The two events are dependent because there are fewer sweets to choose from when she picks the second time.
10.2 Identities
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10.2 Identities (EMCJX)
The addition rule (also called the sum rule) for any \(\text{2}\) events, \(A\) and \(B\) is \[P(A \text{ or } B) = P(A) + P(B)  P(A \text{ and } B)\] This rule relates the probabilities of \(\text{2}\) events with the probabilities of their union and intersection.
The addition rule for \(\text{2}\) mutually exclusive events is \[P(A \text{ or } B) = P(A) + P(B)\] This rule is a special case of the previous rule. Because the events are mutually exclusive, \(P(A \text{ and } B) = 0\).
The complementary rule is \[P(\text{not } A) = 1  P(A)\] This rule is a special case of the previous rule. Since \(A\) and \((\text{not } A)\) are complementary, \(P(A \text{ or } (\text{not } A)) = 1\).
The product rule for independent events \(A\) and \(B\) is:
\[P(A\text{ and } B) = P(A) \times P(B)\]If two events \(A\) and \(B\) are dependent then: \[P(A\text{ and } B) \ne P(A) \times P(B)\]
Just because two events are mutually exclusive does not necessarily mean that they are independent. To test whether events are mutually exclusive, always check that \(P(A \text{ and } B) = 0\). To test whether events are independent, always check that \(P(A \text{ and } B) = P(A) \times P(B)\). See the exercises below for examples of events that are mutually exclusive and independent in different combinations.
Worked example 1: Dependent and independent events
Write down which of the following events are dependent and which are independent:
 The student council chooses a head student and then a deputy head student.
 A bag contains blue marbles and red marbles. You take a red marble out of the bag and then throw it back in again before you take another marble out of the bag.
Ask the question: Did the available choices change for the second event because of the first event?
 Yes, because after selecting the head student there are fewer council members available to choose for the deputy head student position. Therefore the two events are dependent.
 No, because when you throw the first marble back into the bag, there are the same number and colour composition of choices for the second marble. Therefore the two events are independent.
Worked example 2: Independent and dependent events
A bag contains \(\text{3}\) yellow and \(\text{4}\) black beads. We remove a random bead from the bag, record its colour and put it back into the bag. We then remove another random bead from the bag and record its colour.
 What is the probability that the first bead is yellow?
 What is the probability that the second bead is black?
 What is the probability that the first bead is yellow and the second bead is black?
 Are the first bead being yellow and the second bead being black independent events?
Probability of a yellow bead first
Since there is a total of \(\text{7}\) beads, of which \(\text{3}\) are yellow, the probability of getting a yellow bead is \[P(\text{first bead yellow}) = \frac{3}{7}\]
Probability of a black bead second
The problem states that the first bead is placed back into the bag before we take the second bead. This means that when we draw the second bead, there are again a total of \(\text{7}\) beads in the bag, of which \(\text{4}\) are black. Therefore the probability of drawing a black bead is \[P(\text{second bead black}) = \frac{4}{7}\]
Probability of yellow first and black second
When drawing two beads from the bag, there are \(\text{4}\) possibilities. We can get
 a yellow bead and then another yellow bead;
 a yellow bead and then a black bead;
 a black bead and then a yellow bead;
 a black bead and then another black bead.
We want to know the probability of the second outcome, where we have to get a yellow bead first. Since there are \(\text{3}\) yellow beads and \(\text{7}\) beads in total, there are \(\frac{3}{7}\) ways to get a yellow bead first. Now we put the first bead back, so there are again \(\text{3}\) yellow beads and \(\text{4}\) black beads in the bag. Therefore there are \(\frac{4}{7}\) ways to get a black bead second if the first bead was yellow. This means that there are \[\frac{3}{7}\times\frac{4}{7}=\frac{12}{\text{49}}\] ways to get a yellow bead first and a black bead second. So, the probability of getting a yellow bead first and a black bead second is \(\frac{12}{49}\).
Dependent or independent?
According to the definition, events are independent if and only if \[P(A\text{ and }B) = P(A) \times P(B)\]
In this problem:
 \(P(\text{first bead yellow}) = \frac{3}{7}\)
 \(P(\text{second bead black}) = \frac{4}{7}\)
 \(P(\text{first bead yellow and second bead black}) = \frac{12}{49}\)
Worked example 3: Independent and dependent events
In the previous example, we picked a random bead and put it back into the bag before continuing. This is called sampling with replacement. In this worked example, we will follow the same process, except that we will not put the first bead back into the bag. This is called sampling without replacement.
So, from a bag with \(\text{3}\) red and \(\text{5}\) green beads, we remove a random bead and record its colour. Then, without putting back the first bead, we remove another random bead from the bag and record its colour.
 What is the probability that the first bead is red?
 What is the probability that the second bead is green?
 What is the probability that the first bead is red and the second bead is green?
 Are the first bead being red and the second bead being green independent events?
Count the number of outcomes
We will examine the number ways in which we can get the different possible outcomes when removing \(\text{2}\) beads. The possible outcomes are
 a red bead and then another red bead (RR);
 a red bead and then a green bead (RG);
 a green bead and then a red bead (GR);
 a green bead and then another green bead (GG).
