0580 · 8.3Probability of combined events

Sample spaces and combined events

0:00Taking a coloured counter from a bag seems straightforward until you take a second one. Putting the first counter back can change the probability of your final result. Choosing one red counter and one blue counter can also happen in two different orders. Combined probability is about keeping track of these possibilities without missing any or counting them twice. By the end of this video, you will understand how sample space diagrams, Venn diagrams and tree diagrams organise combined events.

0:26Each diagram shows a different kind of structure. We will use that structure to decide when to count, when to multiply and when to add. The calculations will follow from the meaning of the events. An outcome is a possible result, while an event is a collection of outcomes satisfying a description. For example, rolling an even number includes three possible die outcomes. When individual outcomes are equally likely, probability equals the number of favourable outcomes divided by the total number of possible outcomes.

0:54That counting rule needs equal likelihood, not just a list of possibilities. When a question joins two events with the word and, both requirements must be satisfied. The word or normally means at least one requirement is satisfied, and that covers both together. Exactly one means that one requirement is satisfied but the other is not. These distinctions matter because a small change in wording can select a different collection of outcomes.

1:18Our first exam-style example uses two fair spinners, spun independently. Spinner U has three equal sectors labelled one, two and three. Spinner T has four equal sectors labelled one, two, three and four. Calculate the probability that the sum of their results is five. A sample space is the collection of all possible outcomes. We put the first spinner results down the rows and the second spinner results across the columns.

1:43Each cell represents one particular combination, and we write its sum inside. Three rows multiplied by four columns gives twelve possible combinations. Each combination is equally likely because the sectors are equal and the spins are independent. Independent means that one result does not affect the probability of the other. The grid keeps the results separate before combining them. That is important because different combinations can produce the same sum.

2:09A sum of five comes from 1 + 4, 2 + 3, or 3 + 2. These occupy three different cells in the grid. The probability is therefore 3/12. Simplifying by dividing the numerator and denominator by three gives 1/4. A common mistake is to treat the possible sums as equally likely. Our sums run from two to seven, but some appear in several cells. A sum of two appears only once, whereas five appears three times.

2:37Count equally likely combinations, rather than assuming that every different total has the same chance.

Venn diagrams and overlaps

2:43A Venn diagram is useful when events can overlap within one group. In our second example, a class contains forty students. Twenty play sport, eighteen play music, and eight do both. One student is chosen at random, and we will calculate several probabilities from these details. Let S represent playing sport and M represent playing music. The rectangle represents all forty students, and each circle represents one event. The overlap contains students who satisfy both descriptions.

3:12Start there because those eight students belong to both totals given in the question. Sport only is 20 − 8, giving twelve students. For music only, subtracting eight from eighteen leaves ten. Adding twelve, eight and ten gives thirty students inside the circles. The number outside both is 40 − 30, giving ten. The intersection of S and M means that both events happen. On this diagram, it is the overlap containing eight students.

3:39We say P of S intersection M for the probability of choosing someone who plays both sport and music. The intersection symbol tells us to use the shared region, not the whole of either circle. The probability of both is 8/40, which simplifies to 1/5. For the union, add the three regions inside either circle to obtain thirty. P of S union M is therefore 30/40. Dividing the numerator and denominator by ten gives 3/4.

4:06The union means sport or music or both. Exactly one excludes the overlap, so its favourable count is 12 + 10, giving twenty-two. Its probability is 22/40. Simplifying by dividing both parts by two gives 11/20. Neither means a student outside both circles. Ten students sit there, so the probability of neither is 10/40, which is 1/4. Neither is the complement of the union, so it also equals 1 − P of S union M.

4:351 − 3/4 gives 1/4, which confirms the diagram. Adding the two circle totals directly would count the overlap twice. Every student who does both is included in the sport total and again in the music total. We correct this by subtracting the overlap once. This explains the union rule rather than leaving it as something to memorise. P of S union M = P of S + P of M − P of S intersection M.

5:03Here, 20/40 + 18/40 − 8/40 gives 30/40. That again simplifies to 3/4. If events cannot happen together, their overlap is zero, so their probabilities can be added directly. Events that cannot happen together are called mutually exclusive. For one chosen student, sport only and music only cannot both be true, so those two events are mutually exclusive. Events S and M are not mutually exclusive here. The eight students in the overlap show that both descriptions can apply.

