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Renormalising inside an event, reading a contingency table both ways, and keeping the two directions of conditioning apart.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will compute a conditional probability from a joint probability and a marginal one, read both directions of conditioning off a contingency table, and restrict a sample space to the outcomes a piece of information leaves standing. You will also say, of a sentence written in words, which conditional probability it actually states and whether it has quietly swapped the two.
A probability has been a measure over one fixed sample space, and every rule so far has been about that space. Conditioning replaces it with a smaller one, so nothing new has to be proved here: the same three axioms are applied inside $B$ instead of inside $\Omega$.
$P(A \mid B)$ is read the probability of $A$ given $B$. $B$ is the conditioning event, and it must have positive probability for the expression to mean anything. The restricted sample space is $B$ itself: conditioning is not a correction applied to an answer, it is a new sample space with a new total.
For $P(B) > 0$,
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$$
Read it as a rescaling. You are told the outcome lies in $B$, so $B$ becomes the whole world; the part of $A$ that survives is $A \cap B$; and dividing by $P(B)$ makes the new world weigh $1$ again. Everything outside $B$ is gone, not reduced.
$P(\cdot \mid B)$ is itself a probability measure, obeying all three axioms, which is why every rule already proved applies inside it: $P(A^{c} \mid B) = 1 - P(A \mid B)$, and inclusion-exclusion holds conditionally too. What does not transfer is any relation between $P(A \mid B)$ and $P(A \mid B^{c})$ — those live in different worlds.
$P(A \mid B)$ and $P(B \mid A)$ are different numbers, and the difference is the whole of Bayes' theorem. Reading one for the other is the error behind almost every famous misuse of probability: a test that is right ninety-nine times in a hundred does not make a positive result ninety-nine per cent likely to be true.
Another way: picture
A Venn diagram with the region outside $B$ cut away with scissors. What is left is $B$, with $A \cap B$ shaded inside it. The conditional probability is the shaded share of what remains on the table — and the discarded piece takes no part in the answer.
Another way: steps
A hundred people, sorted two ways.
| Cyclist | Not a cyclist | Total | |
|---|---|---|---|
| Commuter | 24 | 16 | 40 |
| Not a commuter | 6 | 54 | 60 |
| Total | 30 | 70 | 100 |
$P(\text{cyclist} \mid \text{commuter}) = 24/40 = 0.6$ — divide along the row. $P(\text{commuter} \mid \text{cyclist}) = 24/30 = 0.8$ — divide down the column. Same cell, different totals, different answers, and neither is the unconditional $P(\text{cyclist}) = 0.3$.
Reading a contingency table is the whole of conditioning in miniature: the cell is the intersection, and the margin you divide by is the world you were put in.
Swapping the two directions. $P(A \mid B)$ and $P(B \mid A)$ are different numbers. This is the error behind the prosecutor's fallacy, the medical-test surprise and most newspaper statistics.
Dividing by the wrong total. In a table, the denominator is the margin of the event you conditioned on. Dividing by the grand total gives the joint probability again, not a conditional one.
Subtracting instead of dividing. Conditioning is a rescaling. Nothing is taken away from the numerator.
Conditioning on an event of probability zero. $P(A \mid B)$ is undefined when $P(B) = 0$; the formula divides by zero and the phrase given $B$ has nothing to restrict to. Continuous models need a separate treatment, which comes later in this course.
Two cards are drawn from $52$ without replacement. Given that the first is an ace, what is the probability the second is?
Name the conditioning event.
The world is now the $51$ cards left, of which $3$ are aces.
The deck itself has changed.
So the answer is $3/51 = 1/17$, which is smaller than $4/52$.
Information moved the probability.
One fair die. Given the roll is even, what is the probability it exceeds $3$?
The conditioning event has three outcomes.
The world is $\{2, 4, 6\}$; two of those exceed $3$.
Restrict, then count inside.
So the answer is $2/3$, against an unconditional $1/2$.
Conditioning raised it.
$P(A \mid B) = 0.2 / 0.5 = 0.4$.
Divide by what you were told.
$P(B \mid A) = 0.2 / 0.4 = 0.5$.
Same numerator, other denominator.
Both equal the unconditional probabilities, which is a hint about the next lesson.
Of $100$ students, $50$ study music, $40$ play a sport and $20$ do both. Give the three conditional probabilities.
| Probability | |
|---|---|
| Plays a sport, given that they study music | |
| Studies music, given that they play a sport | |
| Plays a sport, given that they do not study music |
$P(A \cap B) = 1/25$ and $P(B) = 1/10$. What is $P(A \mid B)$?
Answer:
Is this sound: most professional basketball players are tall, but most tall people are not professional basketball players?
In a group of $100$ people, $40$ own a bicycle, $50$ own a car and $10$ own both. Give the two conditional probabilities.
Owns a car given a bicycle: p. Owns a bicycle given a car: q.
A family has two children, and the four ordered possibilities are equally likely. You are told that at least one of them is a boy. Mark which possibilities survive that information, then give the two counts.
| Still possible? | How many | |
|---|---|---|
| Elder a girl, younger a girl | — | |
| Elder a girl, younger a boy | — | |
| Elder a boy, younger a girl | — | |
| Elder a boy, younger a boy | — | |
| Possibilities that survive | — | |
| Of those, both children a boy | — |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A bag holds $40$ tokens, $20$ of them wooden, and $7$ of the wooden ones are also painted. One token is drawn, and you are told it is wooden. The marks are one twentieth apart. Put the marker at the probability that it is painted.
0 |——————————| 1
Mark the position with a cross, then write the value:
You can restrict a sample space to what you were told, divide by the right total, and keep the two directions of conditioning apart. Say in your own words why the probability of a positive test given illness is a different number from the probability of illness given a positive test.
9. Your turn: $P(A) = 0.4$, $P(B) = 0.5$, $P(A \cap B) = 0.2$; find both conditionals, step 3