the disease is. When it is rare, most positives come from the large group of the healthy, so the chance of being ill is far below the accuracy. This is base-rate neglect.

• Wrong Still about $9\%$⁠: the test hardly matters

  Feedback

  The test matters a great deal: its error rate falls tenfold, from $10\%$⁠ to $1\%$⁠, and the false positives with it.

• Wrong About $80\%$⁠

  No feedback of its own; the learner sees the standard message

  Not yet. Look again and try once more.

Once the prediction is locked in, the Scene changes

Test accuracy

99 %fixed

Sort the results

Positives sortedfixed

The Scene then shows: 50 · Per cent chance a positive is ill, by Bayes' rule/nodes/crowd/outputs/posterior

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

How common the disease is

0.1 %fixed

Test accuracy

99 %fixed

Retest the positives

offfixed

Sort the results

Test results

Accepted within

± 0.3

Computed from the Scene at this step’s setting

Per cent chance a positive is ill, by Bayes' rule/nodes/crowd/outputs/posterior

Wrong answers it recognises

• 99 ± 0.3

  Misconception: Takes a test's accuracy as the chance that a positive result is right: 90% accurate, so a positive means 90% ill.

  Feedback from the catalogue

  Write $D$⁠ for “ill” and $+$⁠ for “positive”. Accuracy says what the test does to the ill and to the healthy: $P(+ \mid D)$⁠. The question of someone who tested positive is the reverse: $P(D \mid +)$⁠. The answer also depends on how common the disease is. When it is rare, most positives come from the large group of the healthy, so the chance of being ill is far below the accuracy. This is base-rate neglect.

• 0.1 ± 0.3

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback from the catalogue

  The condition is new information, and you have to use it. “Given $B$⁠” says: cross out every outcome not in $B$⁠, and count again inside what is left. Sometimes the chance comes out the same, and then the events are independent, but you find that out only after counting.

Hint

Worked solution

The Scene at this step

How common the disease is

2 %fixed

Test accuracy

95 %fixed

Retest the positives

offfixed

Sort the results

Test results

Accepted within

± 0.3

Computed from the Scene at this step’s setting

Per cent chance a positive is ill, by Bayes' rule/nodes/crowd/outputs/posterior

Wrong answers it recognises

• 95 ± 0.3

  Misconception: Takes a test's accuracy as the chance that a positive result is right: 90% accurate, so a positive means 90% ill.

  Feedback from the catalogue

  Write $D$⁠ for “ill” and $+$⁠ for “positive”. Accuracy says what the test does to the ill and to the healthy: $P(+ \mid D)$⁠. The question of someone who tested positive is the reverse: $P(D \mid +)$⁠. The answer also depends on how common the disease is. When it is rare, most positives come from the large group of the healthy, so the chance of being ill is far below the accuracy. This is base-rate neglect.

• 1.9 ± 0.3

  Misconception: Confuses the chance of A and B with the chance of A given B: divides by all the outcomes instead of by the outcomes in B, or the reverse.

  Feedback from the catalogue

  “$A$⁠ and $B$⁠” and “$A$⁠ given $B$⁠” count the same outcomes, those in both events, but divide by different things. For “and” you divide by all the outcomes. For “given $B$⁠” you already know $B$⁠ happened, so you divide only by the outcomes in $B$⁠: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$P(A \mid B) =$$\frac{P(A \cap B)}{P(B)}$⁠. The condition shrinks the denominator, so the conditional chance is the larger one.

• 2 ± 0.3

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback from the catalogue

  The condition is new information, and you have to use it. “Given $B$⁠” says: cross out every outcome not in $B$⁠, and count again inside what is left. Sometimes the chance comes out the same, and then the events are independent, but you find that out only after counting.

Hint

Worked solution

The Scene at this step

How common the disease is

2 %

Test accuracy

95 %

Retest the positives

off

Sort the results

Positives sorted

Choices

Shown in a random order.

• Correct $\dfrac{P(+ \mid D) \cdot P(D)}{P(+)}$⁠

• Wrong $P(+ \mid D)$⁠

  Misconception: Treats the chance of A given B as the chance of B given A.

  Feedback from the catalogue

  $P(A \mid B)$⁠ and $P(B \mid A)$⁠ share a numerator, the outcomes in both events, but their denominators differ: once all the outcomes in $B$⁠, once all the outcomes in $A$⁠. They are equal only when $A$⁠ and $B$⁠ are the same size. “Most ill people test positive” is not “most people who test positive are ill”, and mixing them up is called the prosecutor’s fallacy.

• Wrong $P(+ \mid D) \cdot P(D)$⁠

  Misconception: Confuses the chance of A and B with the chance of A given B: divides by all the outcomes instead of by the outcomes in B, or the reverse.

  Feedback from the catalogue

  “$A$⁠ and $B$⁠” and “$A$⁠ given $B$⁠” count the same outcomes, those in both events, but divide by different things. For “and” you divide by all the outcomes. For “given $B$⁠” you already know $B$⁠ happened, so you divide only by the outcomes in $B$⁠: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$P(A \mid B) =$$\frac{P(A \cap B)}{P(B)}$⁠. The condition shrinks the denominator, so the conditional chance is the larger one.

• Wrong $\dfrac{P(D)}{P(+ \mid D)}$⁠

  No feedback of its own; the learner sees the standard message

  Not yet. Look again and try once more.

Hint

Worked solution

The Scene at this step

How common the disease is

2 %fixed

Test accuracy

95 %fixed

Retest the positives

onfixed

Sort the results

Test results

Accepted within

± 2

Computed from the Scene at this step’s setting

Per cent chance a positive is ill, by Bayes' rule/nodes/crowd/outputs/posterior

Wrong answers it recognises

• 27.9 ± 1

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback

  $27.9\%$⁠ is the chance after one test. The second test is more information, and it filters out almost all the false positives.

• 99.75 ± 1

  Misconception: Takes a test's accuracy as the chance that a positive result is right: 90% accurate, so a positive means 90% ill.

  Feedback from the catalogue

  Write $D$⁠ for “ill” and $+$⁠ for “positive”. Accuracy says what the test does to the ill and to the healthy: $P(+ \mid D)$⁠. The question of someone who tested positive is the reverse: $P(D \mid +)$⁠. The answer also depends on how common the disease is. When it is rare, most positives come from the large group of the healthy, so the chance of being ill is far below the accuracy. This is base-rate neglect.

Hint

Worked solution

The Scene at this step

How common the disease is

2 %fixed

Test accuracy

95 %fixed

Retest the positives

onfixed

Sort the results

Positives sorted

Accepted within

± 1

A fixed value in the question

Wrong answers it recognises

• 95 ± 1

  Misconception: Takes a test's accuracy as the chance that a positive result is right: 90% accurate, so a positive means 90% ill.

  Feedback from the catalogue

  Write $D$⁠ for “ill” and $+$⁠ for “positive”. Accuracy says what the test does to the ill and to the healthy: $P(+ \mid D)$⁠. The question of someone who tested positive is the reverse: $P(D \mid +)$⁠. The answer also depends on how common the disease is. When it is rare, most positives come from the large group of the healthy, so the chance of being ill is far below the accuracy. This is base-rate neglect.

• 27.9 ± 0.5

  Misconception: Answers with the plain chance of A when asked for the chance of A given B, as if the condition told you nothing.

  Feedback from the catalogue

  The condition is new information, and you have to use it. “Given $B$⁠” says: cross out every outcome not in $B$⁠, and count again inside what is left. Sometimes the chance comes out the same, and then the events are independent, but you find that out only after counting.