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How do you handle conditional probability and tell independent from dependent events?

Formal notation for independent and conditional events; the multiplication law for independent events; the conditional probability formula; dependent events such as selection without replacement.

A focused answer to Edexcel GCSE Statistics on conditional probability and independence, covering the formal notation for independent and conditional events, the multiplication law, the conditional probability formula, and dependent events such as selection without replacement.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Independent events and notation
  3. Conditional probability
  4. Dependent events and selection without replacement
  5. Reading conditional probabilities from a two-way table
  6. Testing for independence
  7. Venn diagrams for conditional probability

What this dot point is asking

Edexcel codes 3p.08 and 3p.09 require you to know and apply the formal notation for independent events and for conditional probability, the multiplication law for independent events, and the conditional probability formula. The headline application is selection without replacement, where the probabilities change after the first selection, making the events dependent.

Independent events and notation

For independent events the multiplication law applies directly:

Independence typically arises with replacement or with genuinely separate experiments (a coin and a dice). The notation P(AB)P(A \mid B), read "the probability of AA given BB", is the formal way Edexcel writes conditional probability.

Conditional probability

Rearranged, this gives the general multiplication law P(A and B)=P(A)×P(BA)P(A \text{ and } B) = P(A) \times P(B \mid A), which works even when the events are dependent. The simplest place to see conditional probability is a two-way table: "given that the person is a woman" means you only look at the women's row, and divide within it.

Dependent events and selection without replacement

When an item is selected and not replaced, the totals change, so the second selection is dependent on the first. On a tree diagram, the second set of branches has different probabilities depending on what was picked first. For example, drawing two counters from a bag without replacement: if the first is red, there is one fewer red and one fewer counter overall for the second draw. You multiply along the branches using these adjusted probabilities, exactly as P(A)×P(BA)P(A) \times P(B \mid A).

Reading conditional probabilities from a two-way table

A two-way table makes conditional probability concrete. To find P(passwoman)P(\text{pass} \mid \text{woman}), you ignore the men entirely: take the number of women who passed and divide by the total number of women. To find the unconditional P(pass)P(\text{pass}), you divide total passes by the grand total. Keeping clear which total is the denominator is the key skill.

Testing for independence

You can use probabilities from a table to test whether two events are independent. Events AA and BB are independent if P(AB)=P(A)P(A \mid B) = P(A), or equivalently if P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B). So compare the conditional probability with the unconditional one: if "the probability of passing given you are a woman" equals "the probability of passing" overall, gender and passing are independent; if they differ, the events are dependent (gender is associated with passing). This is a common higher-mark task, and it links conditional probability to the idea of association from the correlation topics.

Venn diagrams for conditional probability

A Venn diagram is often the clearest tool for conditional probability. Once the numbers in each region (only AA, only BB, both, neither) are filled in, P(BA)P(B \mid A) is read as "the number in both, divided by the total number in AA". For example, if AA contains 3030 people in total and 1212 of them are also in BB, then P(BA)=1230=0.4P(B \mid A) = \frac{12}{30} = 0.4. Drawing the Venn diagram first turns an abstract conditional probability into a simple counting problem, and the same diagram answers "and", "or" and "not" questions too.

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 1ST0 20214 marksA box contains 77 apples, of which 33 are bruised. Two apples are taken at random without replacement. Find the probability that both apples are bruised.
Show worked answer →

Without replacement the events are dependent, so the second probability changes after the first apple is removed.

First bruised: P=37P = \frac{3}{7}. Given the first was bruised, 22 bruised remain out of 66: P=26=13P = \frac{2}{6} = \frac{1}{3}.

P(both bruised)=37×26=642=17P(\text{both bruised}) = \frac{3}{7} \times \frac{2}{6} = \frac{6}{42} = \frac{1}{7}.

Markers reward recognising dependence (without replacement), reducing both numerator and denominator for the second pick, and the answer 17\frac{1}{7}.

Edexcel 1ST0 20224 marksA two-way table shows 100100 people by gender and whether they passed a test: 4040 men passed, 1010 men failed, 3535 women passed, 1515 women failed. (a) Find the probability a randomly chosen person passed. (b) Given that a person is a woman, find the probability she passed.
Show worked answer →

(a) Total passes =40+35=75= 40 + 35 = 75, out of 100100, so P(pass)=75100=0.75P(\text{pass}) = \frac{75}{100} = 0.75.

(b) This is conditional on being a woman. There are 35+15=5035 + 15 = 50 women, of whom 3535 passed: P(passwoman)=3550=0.7P(\text{pass} \mid \text{woman}) = \frac{35}{50} = 0.7.

Markers reward 0.750.75 for the unconditional probability, and restricting to the 5050 women to get the conditional probability 0.70.7.

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