Conditional probability — P(A - B) = P(A∩B) - P(A): Question bank

Notation reminder, so nothing is used before it is named:

$A \cap B$ ::: "both $A$ and $B$ " happen — the overlap of the two events.
$A^{c}$ ::: "not $A$ " — every outcome outside $A$ (its complement).
$B^{c}$ ::: "not $B$ ".
$P (A ∣ B)$ ::: probability of $A$ given $B$ , defined only when $P (B) > 0$ ; read the bar as "given".
$n (\cdot)$ ::: the count of outcomes in an event (used only in the equally-likely picture below).

True or false — justify

P (A ∣ B) = P (B ∣ A)

always.

False. Both share the numerator

P (A \cap B)

, but the first divides by

P (B)

and the second by

P (A)

(each with its own positivity requirement); unless

P (A) = P (B)

these differ.

A

and

B

are independent then

P (A ∣ B) = P (A)

(assuming

P (B) > 0

True. Independence means knowing

B

tells you nothing about

A

, so the conditioned probability equals the unconditioned one — this is the very definition via Independence of events.

P (A ∣ B) + P (A^{c} ∣ B) = 1

always (when

P (B) > 0

True. With

B

fixed as the universe,

A

and

A^{c}

split that universe completely, so their conditional probabilities must sum to 1.

P (A ∣ B) + P (A ∣ B^{c}) = 1

always.

False. You may not change the event after the bar;

B

and

B^{c}

are two different universes (each needing its own positive probability), so these two numbers have no reason to add to 1.

B \subseteq A

(every

B

-outcome is also in

A

) and

P (B) > 0

then

P (A ∣ B) = 1

True. Then

A \cap B = B

, so

P (A ∣ B) = P (B) / P (B) = 1

— once you're inside

B

A

is guaranteed.

A \subseteq B

and

P (B) > 0

then

P (A ∣ B) = 1

False. Here

A \cap B = A

, so

P (A ∣ B) = P (A) / P (B)

, which is generally less than 1; being inside the bigger set

B

does not force the smaller

A

P (A ∣ B)

can be larger than

P (A)

(when

P (B) > 0

True. Conditioning can raise a probability: if

B

makes

A

more likely (positive association),

P (A ∣ B) > P (A)

P (A ∣ B)

is always at least as large as

P (A \cap B)

(when

P (B) > 0

True. Dividing by

P (B) \leq 1

can only keep it the same or enlarge it, since

P (A ∣ B) = P (A \cap B) / P (B) \geq P (A \cap B)

A

and

B

are mutually exclusive (never both happen) and

P (B) > 0

, then

P (A ∣ B) = 0

True. Mutually exclusive means

A \cap B = \emptyset

, so

P (A \cap B) = 0

and the ratio is

0

— inside

B

A

can never occur.

Mutually exclusive events with positive probability are independent.

False. If they exclude each other,

P (A ∣ B) = 0 \neq = P (A)

, so knowing

B

changes

A

's chance — that is dependence, not independence.

Spot the error

P (A ∣ B) = \frac{P ( A \cap B )}{P ( A )}

Wrong denominator: you must divide by the given event

P (B)

(which must be positive), not

P (A)

. Dividing by

P (A)

silently computes

P (B ∣ A)

instead.

"A test is

99%

accurate, I tested positive, so I'm

99%

likely sick."

That

99%

P (+ ∣ sick)

, the reverse of what you want,

P (sick ∣ +)

. Flipping the bar without Bayes' Theorem ignores how rare the disease is (false positives can dominate).

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

, so all events multiply like this."

Only when

A, B

are independent. The general Multiplication rule of probability is

P (A \cap B) = P (B) P (A ∣ B)

(for

P (B) > 0

); the plain product is a special case.

"Since

P (A ∣ B) = 0.7

and

P (A ∣ B^{c}) = 0.7

, we learn

P (A) = 1.4

You cannot add conditionals like that. The correct combination is the Law of total probability:

P (A) = P (A ∣ B) P (B) + P (A ∣ B^{c}) P (B^{c})

, a weighted average that stays

\leq 1

P (B) = 0

, so

P (A ∣ B) = \frac{P ( A \cap B )}{0} = \infty

Not infinity — it's undefined. You cannot condition on an impossible event; the formula requires

P (B) > 0

and simply does not apply otherwise.