Tree diagrams with replacement

5:35Tree diagrams organise events that happen in stages. Our third example uses a bag containing three red counters and two blue counters. A counter is chosen at random, replaced, and then another is chosen at random. Calculate the probability of two red counters and the probability of exactly one red counter. Replacement means putting the selected counter back before the next draw. The bag therefore contains the original three red counters and two blue counters each time.

6:02The probability of red is 3/5, and blue has probability 2/5. Replacing the counter and mixing the bag makes the second draw independent of the first. On a tree diagram, write outcomes at the ends of branches and probabilities beside the branches. Each new split shows the possibilities available at that stage. The probabilities leaving any one split must add to one. Here, 3/5 + 2/5 = 1 because red and blue cover every possibility.

6:31Following a complete path means that every event along it must happen. We multiply the probabilities along that path. The first probability selects a fraction of all trials, and the next selects a fraction of those remaining trials. Taking a fraction of a fraction is multiplication, which explains the rule. For two red counters, follow the red branch at both stages. Multiply 3/5 by 3/5. The numerators give nine, and the denominators give twenty-five.

6:58The probability of two red counters is 9/25. Exactly one red counter can happen in two orders. The first draw could be red and the second blue. Alternatively, the first could be blue and the second red. These are different complete paths, and they cannot both describe the same pair of draws. Red then blue has probability 3/5 × 2/5, giving 6/25. For blue then red, 2/5 × 3/5 also gives 6/25.

7:28Add the probabilities of these mutually exclusive paths. Adding the two equal products gives 12/25. The instruction to multiply for and, then add for or, needs careful interpretation. Multiplication follows one complete path using the probabilities appropriate to that path. Addition combines separate paths that cannot happen together. For overlapping events in a Venn diagram, simply adding probabilities would count shared outcomes twice.

Tree diagrams without replacement

7:55Our fourth example uses the same bag, but the first counter is not replaced. Calculate the probability of exactly one red counter in two draws. The first-stage probabilities stay unchanged because the original bag still contains five counters. The second-stage probabilities must reflect which counter was removed. If the first counter is red, two red counters and two blue counters remain. There are four counters altogether, so both second-stage probabilities are 2/4.

8:23If the first counter is blue, three red counters and one blue counter remain. Those branch probabilities are 3/4 and 1/4. Both the total number and the relevant colour count need checking after removal. Subtracting one from the denominator without adjusting the removed colour would give an incorrect probability. Each second-stage split still adds to one. The different probabilities show that the second draw now depends on the first result.

8:49Red then blue has probability 3/5 × 2/4, giving 6/20. The reverse order uses 2/5 × 3/4, also giving 6/20. Add these to obtain 12/20. Dividing both parts by four gives 3/5. Three of the final products are 6/20, and the last is 2/20. Together they make 20/20, which is one, so no path is missing. Now find the probability of at least one red counter. The only path with no red is blue then blue, with probability 2/20.

9:24Subtracting that from one gives 18/20, which simplifies to 9/10. The probability differs from the replacement example because the contents of the bag change. After drawing red, blue makes up a larger fraction of the remaining counters. Removing blue similarly increases the proportion of red counters. Opposite colours therefore become more likely than they were with replacement. Watch especially for omitting one order when a question says one of each colour.

9:50Another common error is keeping the original probabilities after a counter has been removed. A third is treating all complete tree paths as equally likely. Count paths only when their probabilities really are equal; otherwise, calculate each relevant product.

Choosing and checking probability methods

10:04Cambridge questions may ask you to complete a diagram and calculate a probability from it. First identify the precise event, including words such as both, neither, exactly one or at least one. Then select the relevant regions or complete paths. Show their counts or products before writing the final probability. These methods must work without a calculator on Paper two and can also appear on calculator Paper four. Keep fractions exact whenever practical, and simplify your final answer.

10:32If a decimal answer is not exact, normally give three significant figures unless instructed otherwise. Do not round intermediate probabilities because that can affect the final result. The central idea is to organise all possibilities before calculating. A sample space diagram displays combinations, a Venn diagram displays overlaps, and a tree diagram displays successive stages. Multiply along a complete tree path and add mutually exclusive alternatives. Above all, check whether replacement changes the probabilities and whether the wording includes or excludes the overlap.