Probability · Combined events & tree diagrams

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Probability without replacement

Why the second pick happens in a smaller, changed bag — so the total always drops by one but a colour's count drops only if that colour was taken — and how that turns "one of each", "at least one" and the reverse quadratic questions.

Probability · Combined events & tree diagrams

Probability without replacement

Why it works

Without replacement means the first thing you take is not put back before you take the second. That one sentence is the whole topic — but it has two consequences, and almost everybody remembers only one of them.

Work with a real bag: 5 red counters and 3 blue counters,

8

in total. Two counters are taken, one after the other, without replacement.

Why you multiply along a path. Label the counters

R_1,\dots,R_5

and

B_1,B_2,B_3

, and think of the two picks as an ordered pair. There are

8

counters you could take first and, whichever one it was,

7

left to take second, so there are

8 \times 7 = 56 \text{ equally likely ordered pairs.}

How many of those are red-then-red? There are

5

reds you could take first, and then

4

reds still sitting in the bag, so

5 \times 4 = 20

pairs. Hence

P(\text{both red}) = \frac{5 \times 4}{8 \times 7} = \frac{20}{56} = \frac{5}{14}.

Now split that fraction up:

\frac{5\times 4}{8 \times 7} = \frac{5}{8} \times \frac{4}{7}

. Multiplying along the branches isn't a new rule at all — it is that counting fraction, written as two fractions.

What changes on the second pick. Compare the

8 \times 7

with the

5 \times 4

The total goes down by one every single time, because one counter has

left the bag whatever colour it was. The denominator always runs

8 \to 7

A colour's count goes down by one **only for the colour you actually

took. Take a red and the reds run $5 \to 4$ while the blues stay on $3$ . Take a blue and the blues run $3 \to 2$ while the reds stay on $5$ .

The trap, made concrete. By far the most common mistake is to change the denominator and leave the numerator alone:

$\underbrace{\frac{5}{8}\times\frac{5}{7} = \frac{25}{56}}_{\text{wrong}} \qquad\text{instead of}\qquad \underbrace{\frac{5}{8}\times\frac{4}{7} = \frac{20}{56}}_{\text{right}}$

That's $0.446$ against $0.357$ — nowhere near each other, so this is a genuinely wrong answer, not a rounding wobble. And it feels right: the total went down, so it seems the "without replacement" instruction has been obeyed. But read $\frac{5}{7}$ out loud: it claims that 5 of the 7 counters still in the bag are red, and after a red has been removed there are only $4$ . It describes a bag that never existed. In the counting picture it allows the red taken first to be picked again second — replacement wearing a disguise.

Here is a check that catches it instantly. On any branch of the tree, the probabilities coming out of that branch must have the same denominator and their numerators must add up to it. After a red is taken: $\frac{4}{7}$ red and $\frac{3}{7}$ blue, and $4 + 3 = 7$ ✓. The wrong version gives $\frac{5}{7}$ and $\frac{3}{7}$ , whose numerators add to $8$ — a set of probabilities summing to more than $1$ . The working was doomed before it reached an answer.

The tree is lopsided.** With replacement, the second-pick probabilities are identical on every branch. Without replacement they differ branch by branch, because a different bag survives:

First pick	Bag left behind	Second pick	Path probability
red $\left(\frac{5}{8}\right)$	4 red, 3 blue	red $\frac{4}{7}$ · blue $\frac{3}{7}$	RR $\frac{20}{56}$ · RB $\frac{15}{56}$
blue $\left(\frac{3}{8}\right)$	5 red, 2 blue	red $\frac{5}{7}$ · blue $\frac{2}{7}$	BR $\frac{15}{56}$ · BB $\frac{6}{56}$

Notice

\frac{5}{7}

is correct on the lower branch — a blue left, so all

5

reds are still there. The same fraction is right in one place and wrong in another; only the bag decides. The four paths total

\frac{56}{56} = 1

✓.

"One of each" needs TWO branches. "One red and one blue" doesn't say which came first, and red-then-blue and blue-then-red are different ordered pairs — so both paths count:

P(\text{one of each}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}.

Using one branch alone halves the answer. The two products happen to be equal here (

5\times3

and

3\times5

are the same numbers reordered), so doubling one of them is a legitimate shortcut — but only for two different colours. Never double a "both the same" calculation.

"At least one" is easiest backwards. "At least one blue" covers RB, BR and BB — three paths to add. Its opposite is "no blues at all", which is both red — a single path:

P(\text{at least one blue}) = 1 - \frac{20}{56} = \frac{36}{56} = \frac{9}{14}.

(Directly:

15 + 15 + 6 = 36

✓.) The opposite of "at least one blue" is not "both blue" — that is the opposite of "at least one red".

Three picks. Nothing new: three fractions, totals

8, 7, 6

, and each colour's count dropping each time that colour is used.

P(\text{three reds}) = \frac{5}{8}\times\frac{4}{7}\times\frac{3}{6} = \frac{60}{336} = \frac{5}{28}

.

Going backwards — why a quadratic appears. Suppose the bag holds

n

red and

k

blue, with

n

unknown. Then

n

sits in the numerator and inside the total, and both drop by one on the second pick:

P(\text{both red}) = \frac{n}{n+k}\times\frac{n-1}{n+k-1} = \frac{n(n-1)}{(n+k)(n+k-1)}.

Set that equal to the probability the question gives you and clear the fractions. Each side is then a product of two brackets containing

n

, so each side produces an

n^2

— and they don't cancel. The quadratic is the price of the numerator and the denominator both moving: it is the algebraic fingerprint of "without replacement".