10.03 Multi-stage experiments

Harder probability displays

As we saw in the previous chapter, a tree diagram can be used to represent the outcomes of any multi-stage experiment. As the number of outcomes increases, however, it can become difficult to construct neatly and interpret. One additional type of display that you may like to use for experiments with two events occurring is a two-dimensional display with the possible outcomes of each event listed on each axis. Within the display (which you may hear referred to as a table or an array) you have the option to list the pair of outcomes or perform a calculation depending on the questions you have been asked. These diagrams are particularly useful for experiments involving dice. 

Worked example

Example 1

A six-sided dice is rolled and the following spinner is spun. The sum of the results is then recorded.

 

a)  Use a table to represent all possible outcomes of this experiment.

Do: we need to construct a table with five columns for the spinner and six columns for the dice. Within this we add the values shown on each object.

		Spinner
		1	2	3	4	5
Dice	1	2	3	4	5	6
	2	3	4	5	6	7
	3	4	5	6	7	8
	4	5	6	7	8	9
	5	6	7	8	9	10
	6	7	8	9	10	11

b)  Determine the probability that an even number was spun on the spinner and a sum that is a multiple of four was obtained.

Think:  On the spinner, there are two even numbers, $2$2 and $4$4. We now look down those columns for any sums that are a multiple of four. The only possible sums are those that equal $4$4 and $8$8. 

Do:  $\frac{3}{30}$330​

c)  Determine the probability the sum was greater than $7$7 or a multiple of four.

Think:  Remember that the instruction of "or" means that we only need one of the two possible events to have occurred. I've highlighted these possibilities in the table below.

		Spinner
		1	2	3	4	5
Dice	1	2	3	4	5	6
	2	3	4	5	6	7
	3	4	5	6	7	8
	4	5	6	7	8	9
	5	6	7	8	9	10
	6	7	8	9	10	11

Do:  $\frac{13}{30}$1330​

Practice question

Question 1

Dependent events 

Independent events in multi-stage experiments do not influence the outcome of each other. For example, the events of rolling a dice and tossing a coin are independent. We say that two events are dependent if the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed. The most common type of question you will see involving dependent events is when a selection is made, such as a coloured marble from a bag, and it is not replaced before a second marble is drawn. As there are less marbles available to select, the probabilities associated with the second draw cannot be the same as in the first draw.

Worked example

Example 2

Three cards are chosen at random from a deck of $52$52 cards without replacement. What is the probability of choosing $3$3 kings?

Think: The first important thing to realise is that when we solve probability problems, no two events can occur at exactly the same time. We always have to consider events as occurring one after each other. So, this is a multi-stage experiment where we draw one card, then another, then the last. 

Do: On the first draw we have $52$52 cards, and there are $4$4 kings in the pack, so the chance of drawing a king is $\frac{4}{52}$452​. 

On the second draw, we have selected a king from the pack and put it aside. So now we have $51$51 cards, and $3$3 kings still in the pack, and the probability of choosing a king is $\frac{3}{51}$351​. On the third draw, because we have already removed two kings, we have $50$50 cards and just $2$2 kings to choose from and$P(king)$P(king)$=\frac{2}{50}$=250​.  

This results in the following probability of $3$3 kings being selected:

$\text{P(3 Kings) }$P(3 Kings)	$=$=	$\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}$452​×351​×250​
	$=$=	$\frac{1}{5525}$15525​

Reflect: You might be thinking, "but we didn't draw a tree diagram or any kind of display for this problem - is this ok?" If a question doesn't direct you create a certain diagram, it is your choice how to approach the problem. It is very important that your working shows your thought process, however, not just an answer. In this case, that was done by showing the three probabilities that were multiplied to give the final answer of $\frac{1}{5525}$15525​. 

For a question like this, it might be helpful to draw a few boxes marking the cards we are interested in, then writing the probabilities for each one. 

The practice question below requires you to adjust the probabilities assigned to the branches of your tree diagram in a "without replacement" question. It is important that you don't simplify the initial probabilities on your branches. This because the numbers within each fraction represent the actual number of objects available to be selected, and after the first selection, we will reduce the relevant numerators and denominators by $1$1.

Practice question

Question 2

With or without replacement?

One of the most difficult tasks in multi-stage experiments is to determine whether the subsequent stages of the experiment are dependent or independent, and so whether the probabilities assigned at each stage stay the same or change. 

One specific case you need to be aware of is that when we choose people (or things) from large populations. You may think that in the case of selecting people, we should not replace the first person before selecting the next. This would make the selections dependent and that we should change the probabilities. However, in practice, we realise that the probabilities will be changed in such an insignificant way that it will make only a minuscule difference to our final answer. And so, we consider these selections to be independent. 

Consider the following problem: a nationwide survey found that $64%$64% of people in a small country town have unreliable internet access.  If $3$3 people are selected at random, what is the probability that all three have unreliable internet access?

In this case, we haven't even been told how many people live in the town, so we assume that it is a large enough population that choosing $3$3 people will not affect the probabilities significantly. And so, we calculate our answer as 

$P(I)\times P(I)\times P(I)$P(I)×P(I)×P(I)$=$=$0.64\times0.64\times0.64=0.262144$0.64×0.64×0.64=0.262144 

Practice questions

Question 3

Question 4