IB Mathematics Analysis and Approaches Topic SL…

Conditional Probability And Independence: When One Event Changes Another

Learning a single fact can change the odds of everything else, like how the chance a stranger carries an umbrella jumps once you learn it is raining.
Conditional probability measures exactly how much the probability of one event shifts once you know another has happened.
In 4.5 and 4.6 you handled this with tree diagrams and tables, and here you make it formal with a single formula.
From that formula you get a clean test for whether two events are independent, meaning one tells you nothing about the other.

Deriving The Conditional Probability Formula

Write $P(A \mid B)$ for the probability of $A$ given that $B$ has happened.
The key idea is that once you know $B$ occurred, $B$ becomes your entire world, so the outcomes outside $B$ are no longer possible.
Restrict attention to the outcomes inside $B$, which is your new, smaller sample space.
Among those, the ones that also give $A$ are exactly the outcomes in $A \cap B$.
Probability is favourable over total, so divide the chance of $A \cap B$ by the chance of the new whole, $B$. $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) \neq 0$$

Note

This formula is given in the formula booklet under Topic 4, so you apply it rather than recall it.
Multiply both sides by $P(B)$ for the multiplication rule $P(A \cap B) = P(B)\,P(A \mid B)$, which is how you fill in a tree diagram.

Example question

A survey of 100 people records whether each likes coffee and whether each likes tea.

	Likes tea	No tea	Total
Likes coffee	30	30	60
No coffee	15	25	40
Total	45	55	100

Find (a) $P(\text{coffee} \mid \text{tea})$ and (b) $P(\text{tea} \mid \text{coffee})$.

Solution

(a): condition on tea, so the new whole is the 45 tea-likers, of whom 30 also like coffee.
$P(\text{coffee} \mid \text{tea}) = \dfrac{30}{45} = \dfrac{2}{3} = 0.667$ (3 s.f.).
(b): condition on coffee, so the new whole is the 60 coffee-likers, of whom 30 also like tea.
$P(\text{tea} \mid \text{coffee}) = \dfrac{30}{60} = \dfrac{1}{2} = 0.5$.
The two answers differ, which shows $P(A \mid B)$ and $P(B \mid A)$ are generally not equal.

Independence: When Knowing One Event Tells You Nothing

Two events are independent if learning that one happened does not change the probability of the other.
Written with the conditional probability, that is simply the statement below. $$P(A \mid B) = P(A)$$
Substitute this into the conditional formula: if $P(A \mid B) = P(A)$, then $\dfrac{P(A \cap B)}{P(B)} = P(A)$.
Multiplying up gives the clean symmetric test. $$P(A \cap B) = P(A)\,P(B)$$
1. $P(A \mid B) = P(A)$: knowing $B$ does not change $A$.
2. $P(B \mid A) = P(B)$: knowing $A$ does not change $B$.
3. $P(A \cap B) = P(A)\,P(B)$: the joint probability factorises.

Note

The formula booklet gives the independence condition as $P(A \cap B) = P(A)\,P(B)$, together with $P(A \mid B) = P(A) = P(A \mid B')$.
The factorising form is usually the fastest to test, because you only need three probabilities.

Example question

Using the same survey of 100 people, let $A$ be "likes coffee" and $B$ be "likes tea". Determine whether $A$ and $B$ are independent.

Solution

Read the probabilities: $P(A) = \tfrac{60}{100} = 0.6$, $P(B) = \tfrac{45}{100} = 0.45$, $P(A \cap B) = \tfrac{30}{100} = 0.3$.
Compare $P(A \cap B)$ with $P(A)\,P(B) = 0.6 \times 0.45 = 0.27$.
Since $0.3 \neq 0.27$, the joint probability does not factorise.
Therefore $A$ and $B$ are not independent, since knowing someone likes tea raises their chance of liking coffee from 0.6 to 0.667.

Independent Is Not The Same As Mutually Exclusive

These two ideas are the most confused pair in the topic, because both sound like "the events do not go together".
They are different, and often opposite.

