Probability in several steps

When something happens in several steps you draw a tree diagram.

A bag with 3 red and 2 blue marbles. You draw two, without putting them back.

            ┌─ red 2/4 ──→ RR:  3/5 · 2/4 = 6/20
   red 3/5 ─┤
            └─ blue 2/4 ─→ RB:  3/5 · 2/4 = 6/20
start ──┤
            ┌─ red 3/4 ──→ BR:  2/5 · 3/4 = 6/20
  blue 2/5 ─┤
            └─ blue 1/4 ─→ BB:  2/5 · 1/4 = 2/20

Two rules, and they are all you need:

Rule	When
Multiply along a branch	the events happen one after the other («and»)
Add between branches	several different routes give the same thing («or»)

A check: all the final results should add up to 1. Here: 6/20 + 6/20 + 6/20 + 2/20 = 20/20 = 1. ✓

That check takes five seconds and catches nearly every arithmetic slip.

With or without replacement is the decisive difference:

	With replacement	Without replacement
The denominator	unchanged every time	decreases by 1
The events are	independent	dependent
Example	dice throws, coin throws	drawing marbles, dealing cards

In the example above the denominator changed from 5 to 4 — that is what «without replacement» means in practice.

A conditional probability is written $P(B \mid A)$ and means «the probability of B, given that A has already happened». The multiplication rule then becomes:

$P(A \text{ and } B) = P(A) \cdot P(B \mid A)$

If the events are independent then $P(B\mid A) = P(B)$, and the rule simplifies to an ordinary product.

The complement rule saves a lot of work:

$P(\text{at least one}) = 1 - P(\text{none})$

An example: you throw three dice. The probability of at least one six?

Computing it directly requires summing «exactly one», «exactly two» and «exactly three». Computing the opposite requires one step:

$1 - \left(\frac{5}{6}\right)^3 = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.42$

The rule to remember: when the question contains «at least one» — compute «none» and subtract from 1.

The most common source of error is forgetting that the denominator changes without replacement. The second most common is adding when you should multiply. The tree diagram protects against both, since you see the steps in front of you.

from fractions import Fraction as F
import random

# Without replacement: the denominator decreases
red, blue = 3, 2
routes = {
    "RR": F(red, 5) * F(red - 1, 4),
    "RB": F(red, 5) * F(blue, 4),
    "BR": F(blue, 5) * F(red, 4),
    "BB": F(blue, 5) * F(blue - 1, 4),
}
for name, p in routes.items():
    print(f"  {name}: {p} = {float(p):.3f}")
print("the sum:", sum(routes.values()))       # 1  ← the check

print("at least one red:", 1 - routes["BB"], "=", float(1 - routes["BB"]))   # 9/10 = 0.9
print("exactly one red:", routes["RB"] + routes["BR"])                       # 3/5

# With replacement: the denominator is unchanged
with_repl = {"RR": F(3, 5) ** 2, "RB": F(3, 5) * F(2, 5),
             "BR": F(2, 5) * F(3, 5), "BB": F(2, 5) ** 2}
print("with replacement, both red:", with_repl["RR"], "against without:", routes["RR"])
# 9/25 = 0.36  against  3/10 = 0.30

# The complement rule: at least one six on three dice
print("at least one six:", 1 - F(5, 6) ** 3, "=", float(1 - F(5, 6) ** 3))   # 91/216 = 0.421

# Simulate and check
def simulate(n=200_000, seed=0):
    rng = random.Random(seed)
    hits = 0
    for _ in range(n):
        bag = ["R"] * 3 + ["B"] * 2
        rng.shuffle(bag)
        if bag[0] == "R" and bag[1] == "R":
            hits += 1
    return hits / n

print(f"simulated: {simulate():.4f}  theoretical: {float(routes['RR']):.4f}")
# simulated: 0.3001  theoretical: 0.3000

The simulation at the bottom is worth running: when a tree diagram feels uncertain you can always check it by letting the computer draw marbles two hundred thousand times.