6. Compound probability · Probability

Definition

Compound probability

We speak of compound probability when we need to compute the probability that two or more events happen, where the events may come from different experiments or from the same experiment repeated several times. That is, events that follow one another in time.

The key idea is:

\boxed{\;\text{In compound probability, you }\textbf{multiply}\text{ the individual probabilities.}\;}

Formally, if $A$ is the first event and $B$ the second:

P(A \cap B) \;=\; P(A) \cdot P(B \mid A).

And if $A$ and $B$ are independent ($P(B \mid A) = P(B)$), the formula simplifies to:

P(A \cap B) \;=\; P(A) \cdot P(B).

This is the same multiplication rule from Section 5, viewed now from the angle of events chained over time: first $A$ happens, then $B$.

The 4 possible situations

To compute a compound probability correctly, you first need to identify which situation you have. Two axes define 4 scenarios:

Which formula applies depending on the experiment and on the dependence between events.
	Independent $P(B\mid A)=P(B)$	Dependent $P(B\mid A)\neq P(B)$
Two different experiments	$P(A\cap B) = P(A)\cdot P(B)$. E.g. tossing a coin and rolling a die.	Uncommon at 1st BTL level. Different experiments are typically independent.
The same experiment repeated	$P(A\cap B) = P(A)\cdot P(B)$. E.g. rolling the same die twice; with replacement (urn, deck).	$P(A\cap B) = P(A)\cdot P(B\mid A)$. E.g. drawing two balls from an urn without replacement.

The key question: "is there replacement?"

When extracting objects (balls, cards, parts…) one after another, the critical question is whether you put them back before the next draw:

· With replacement → the experiment restarts identical → independent.

· Without replacement → the urn changes between draws → dependent.

If the wording is not explicit, the context usually clarifies: "draw two balls" (without), "toss a coin twice" (with, by construction).

Examples

Example 1 — Coin and die (different experiments, independent)

A fair coin is tossed and a fair 6-sided die is rolled. What is the probability of getting heads and a 5?

The two experiments are independent (the result of one does not influence the other), so:

P(\text{heads},\,5) \;=\; P(\text{heads}) \cdot P(5) \;=\; \frac{1}{2} \cdot \frac{1}{6} \;=\; \boxed{\;\frac{1}{12}\;\approx\;0{,}083.\;}

Example 2 — A die rolled twice (same experiment, independent)

A fair die is rolled twice. Compute:

a) Probability of rolling $5$ on the first toss and $5$ on the second:

P(5,5) \;=\; \frac{1}{6} \cdot \frac{1}{6} \;=\; \boxed{\;\frac{1}{36}.\;}

b) Probability of rolling the same number both times (whichever number):

There are 6 ways to "make a pair" (1,1), (2,2), …, (6,6), each with probability $\tfrac{1}{36}$. Since they are mutually exclusive, we add them:

P(\text{2 equal}) \;=\; P(1,1) + P(2,2) + \cdots + P(6,6) \;=\; 6 \cdot \frac{1}{36} \;=\; \boxed{\;\frac{1}{6}.\;}

Idea: there are 36 ordered pairs $(i,j)$ and 6 of them are "diagonal" $(i,i)$.

Example 3 — Urn with 5 balls (3 red, 2 black)

An urn contains 3 red balls and 2 black balls. We draw 2 balls one after the other. We compare the two key situations:

a) With replacement (after the first ball, we return it to the urn and shake):

The urn returns to 5 balls before the 2nd draw. Therefore the two draws are independent:

P(\text{red, red}) \;=\; \frac{3}{5} \cdot \frac{3}{5} \;=\; \frac{9}{25} \;=\; 0{,}36.

b) Without replacement (we don't return it):

After taking out the first red, the urn has 4 balls left and only 2 reds. The conditional probability $P(B\mid A)$ changes:

P(\text{red, red}) \;=\; \underbrace{\frac{3}{5}}_{P(A)} \cdot \underbrace{\frac{2}{4}}_{P(B\mid A)} \;=\; \frac{6}{20} \;=\; \boxed{\;\frac{3}{10} \;=\; 0{,}3.\;}

The probability drops without replacement because the first draw has "used up" one red — the urn becomes less favourable for the second.

Check: $\tfrac{3}{10} + \tfrac{3}{10} + \tfrac{3}{10} + \tfrac{1}{10} = 1$ ✓.

Strategy: the tree diagram

Method

For compound probability problems with 2 or 3 stages, draw a tree diagram:

1. Each level of the tree represents one stage (1st draw, 2nd draw, …).

2. On each branch, write the probability conditional on the path so far.

3. The probability of a complete path is the product of its branch probabilities.

4. For a question like "what is the probability of…", locate every path that satisfies it and add their products (different paths are mutually exclusive events).

The tree is the natural way to combine the multiplication rule (along each path) with the addition rule (between paths). It connects directly to the law of total probability.

Exercises

1 Coin and die — variants

A fair coin and a fair 6-sided die are tossed. Compute the probability of:

a) Getting tails and an even number.

Independent: $P(\text{tails}) = \tfrac{1}{2}$, $P(\text{even}) = \tfrac{3}{6} = \tfrac{1}{2}$.

P(\text{tails},\text{even}) = \tfrac{1}{2}\cdot\tfrac{1}{2} = \boxed{\;\tfrac{1}{4}.\;}

b) Getting heads and a number greater than 4.

