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Chaining conditional probabilities along the stages of a draw, and reading a tree as products along paths and sums across them.
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 the probability of a sequence of events by multiplying along the stages, conditioning each factor on everything already assumed, and you will draw the tree that makes those stages visible. You will also decide how many paths a described event covers, and add across them.
$P(A \mid B) = P(A \cap B) / P(B)$ was a way of computing a conditional probability from a joint one. Multiply both sides by $P(B)$ and it becomes a way of computing a joint probability from a conditional one. That is the whole of this lesson, and it is why no new axiom is needed for it.
A stage is one step of an experiment performed in order, and a tree draws the stages left to right with one branch per possibility. A path is a route from the root to an end, and its probability is the product of the branches along it. Without replacement means an item taken at one stage is not available at the next.
$$P(A \cap B) = P(A) \, P(B \mid A) = P(B) \, P(A \mid B),$$ and the chain continues for any number of stages:
$$P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1) \, P(A_2 \mid A_1) \, P(A_3 \mid A_1 \cap A_2) \cdots$$
Each factor is conditioned on everything already assumed to have happened, which is what makes the rule exact rather than approximate. Drawing without replacement is the standard case: both the favourable count and the total fall as you go, so every factor is different.
The picture is a tree. Each level is a stage, each branch carries a conditional probability, and the probability of a path is the product of the branches along it. An event that several paths reach is the sum of those products — and deciding how many paths the words cover is usually the only judgement in the problem.
Another way: picture
A tree drawn left to right. From the root, two branches labelled with $P(A)$ and $P(A^{c})$; from each of those, two more labelled with conditional probabilities. Multiply along a path, add across paths. The branch probabilities at each fork add to one; the path probabilities at the far right add to one too.
Another way: steps
Two draws from a box of five tickets, two of them winning.
| With replacement | Without replacement | |
|---|---|---|
| First wins | $2/5$ | $2/5$ |
| Second wins, given the first did | $2/5$ | $1/4$ |
| Both win | $4/25 = 0.16$ | $2/20 = 0.1$ |
The first row is identical and every later row differs. With replacement is the special case in which the conditioning does nothing — which is exactly the definition of independence, and the subject of the next lesson.
Reusing the first stage's probability at the second. The commonest error in the lesson, and it silently converts a problem about drawing without replacement into one about drawing with it.
Changing the numerator and forgetting the denominator. Taking a winning ticket out removes one from the winners and one from the box.
Multiplying $P(A)$ by $P(B)$. That is the product rule for independent events. Using it here assumes the very thing the question is about.
Adding along a path or multiplying across paths. Multiply along, add across. Getting these the wrong way round usually produces a number outside $[0, 1]$, which is at least a loud failure.
From a full deck, $P(\text{first ace}) = 4/52$.
The deck as it stands.
Given that, $P(\text{second ace}) = 3/51$; given both, $P(\text{third ace}) = 2/50$.
Both counts fall each time.
So the product is $\dfrac{4}{52} \cdot \dfrac{3}{51} \cdot \dfrac{2}{50} = \dfrac{1}{5525}$.
Multiply along the path.
A box holds $3$ red and $2$ blue. Two are drawn. What is $P(\text{one of each})$?
The words cover two paths.
Red then blue: $\dfrac{3}{5} \cdot \dfrac{2}{4} = \dfrac{3}{10}$. Blue then red: $\dfrac{2}{5} \cdot \dfrac{3}{4} = \dfrac{3}{10}$.
Multiply along each.
Add: $\dfrac{3}{5} = 0.6$.
Add across paths.
First green: $4/10$.
The bag as it stands.
Second green given the first was: $3/9$.
Both counts have fallen.
So $\dfrac{4}{10} \cdot \dfrac{3}{9} = \dfrac{2}{15}$.
A box holds $5$ tickets, $2$ of them winning. Two are drawn one after the other without replacement. Give the three probabilities.
| Probability | |
|---|---|
| The first ticket wins | |
| The second wins, given that the first did | |
| Both tickets win |
$P(A) = 7/10$ and $P(B \mid A) = 3/10$. What is $P(A \cap B)$?
Answer:
A bag holds $10$ marbles, $5$ of them blue. Three are drawn without replacement. Mark the two lines of this working that use the wrong probability.
This task has no paper form; do it on a device.
A tray holds $5$ eggs, $2$ of them cracked. Two are taken out one after the other. Give the probability that the first is cracked, and the probability that the second is cracked given that the first was.
First cracked: p. Second cracked given the first was: q.
A deck of $20$ cards holds $5$ aces. Two cards are dealt without replacement, and nobody looks at the first. What is the probability that the *second* card is an ace?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A box holds $8$ items, some faulty. Match each event to the product that gives its probability.
| $P(F_1) \, P(F_2 \mid F_1)$ | $P(S_1) \, P(S_2 \mid S_1)$ | $P(F_1) \, P(S_2 \mid F_1)$ | $P(F_1) \, P(S_2 \mid F_1) + P(S_1) \, P(F_2 \mid S_1)$ | |
|---|---|---|---|---|
| Both items are faulty | ||||
| Neither item is faulty | ||||
| The first is faulty and the second is sound | ||||
| Exactly one of the two is faulty |
You can chain conditional probabilities through several stages, read a tree as products along paths and sums across them, and say what each factor is conditioned on. Say in your own words what changes between the first and second factor when a draw is made without replacement.
9. Your turn: a bag holds $4$ green and $6$ yellow; two are drawn without replacement; find the probability that both are green, step 3