"Two kings without replacement:

P = \frac{4}{52} \cdot \frac{4}{52}

The second draw must be conditioned on the first: after removing one king, only

3

kings remain in

51

cards, so it's

\frac{4}{52} \cdot \frac{3}{51}

"In a tree, I multiply along a branch AND across the top to combine two paths."

Multiply along a branch (chaining conditionals), but add across separate branches when combining different paths to the same outcome.

Why questions

Why divide by

P (B)

instead of

P (A)

Because

B

is the new universe you now live in; rescaling by

P (B)

(which is why we insist

P (B) > 0

) forces the probabilities inside

B

to sum to

1

(indeed

P (B ∣ B) = 1

Why must

P (B) > 0

for

P (A ∣ B)

to exist?

Conditioning means "assume

B

happened"; if

B

has probability

0

it can't have happened, and dividing by

0

is undefined.

Why does the "equally likely" derivation still give a formula that works when outcomes are not equally likely?

Writing the total number of equally-likely outcomes as

N

, we start from

\frac{n ( A \cap B )}{n ( B )}

and divide both counts by

N

; this converts each count into a probability, giving

\frac{P ( A \cap B )}{P ( B )}

, a ratio form that no longer mentions equal likelihood.

Why can conditioning make an event more likely?

B

"lives near"

A

(their overlap is a big slice of

B

), then shrinking the universe to

B

concentrates the

A

-outcomes, pushing the fraction above

P (A)

Why is independence symmetric — if

P (A ∣ B) = P (A)

, must

P (B ∣ A) = P (B)

Both are equivalent to the single condition

P (A \cap B) = P (A) P (B)

, which treats

A

and

B

symmetrically, so one holding forces the other (when both probabilities are positive).

Why does

P (A ∣ B) = P (A)

still allow

A

and

B

to overlap?

Independence forbids

B

from changing

A

's chance, not from co-occurring; overlap of size exactly

P (A) P (B)

is precisely the "no information" case.

Why can't you read

P (A ∣ B)

straight off a Venn overlap without dividing?

The raw overlap

P (A \cap B)

is measured against the whole sample space; conditioning re-measures it against just

B

, which requires the division by

P (B)

Edge cases

P (B) = 1

(B is certain), what is

P (A ∣ B)

It equals

P (A)

, since the "universe"

B

is already everything — conditioning on a sure event changes nothing.

A = B

(same event,

P (A) > 0

), what is

P (A ∣ B)

P (A ∣ A) = P (A \cap A) / P (A) = P (A) / P (A) = 1

— given an event, that event certainly holds.

P (A) = 0

but

P (B) > 0

, what is

P (A ∣ B)

P (A \cap B) \leq P (A) = 0

, so the numerator is

0

and

P (A ∣ B) = 0

; an impossible

A

stays impossible inside any universe.

A

and

B

are independent and

P (A) = 1

, is

P (A ∣ B) = 1

Yes:

P (A ∣ B) = P (A) = 1

; a certain event remains certain, and independence is trivially satisfied.

Can

P (A ∣ B) = P (A)

hold while

A

and

B

are mutually exclusive?

Only if

P (A) = 0

, because exclusivity forces

P (A ∣ B) = 0

, which can equal

P (A)

merely when

A

itself is impossible.

What happens to

P (2nd king ∣ 1st king)

with replacement instead of without?

The universe resets to a full

52

-card deck, so it becomes

4/52

— the draws are then independent and the condition stops mattering.

If a tree branch has

P (B)

tiny but

P (A ∣ B)

huge, is the path

P (A \cap B)

large?

Not necessarily — the path probability is the product

P (B) P (A ∣ B)

, so a tiny

P (B)

can shrink even a near-certain conditional.

Recall One-line survival guide

Divide by the event to the right of the bar — the given event $B$ — and only when $P (B) > 0$ ; you may swap what's left of the bar (complement law $P (A ∣ B) + P (A^{c} ∣ B) = 1$ ) but never what's right of it.

Connections

Independence of events — the $P (A ∣ B) = P (A)$ boundary that several traps probe.
Multiplication rule of probability — why the second card is $3/51$ , not $4/52$ .
Bayes' Theorem — the correct way to flip $P (+ ∣ sick)$ into $P (sick ∣ +)$ .
Law of total probability — why $P (A ∣ B) + P (A ∣ B^{c}) \neq = 1$ but a weighted sum recovers $P (A)$ .
Sample space and events — the universe we shrink when conditioning.
Tree diagrams — multiply along branches, add across them.