$P(\text{heads}) = \tfrac{1}{2}$, $P(>\!4) = P(\{5,6\}) = \tfrac{2}{6} = \tfrac{1}{3}$.

P = \tfrac{1}{2}\cdot\tfrac{1}{3} = \boxed{\;\tfrac{1}{6}.\;}

c) Neither heads nor a 1.

"Neither heads nor 1" = (tails) and (not 1). Both independent:

P(\text{tails}, \neq\!1) = \tfrac{1}{2}\cdot\tfrac{5}{6} = \boxed{\;\tfrac{5}{12} \approx 0{,}417.\;}

2 A die rolled twice

A fair die is rolled twice. Compute:

a) $P(3,4)$ — exactly a 3 first and a 4 second.

P(3,4) = \tfrac{1}{6}\cdot\tfrac{1}{6} = \boxed{\;\tfrac{1}{36}.\;}

b) $P(\text{sum} = 7)$.

Ordered pairs with sum 7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — 6 pairs, each with probability $\tfrac{1}{36}$:

P(\text{sum}=7) = 6\cdot\tfrac{1}{36} = \boxed{\;\tfrac{1}{6}.\;}

c) $P(\text{first is odd and second is even})$.

Independent: $P(\text{odd}) = \tfrac{3}{6}=\tfrac{1}{2}$, $P(\text{even}) = \tfrac{1}{2}$.

P = \tfrac{1}{2}\cdot\tfrac{1}{2} = \boxed{\;\tfrac{1}{4}.\;}

d) $P(\text{neither roll is a 6})$.

"Neither" = "no 6 on the 1st" and "no 6 on the 2nd". Independent:

P = \tfrac{5}{6}\cdot\tfrac{5}{6} = \boxed{\;\tfrac{25}{36}\approx 0{,}694.\;}

3 Urn — with and without replacement

An urn contains 4 white balls and 3 black balls. We draw 2 balls consecutively.

a) With replacement: $P(\text{both white})$.

Each draw is independent; the urn always has 4 whites among 7 balls:

P = \tfrac{4}{7}\cdot\tfrac{4}{7} = \boxed{\;\tfrac{16}{49}\approx 0{,}327.\;}

b) Without replacement: $P(\text{both white})$.

After the 1st white, there are 3 whites among 6 balls:

P = \tfrac{4}{7}\cdot\tfrac{3}{6} = \tfrac{12}{42} = \boxed{\;\tfrac{2}{7}\approx 0{,}286.\;}

Without replacement gives lower probability: the first draw "consumed" a white.

c) Without replacement: $P(\text{one white and one black, in any order})$.

There are two compatible paths: WB and BW. By the multiplication and addition rules:

P(\text{WB}) = \tfrac{4}{7}\cdot\tfrac{3}{6} = \tfrac{12}{42}, \qquad P(\text{BW}) = \tfrac{3}{7}\cdot\tfrac{4}{6} = \tfrac{12}{42}.

P(\text{one of each}) = \tfrac{12}{42} + \tfrac{12}{42} = \tfrac{24}{42} = \boxed{\;\tfrac{4}{7}\approx 0{,}571.\;}

4 Spanish deck — two cards in a row

From a Spanish deck of 40 cards (10 coins, 10 cups, 10 swords, 10 clubs), we draw 2 cards in a row without replacement.

a) $P(\text{both are coins})$.

P = \tfrac{10}{40}\cdot\tfrac{9}{39} = \tfrac{90}{1560} = \boxed{\;\tfrac{3}{52}\approx 0{,}058.\;}

b) $P(\text{1st is a coin and 2nd is a cup})$.

After drawing a coin, 39 cards remain with all 10 cups intact:

P = \tfrac{10}{40}\cdot\tfrac{10}{39} = \tfrac{100}{1560} = \boxed{\;\tfrac{5}{78}\approx 0{,}064.\;}

c) $P(\text{neither card is a coin})$.

"No coins" = no coin on the 1st and no coin on the 2nd. There are 30 non-coins:

P = \tfrac{30}{40}\cdot\tfrac{29}{39} = \tfrac{870}{1560} = \boxed{\;\tfrac{29}{52}\approx 0{,}558.\;}

d) $P(\text{at least one coin})$.

By the complement of (c):

P(\text{at least one coin}) = 1 - P(\text{no coins}) = 1 - \tfrac{29}{52} = \boxed{\;\tfrac{23}{52}\approx 0{,}442.\;}

Trick: "at least one" is often easier via the complement "none".

5 Three stages — coin tossed 3 times

A fair coin is tossed three times. Each toss is independent of the others.

a) $P(\text{three heads in a row})$.

P = \tfrac{1}{2}\cdot\tfrac{1}{2}\cdot\tfrac{1}{2} = \boxed{\;\tfrac{1}{8}=0{,}125.\;}

b) $P(\text{exactly two heads})$.

The 8 sequences are equiprobable ($\tfrac{1}{8}$ each). Those with exactly 2 heads: HHT, HTH, THH — there are 3.

P(\text{2 heads}) = 3 \cdot \tfrac{1}{8} = \boxed{\;\tfrac{3}{8}=0{,}375.\;}

c) $P(\text{at least one head})$.

By the complement "no heads" (= 3 tails in a row):

P(\text{at least 1 head}) = 1 - \left(\tfrac{1}{2}\right)^3 = 1 - \tfrac{1}{8} = \boxed{\;\tfrac{7}{8}=0{,}875.\;}