0580 · 8.4Conditional probability

Conditional probability as a filter

0:00Imagine choosing a student at random from your school. You are then told that the student belongs to the swimming club. That information changes the group you should consider when calculating probabilities. Conditional probability is about using information like this correctly. We will explore the same idea through Venn diagrams, tables and trees. The key is to recognise which outcomes are still possible after receiving the information. A student outside the swimming club is no longer a possible choice.

0:25We therefore calculate using swimming club members, rather than everyone in the school. The information changes our calculation because it narrows down the possible outcomes. You need one familiar starting idea. When individual outcomes are equally likely, probability equals the count of favourable outcomes divided by the total number of possibilities. Favourable outcomes are those that satisfy the question. For conditional probability, both counts must come from the group allowed by the information given.

0:52A conditional probability measures how likely an event is, given that something else has already happened. The condition is the information we are told to accept as true. Questions often introduce it using the words given that. They might instead say that someone is selected from a particular group. That wording also restricts the possible outcomes. Think of the condition as a filter. First, keep only the outcomes that satisfy that condition.

1:16Next, identify which of those remaining outcomes also satisfy the event you want. Finally, compare that favourable part with the whole filtered group. You do not need special conditional probability notation for this syllabus.

Conditional probability with Venn diagrams

1:29Our first exam-style example uses forty students. Twelve belong only to the reading club, eight belong to both clubs, and ten belong only to the swimming club. Another ten belong to neither club. One student is selected at random, and we are told that this student belongs to the swimming club. Calculate the probability that this student also belongs to the reading club. The condition restricts us to the swimming circle, including the overlap with reading.

1:53There are eight students in the overlap. The other ten swimmers do not belong to the reading club, so they are not favourable outcomes. Add eight and ten to find eighteen students in the swimming club. Of these, eight also belong to the reading club. The required probability is therefore 8/18. Dividing the numerator and denominator by two gives 4/9. The denominator is eighteen because we already know the chosen student is a swimmer.

2:17Using 8/40 would answer a different question. It gives the probability that a randomly selected student from the whole group belongs to both clubs. Our question already guarantees membership of the swimming club. Students outside that circle must therefore be excluded from the denominator, even though they appear in the diagram. Now reverse the information and calculate the probability of swimming, given membership of the reading club. The reading group contains 12 + 8, giving twenty students.

2:43Eight of those students also swim, so the probability is 8/20. Simplifying gives 2/5, which differs from our previous answer because the condition has changed. The overlap remains the favourable region in both questions, but the denominator changes. This is why reversing the wording can change a conditional probability. Also, belonging to both clubs does not mean one activity caused the other. Conditional probability describes how information affects the possible outcomes, not whether one event causes another.

3:12Some Venn diagrams show probabilities instead of counts, and the method stays the same. Suppose the reading-only region holds zero point three, and the overlap holds zero point one five. Swimming alone has zero point two five, leaving zero point three outside both circles. Given swimming, the restricted total is 0.15 + 0.25, which is zero point four. Dividing zero point one five by zero point four gives 3/8.

Conditional probability from tables

3:37Our second exam-style example uses a table of sixty students. Thirty travel by bus, and thirty walk. Of the bus travellers, six are late and twenty-four are on time. The walkers split into four late and twenty-six on time. A student is selected at random from those who are late. Calculate the probability that the selected student travels by bus. The phrase from those who are late tells us the condition.

4:00Only the late column remains relevant, so neither the whole table nor the bus row supplies our denominator. Within that column, the bus and late cell supplies the favourable count. There are 6 + 4, giving ten late students altogether. Six of these travel by bus. The probability is 6/10, which simplifies to 3/5. We used the column total because the condition was being late. A row total would be appropriate only if the condition restricted us to that row.

4:26For comparison, calculate the probability of being late, given that the student travels by bus. The relevant group now contains 6 + 24, giving thirty students. Six satisfy the event of being late, so the probability is 6/30. This simplifies to 1/5, demonstrating again that the order of the information matters. Tables and Venn diagrams express the same underlying idea. In a table you restrict attention to the right row or column, and in a Venn diagram to the right circle or region.

4:55Then you count the favourable outcomes within that restricted group, rather than using the original total automatically.

Conditional probability with tree diagrams

5:01Our third exam-style example uses a bag containing three red counters and two blue counters. Two counters are selected at random without replacement. The tree labels red outcomes with R and blue outcomes with B. Each second-stage probability already depends on the first result, because the counter drawn earlier is not returned to the bag. After a red first draw, two counters of each colour remain. The next probabilities are therefore 2/4 for each colour.

5:26If the first counter is blue, three red and one blue remain. That gives 3/4 for red and 1/4 for blue. Each pair of branches adds to one. Begin with the probability that the second counter is red, given that the first one drawn is blue. Removing a blue counter leaves all three reds available. The remaining total is 5 − 1, giving four counters. The answer is 3/4, which we can read directly from the appropriate second-stage branch.

5:52A condition does not always match a single branch. Suppose instead that we know the second counter is red. Calculate the probability that the first counter was blue. This information allows two complete paths through the tree: red then red, or blue then red. We compare the favourable path with everything that satisfies the condition. For red then red, multiply 3/5 by 2/4. This gives 6/20, which simplifies to 3/10.

6:19For blue then red, we multiply 2/5 by 3/4. That result is also 6/20, or 3/10. Add the two allowed path probabilities to get 6/10 for a red second counter. The favourable path is blue then red, with probability 3/10. Dividing 3/10 by 6/10 gives 3/6, which simplifies to one half. Picture ten equally likely shares: six end with red, and three of those six began with blue. Simply giving 3/10 would calculate the probability of blue then red before receiving any information.

6:52Our conditional question guarantees that the second counter is red. We must therefore exclude paths ending in blue and compare only the remaining paths. With trees, add the probabilities of the allowed routes, because different paths need not be equally likely. In an examination, identify the condition before calculating, then show the relevant total and the favourable part. These methods must work without a calculator on Paper Two, and they also apply on calculator Paper Four.

7:19Exact simplified fractions avoid unnecessary rounding. If a decimal approximation is required, follow the stated accuracy, normally three significant figures, and do not round intermediate calculations. The central idea is that given information changes the group of possible outcomes. Venn diagrams show that group as a region, tables as a row or column, and trees as branches or paths. Keep only outcomes satisfying the condition, then identify the favourable part within them.

7:45That reasoning is what makes conditional probability work.