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Welcome to GCSE Edexcel Maths revision.

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Topic M 38: Combined events and tree diagrams.

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This video covers Higher tier.

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It includes the shared content and the labelled Higher extensions.

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A tree diagram shows successive stages.

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Label each branch with its probability and each endpoint with its full outcome.

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Branches leaving one node must have probabilities summing to 1.

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Two draws without replacement

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For a complete path, multiply probabilities along the branches.

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This uses each stage's appropriate probability given the earlier outcome.

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To combine separate mutually exclusive paths, add their probabilities.

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And within a path usually leads to multiplication; either path leads to addition.

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Independent events do not affect each other's probabilities.

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For a fair coin and independent fair die, probability of open bracket head and 6 close bracket equals 1 over 2 multiplied by 1 over 6 equals 1 over 12 .

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With replacement, a random object is returned before the next draw, so the bag's composition is restored.

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Without replacement, totals and favourable counts change.

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Worked example: A bag has 3 red and 2 blue counters.

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Without replacement, probability of open bracket two red close bracket equals 3 over 5 multiplied by 2 over 4 equals 3 over 10 .

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The second denominator is 4, not 5.

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For the same bag, probability of open bracket one red and one blue close bracket equals 3 over 5 multiplied by 2 over 4 plus 2 over 5 multiplied by 3 over 4 equals 3 over 5 .

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Include both orders.

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With replacement, probability of open bracket two red close bracket would be 3 over 5 multiplied by 3 over 5 equals 9 over 25 .

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Do not treat draws as independent unless the situation justifies it.

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To find at least one,

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sometimes use the complement: probability of open bracket at least one success close bracket equals 1 minus probability of open bracket no successes close bracket .

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This can be quicker than listing many paths.

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A two-way table cross-classifies two attributes.

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Row totals, column totals and the grand total must agree; derive missing counts before computing probabilities.

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For a randomly selected person from the whole group, use the grand total as denominator.

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Choosing from an already specified subgroup changes the denominator.

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probability of open bracket A given B close bracket means the probability of A given B has happened.

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Restrict the sample space to B: probability of open bracket A given B close bracket equals the fraction with numerator open bracket probability of open bracket A intersection B close bracket close bracket and denominator open bracket probability of open bracket B close bracket close bracket ,

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end fraction ,

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provided probability of open bracket B close bracket is greater than 0.

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Worked example: Of 30 students, 12 study Spanish and 5 study both Spanish and French.

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Given a Spanish student is selected, probability of open bracket French given Spanish close bracket equals 5 over 12 , not 5 over 30 .

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On a tree,

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later branches already represent probabilities conditional on the route taken. probability of open bracket A intersection B close bracket equals probability of open bracket A close bracket probability of open bracket B given A close bracket ;

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for independent events,

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probability of open bracket B given A close bracket equals probability of open bracket B close bracket .

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Use expected-frequency representations for conditional problems when useful.

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If 1000 items include 100 defective,

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and a test flags 90 defective and 45 non-defective items,

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then probability of open bracket defective given flagged close bracket equals 90 over 135 equals 2 over 3 .

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Do not reverse a condition: probability of open bracket defective given flagged close bracket and probability of open bracket flagged given defective close bracket answer different questions.

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For the example, the latter is 90 over 100 equals 0.9.

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That completes Combined events and tree diagrams.

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Revisit the notes and test yourself on the revision website.