Rukasha Maths

GCSE maths Beta

Conditional probability and dependent trees · Revision guide

Higher · Probability · 1 page in the PDF

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Learning objectives

  • Calculate dependent probabilities for draws without replacement.
  • Use the condition to identify the correct denominator group.

Key ideas

Without replacement, reduce the total and the count of the colour removed.

Multiply probabilities along a path; add mutually exclusive paths for the requested outcome.

Conditioning restricts the sample space to the stated group.

P(A∣B)=P(A∩B)P(B),P(B)>0P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0

In a two-way table, divide by the total of the group specified by the condition.

Worked examples

Example 1

A bag has four red and three blue counters. Draw two at random without replacement. Find the probability of exactly one of each colour.

P(RB)=47×36=27P(RB)=\frac47\times\frac36=\frac27
P(BR)=37×46=27P(BR)=\frac37\times\frac46=\frac27
P(one of each)=P(RB)+P(BR)P(\text{one of each})=P(RB)+P(BR)
27+27=47\frac27+\frac27=\frac47

Example 2

In the same bag, find the probability the second counter is red given the first is red.

remaining counters:3 red out of 6\text{remaining counters}:3\ \text{red out of}\ 6
P(R2∣R1)=36=12P(R_2\mid R_1)=\frac36=\frac12

Watch out

The second draw has a smaller total without replacement. Include both possible orders when finding exactly one of each colour.