General addition rule?

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

Define independent events.

The occurrence of one does not change the probability of the other.

Can mutually exclusive events with non-zero probabilities be independent?

No: P(A\cap B)=0 but P(A)P(B)0.

Why are X=1 and X=2 mutually exclusive for a discrete random variable?

The variable cannot take both values at once.

What must you assume to multiply probabilities of repeated trials?

That the trials are independent.

Mutually exclusive and independent eventsAQA A-Level Maths: Flashcards

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Define mutually exclusive events.

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Define mutually exclusive events.: Events that cannot occur together, so $P(A\cap B)=0$ .
Addition rule for mutually exclusive events?: $P(A\cup B)=P(A)+P(B)$
General addition rule?: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
Define independent events.: The occurrence of one does not change the probability of the other.
Multiplication rule for independent events?: $P(A\cap B)=P(A)\times P(B)$
How do you show two events are independent?: Show that $P(A\cap B)=P(A)P(B)$ .
Can mutually exclusive events with non-zero probabilities be independent?: No: $P(A\cap B)=0$ but $P(A)P(B)>0$ .
If $A$ and $B$ are independent, are $A'$ and $B$ independent?: Yes.
$P(A)=0.3$ , $P(B)=0.5$ , independent. Find $P(A\cup B)$ .: $0.3+0.5-0.15=0.65$
$P(A)=0.2$ , $P(B)=0.7$ , mutually exclusive. Find $P(\text{neither})$ .: $1-0.9=0.1$
How do you find the probability of 'at least one'?: $1-P(\text{none})$
Why are $X=1$ and $X=2$ mutually exclusive for a discrete random variable?: The variable cannot take both values at once.
What must you assume to multiply probabilities of repeated trials?: That the trials are independent.

Exam questions on Mutually exclusive and independent events

Events $A$ and $B$ are mutually exclusive, with $P(A)=0.4$ and $P(B)=0.35$ .
Determine whether $A$ and $B$ are independent.2 marks
Events $A$ and $B$ are independent, with $P(A)=0.6$ and $P(B)=0.5$ .
Find $P(A'\cap B)$ .2 marks
A factory has two machines, $X$ and $Y$ . An item from $X$ is defective with probability $0.04$ and an item from $Y$ is defective with probability $0.06$ , independently of each other. A pair consists of one item from each machine.
Find the probability that exactly one item in a pair is defective.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).