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How do you measure the chance of events, and how do you combine probabilities of related events?

Probability of events, mutually exclusive and independent events, the addition and multiplication laws, conditional probability, and Venn and tree diagrams.

A focused answer to the Edexcel A-Level Mathematics probability content, covering the probability of events, mutually exclusive and independent events, the addition and multiplication laws, conditional probability, and the use of Venn and tree diagrams.

Generated by Claude Opus 4.89 min answer

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What this dot point is asking

Edexcel wants you to find probabilities of single and combined events, use the language of mutually exclusive and independent events, apply the addition and multiplication laws, calculate conditional probabilities, and use Venn diagrams and tree diagrams to model situations.

The answer

The laws of probability

Conditional probability

Diagrams

A Venn diagram shows each event as a region; the overlap is the intersection ABA \cap B and the whole shaded area is the union ABA \cup B. A tree diagram lists outcomes in stages, with each branch labelled by its probability. You multiply along a path to find the probability of a sequence of outcomes, then add the relevant paths to combine several ways of reaching the same overall result.

Mutually exclusive and independent events

Complementary events

The complement of AA, written AA', is the event that AA does not happen, and P(A)=1P(A)P(A') = 1 - P(A). Using the complement is often the quickest route to an answer when the phrase "at least one" appears, because the complement of "at least one" is "none".

Examples in context

Try this

Q1. Events AA and BB satisfy P(A)=0.5P(A) = 0.5, P(B)=0.4P(B) = 0.4 and P(AB)=0.2P(A \cap B) = 0.2. Find P(AB)P(A \cup B). [2 marks]

  • Cue. 0.5+0.40.2=0.70.5 + 0.4 - 0.2 = 0.7.

Q2. Using the values in Q1, find P(AB)P(A \mid B). [2 marks]

  • Cue. 0.20.4=0.5\dfrac{0.2}{0.4} = 0.5, so AA and BB are independent.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksIn a class, P(A)=0.55P(A) = 0.55 study French, P(B)=0.40P(B) = 0.40 study German, and P(AB)=0.15P(A \cap B) = 0.15 study both. Calculate the probability that a student studies French or German, the probability that a student studies neither, and P(AB)P(A \mid B).
Show worked answer →

French or German uses the addition law (M1): P(AB)=0.55+0.400.15=0.80P(A \cup B) = 0.55 + 0.40 - 0.15 = 0.80 (A1).

Neither is the complement (M1): 1P(AB)=10.80=0.201 - P(A \cup B) = 1 - 0.80 = 0.20 (A1).

Conditional probability (M1): P(AB)=P(AB)P(B)=0.150.40=0.375P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.15}{0.40} = 0.375 (A1).

Markers reward the addition law, the complement, and the conditional formula.

Edexcel 20225 marksA bag contains 55 red and 33 blue counters. Two are drawn without replacement. Calculate the probability that both are red and the probability that the counters are different colours.
Show worked answer →

Both red, multiplying along the tree branches (M1): P(RR)=58×47=2056=514P(RR) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14} (A1).

Different colours is red-then-blue or blue-then-red (M1): 58×37+38×57=1556+1556=3056=1528\dfrac{5}{8} \times \dfrac{3}{7} + \dfrac{3}{8} \times \dfrac{5}{7} = \dfrac{15}{56} + \dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28} (A1, A1).

Markers reward conditional branch probabilities, both ordered cases for mixed colours, and adding them.

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