Independent Events

MATConditional Probability & Independence

Two events $A$ and $B$ are independent when the occurrence of one does not change the probability of the other, captured by the product rule $P(A\cap B)=P(A)\,P(B)$ . ISC questions either ask you to verify independence from given probabilities or to use independence to compute the probability that "at least one" of several independent events occurs.

Test / definition of independence

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

A

and

B

are independent iff this holds. Equivalently

P(A\mid B) = P(A)

and

P(B\mid A) = P(B)

(provided the conditioning probabilities are nonzero).

Addition theorem (any two events)

P(A \cup B) = P(A) + P(B) - P(A \cap B)

General rule; for independent events substitute

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

Independence of complements

P(A' \cap B') = P(A')\,P(B') = \big(1-P(A)\big)\big(1-P(B)\big)

A,B

are independent then so are

A',B'

(and

A,B'

and

A',B

). Also equals

1 - P(A\cup B)

by De Morgan.

At least one of independent events occurs

P(\text{at least one}) = 1 - P(A')\,P(B')\,P(C')\cdots

Each

P(\cdot')

is a probability of non-occurrence; valid only when the events are independent. The complement of "at least one" is "none".

To test independence: compute $P(A)\,P(B)$ and compare with the given $P(A\cap B)$ — equal means independent, unequal means dependent.
Independent is NOT the same as mutually exclusive: mutually exclusive events with positive probability are always dependent, since $P(A\cap B)=0 \ne P(A)\,P(B)$ .
For independent events the easiest route to "at least one" is the complement: $1 - \prod P(\text{not occurring})$ .
If $A,B$ are independent, every pairing among $A,A',B,B'$ is independent too — handy for finding $P(A'\cap B')$ .
To find an unknown probability, you may use $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ to first get $P(A\cap B)$ , then check whether it equals $P(A)\,P(B)$ .
For the three-students problem, $P(\text{solved}) = 1 - \left(1-\tfrac12\right)\left(1-\tfrac13\right)\left(1-\tfrac14\right)$ .
Independence is a property of the events, but for ISC verification you confirm it purely via the product condition.

	Independent	Mutually exclusive
Defining test	$P(A\cap B) = P(A)\,P(B)$	$P(A\cap B) = 0$
Knowing $B$ occurred…	tells you nothing: $P(A\mid B) = P(A)$	rules $A$ out: $P(A\mid B) = 0$
$P(A\cup B)$	$P(A) + P(B) - P(A)P(B)$	$P(A) + P(B)$
Can both occur together?	Yes	No
Can events be both at once?	Only if one of them has probability $0$	Only if one of them has probability $0$

The most-punished confusion in the chapter. Mutually exclusive events are strongly *dependent*: if one happens, the other certainly did not.

Worked example · 2 marks

The probability of event

A

\dfrac{1}{3}

and of event

B

\dfrac{1}{2}

. If

A

and

B

are independent, find the probability that neither occurs.

'Neither' is $A' \cap B'$ .
If $A$ and $B$ are independent then so are their complements, so $P(A'\cap B') = P(A')\,P(B')$ .
$P(A') = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ and $P(B') = 1 - \dfrac{1}{2} = \dfrac{1}{2}$ .
$P(A'\cap B') = \dfrac{2}{3}\times\dfrac{1}{2}$ .

Answer

\dfrac{1}{3}

. Note the shortcut this licenses:

P(\text{at least one}) = 1 - P(\text{none})= \dfrac{2}{3}

Where the marks go

Assuming independent means mutually exclusive (or vice versa) — they are different and rarely hold together.
Adding probabilities for "at least one" instead of using $1 -$ (product of complements), which double-counts overlaps.
Using $P(A\cup B) = P(A) + P(B)$ for independent (non-disjoint) events, forgetting to subtract $P(A\cap B) = P(A)\,P(B)$ .
Concluding dependence after a small arithmetic slip; always recompute $P(A)\,P(B)$ exactly with common denominators.

How the board asks it

Give reasonsproduct test for independence2 mkAsked 2026
If $P(A) = \tfrac{3}{5}$ , $P(B) = \tfrac{1}{3}$ and $P(A\cap B) = \tfrac{1}{5}$ , examine whether the events $A$ and $B$ are independent. Give reasons for your answer.
Applicationat least one of independent events occurs1 mkAsked 2024
A problem in Mathematics is given to three students whose chances of solving it are $\tfrac{1}{2}$ , $\tfrac{1}{3}$ and $\tfrac{1}{4}$ respectively. Assuming they work independently, find the probability that the problem will be solved.
Numericalindependence of complements; exactly one occurs2 mkAsked 2023
$A$ and $B$ are independent events with $P(A) = 0.3$ and $P(B) = 0.4$ . Find $P(A\cup B)$ , $P(A'\cap B')$ and the probability that exactly one of the two events occurs.
Numericalrecover $P(A\cap B)$ then apply product test1 mkAsked 2023
For two events $A$ and $B$ , $P(A) = \tfrac{1}{2}$ , $P(B) = \tfrac{7}{12}$ and $P(\text{neither } A \text{ nor } B) = \tfrac{1}{4}$ . Find $P(A\cap B)$ and hence determine whether $A$ and $B$ are independent.
Applicationat least one of independent events occurs1 mkAsked 2025
The probabilities that $A$ and $B$ hit a target are $\tfrac{1}{3}$ and $\tfrac{2}{5}$ respectively. If both fire at the target independently, find the probability that the target is hit by at least one of them.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.