Feature	Mutually exclusive	Independent
Meaning	cannot both happen	one does not affect the other
Key equation	P(A and B) = 0	P(A and B) = P(A) P(B)
On a Venn diagram	no overlap	overlap in general
Knowing B happened	makes A impossible	leaves A unchanged

Common Mistake

If two events are mutually exclusive and both have probability greater than 0, they are not independent.
Knowing one happened tells you the other did not, which is a very strong effect, the opposite of independence.
So never assume disjoint events are independent; check $P(A \cap B) = P(A)\,P(B)$ every time.

See The Sample Space Shrink When You Condition

Conditioning is easiest to feel as a physical shrinking of the sample space.
The moment you know $B$, everything outside $B$ disappears, and the fraction of $A$ inside what remains is the conditional probability.
Independence is then just the special case where that shrinking does not change the fraction, so $P(A \mid B) = P(A)$.

Active recall

Write down the formula for $P(A \mid B)$ and explain what the denominator $P(B)$ represents.
State the multiplication rule that comes from rearranging the conditional probability formula.
Give the three equivalent conditions for two events $A$ and $B$ to be independent.
Two events each have probability greater than zero and are mutually exclusive. Are they independent? Justify your answer.
Using $P(A) = 0.6$, $P(B) = 0.45$, $P(A \cap B) = 0.3$, show whether $A$ and $B$ are independent.

Conditional Probability And Independence: When One Event Changes Another

Learning a single fact can change the odds of everything else, like how the chance a stranger carries an umbrella jumps once you learn it is raining.
Conditional probability measures exactly how much the probability of one event shifts once you know another has happened.
In 4.5 and 4.6 you handled this with tree diagrams and tables, and here you make it formal with a single formula.
From that formula you get a clean test for whether two events are independent, meaning one tells you nothing about the other.

Deriving The Conditional Probability Formula

Write $P(A \mid B)$ for the probability of $A$ given that $B$ has happened.
The key idea is that once you know $B$ occurred, $B$ becomes your entire world, so the outcomes outside $B$ are no longer possible.
Restrict attention to the outcomes inside $B$, which is your new, smaller sample space.
Among those, the ones that also give $A$ are exactly the outcomes in $A \cap B$.
Probability is favourable over total, so divide the chance of $A \cap B$ by the chance of the new whole, $B$. $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) \neq 0$$

Note

This formula is given in the formula booklet under Topic 4, so you apply it rather than recall it.
Multiply both sides by $P(B)$ for the multiplication rule $P(A \cap B) = P(B)\,P(A \mid B)$, which is how you fill in a tree diagram.

Example question

A survey of 100 people records whether each likes coffee and whether each likes tea.

	Likes tea	No tea	Total
Likes coffee	30	30	60
No coffee	15	25	40
Total	45	55	100

Find (a) $P(\text{coffee} \mid \text{tea})$ and (b) $P(\text{tea} \mid \text{coffee})$.

Solution

(a): condition on tea, so the new whole is the 45 tea-likers, of whom 30 also like coffee.
$P(\text{coffee} \mid \text{tea}) = \dfrac{30}{45} = \dfrac{2}{3} = 0.667$ (3 s.f.).
(b): condition on coffee, so the new whole is the 60 coffee-likers, of whom 30 also like tea.
$P(\text{tea} \mid \text{coffee}) = \dfrac{30}{60} = \dfrac{1}{2} = 0.5$.
The two answers differ, which shows $P(A \mid B)$ and $P(B \mid A)$ are generally not equal.

Independence: When Knowing One Event Tells You Nothing

Two events are independent if learning that one happened does not change the probability of the other.
Written with the conditional probability, that is simply the statement below. $$P(A \mid B) = P(A)$$
Substitute this into the conditional formula: if $P(A \mid B) = P(A)$, then $\dfrac{P(A \cap B)}{P(B)} = P(A)$.
Multiplying up gives the clean symmetric test. $$P(A \cap B) = P(A)\,P(B)$$
1. $P(A \mid B) = P(A)$: knowing $B$ does not change $A$.
2. $P(B \mid A) = P(B)$: knowing $A$ does not change $B$.
3. $P(A \cap B) = P(A)\,P(B)$: the joint probability factorises.

Note

The formula booklet gives the independence condition as $P(A \cap B) = P(A)\,P(B)$, together with $P(A \mid B) = P(A) = P(A \mid B')$.
The factorising form is usually the fastest to test, because you only need three probabilities.

Example question

Using the same survey of 100 people, let $A$ be "likes coffee" and $B$ be "likes tea". Determine whether $A$ and $B$ are independent.

Solution

Read the probabilities: $P(A) = \tfrac{60}{100} = 0.6$, $P(B) = \tfrac{45}{100} = 0.45$, $P(A \cap B) = \tfrac{30}{100} = 0.3$.
Compare $P(A \cap B)$ with $P(A)\,P(B) = 0.6 \times 0.45 = 0.27$.
Since $0.3 \neq 0.27$, the joint probability does not factorise.
Therefore $A$ and $B$ are not independent, since knowing someone likes tea raises their chance of liking coffee from 0.6 to 0.667.

Independent Is Not The Same As Mutually Exclusive

These two ideas are the most confused pair in the topic, because both sound like "the events do not go together".
They are different, and often opposite.

Feature	Mutually exclusive	Independent
Meaning	cannot both happen	one does not affect the other
Key equation	P(A and B) = 0	P(A and B) = P(A) P(B)
On a Venn diagram	no overlap	overlap in general
Knowing B happened	makes A impossible	leaves A unchanged

Common Mistake

If two events are mutually exclusive and both have probability greater than 0, they are not independent.
Knowing one happened tells you the other did not, which is a very strong effect, the opposite of independence.
So never assume disjoint events are independent; check $P(A \cap B) = P(A)\,P(B)$ every time.

See The Sample Space Shrink When You Condition

Conditioning is easiest to feel as a physical shrinking of the sample space.
The moment you know $B$, everything outside $B$ disappears, and the fraction of $A$ inside what remains is the conditional probability.
Independence is then just the special case where that shrinking does not change the fraction, so $P(A \mid B) = P(A)$.

Active recall

Write down the formula for $P(A \mid B)$ and explain what the denominator $P(B)$ represents.
State the multiplication rule that comes from rearranging the conditional probability formula.
Give the three equivalent conditions for two events $A$ and $B$ to be independent.
Two events each have probability greater than zero and are mutually exclusive. Are they independent? Justify your answer.
Using $P(A) = 0.6$, $P(B) = 0.45$, $P(A \cap B) = 0.3$, show whether $A$ and $B$ are independent.

IB Mathematics Analysis and Approaches SL 4.11—conditional and Independent Probabilities, Test for Independence Notes

IB Mathematics Analysis and Approaches SL 4.11—conditional and Independent Probabilities, Test for Independence Notes

Conditional Probability And Independence: When One Event Changes Another

Deriving The Conditional Probability Formula

Independence: When Knowing One Event Tells You Nothing

Independent Is Not The Same As Mutually Exclusive

See The Sample Space Shrink When You Condition

Conditional Probability And Independence: When One Event Changes Another

Deriving The Conditional Probability Formula

Independence: When Knowing One Event Tells You Nothing

Independent Is Not The Same As Mutually Exclusive

See The Sample Space Shrink When You Condition

IB Mathematics Analysis and Approaches SL 4.11—conditional and Independent Probabilities, Test for Independence Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 4.11—conditional and Independent Probabilities, Test for Independence Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Conditional Probability And Independence: When One Event Changes Another

Deriving The Conditional Probability Formula

Independence: When Knowing One Event Tells You Nothing

Independent Is Not The Same As Mutually Exclusive

See The Sample Space Shrink When You Condition

Conditional Probability And Independence: When One Event Changes Another

Deriving The Conditional Probability Formula

Independence: When Knowing One Event Tells You Nothing

Independent Is Not The Same As Mutually Exclusive

See The Sample Space Shrink When You Condition