For the first outcome, we have to get a red bead first. Since there are \(\text{3}\) red beads and \(\text{8}\) beads in total, there are \(\frac{3}{8}\) ways to get a red bead first. After we have taken out a red bead, there are now \(\text{2}\) red beads and \(\text{5}\) green beads left. Therefore there are \(\frac{2}{7}\) ways to get a red bead second if the first bead was also red. This means that there are \[\frac{3}{8}\times\frac{2}{7}=\frac{6}{56}=\frac{3}{28}\] ways to get a red bead first and a red bead second. The probability of the first outcome is \(\frac{3}{28}\).
For the second outcome, we have to get a red bead first. As in the first outcome, there are \(\frac{3}{8}\) ways to get a red bead first; and there are now \(\text{2}\) red beads and \(\text{5}\) green beads left. Therefore there are \(\frac{5}{7}\) ways to get a green bead second if the first bead was red. This means that there are \[\frac{3}{8}\times\frac{5}{7}=\frac{15}{56}\] ways to get a red bead first and a green bead second. The probability of the second outcome is \(\frac{15}{56}\).
In the third outcome, the first bead is green and the second bead is red. There are \(\frac{5}{8}\) ways to get a green bead first; and there are now \(\text{4}\) green beads and \(\text{3}\) red beads left. Therefore there are \(\frac{3}{7}\) ways to get a red bead second if the first bead was green. This means that there are \[\frac{5}{8}\times\frac{3}{7}=\frac{15}{56}\] ways to get a red bead first and a green bead second. The probability of the third outcome is \(\frac{15}{56}\).
In the fourth outcome, the first and second beads are both green. Since there are \(\text{5}\) green beads and \(\text{8}\) beads in total, there are \(\frac{5}{8}\) ways to get a green bead first. After we have removed a green bead, \(\text{3}\) red beads and \(\text{4}\) green beads remain in the bag. Therefore there are \(\frac{4}{7}\) ways to get a green bead second if the first bead was also green. This means that there are \[\frac{5}{8}\times\frac{4}{7}=\frac{20}{56}=\frac{5}{14}\] ways to get a green bead first and a green bead second. Therefore the probability of the fourth outcome is \(\frac{5}{14}\).
To summarise, these are the possible outcomes and their probabilities:
 first bead red and second bead red (RR): \(\frac{3}{28}\);
 first bead red and second bead green (RG): \(\frac{15}{56}\);
 first bead green and second bead red (GR): \(\frac{15}{56}\);
 first bead green and second bead green (GG): \(\frac{5}{14}\).
Probability of a red bead first
To determine the probability of getting a red bead on the first draw, we look at all of the outcomes that contain a red bead first. These are
 a red bead and then another red bead (RR);
 a red bead and then a green bead (RG).
The probability of the first outcome is \(\frac{3}{28}\) and the probability of the second outcome is \(\frac{15}{56}\). By adding these two probabilities, we see that the probability of getting a red bead first is \[P(\text{first bead red}) = \frac{3}{28} + \frac{15}{56} = \frac{6}{56} + \frac{15}{56} =\frac{21}{56} =\frac{3}{8}\]
This is the same as in the previous worked example, which should not be too surprising since the probability of the first bead being red is not affected by whether or not we put it back into the bag before drawing the second bead.
Probability of a green bead second
To determine the probability of getting a green bead on the second draw, we look at all of the outcomes that contain a green bead second. These are
 a red bead and then a green bead (RG);
 a green bead and then another green bead (GG).
The probability of the first outcome is \(\frac{15}{56}\) and the probability of the second outcome is \(\frac{5}{14}\). By adding these two probabilities, we see that the probability of getting a green bead second is \[P(\text{second bead green}) = \frac{15}{56} + \frac{5}{14} = \frac{15}{56} + \frac{20}{56} =\frac{35}{56} = \frac{5}{8}\]
Probability of red first and green second
We have already calculated the probability that the first bead is red and the second bead is green (RG). It is \(\frac{15}{56}\).
Dependent or independent?
According to the definition, events are independent if and only if \[P(A\text{ and }B) = P(A) \times P(B)\]
In this problem:
 \(P(\text{first bead red}) = \frac{3}{8}\)
 \(P(\text{second bead green}) = \frac{5}{8}\)
 \(P(\text{first bead red and second bead green}) = \frac{15}{56}\)
Worked example 4: The addition rule for \(\text{2}\) mutually exclusive events
A sample space, \(S\), consists of all natural numbers less than \(\text{16}\). \(A\) is the event of drawing an even number at random. \(B\) is the event of randomly drawing a prime number. Are \(A\) and \(B\) mutually exclusive events? Prove this using the addition rule.
Write down the sample space
The sample space contains all the natural numbers less than \(\text{16}\). \[S = \{1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15\}\]
Write down the events
The even natural numbers less than \(\text{16}\) are \[A = \{2; 4; 6; 8; 10; 12; 14\}\]
The prime numbers less than \(\text{16}\) are \[B = \{2; 3; 5; 7; 11; 13\}\]
We can already see from writing down our event sets that \(A\) and \(B\) share the event \(\text{2}\) and are thus not mutually exclusive. However, the question asked us to prove this using the addition rule so let's go ahead and do that.
Compute the probabilities
The probability of an event is the number of outcomes in the event set divided by the number of outcomes in the sample space. There are \(\text{15}\) outcomes in the sample space.

Since there are \(\text{7}\) outcomes in the \(A\) event set, \(P(A) = \dfrac{n(A)}{n(S)} = \dfrac{7}{15}\).

Since there are \(\text{6}\) outcomes in the \(B\) event set, \(P(B) = \dfrac{n(B)}{n(S)} = \dfrac{6}{15} = \dfrac{2}{5}\).

The event that is a prime number or an even number is the union of the above two event sets. There are \(\text{12}\) elements in the union of the event sets, so \(P(A \text{ or } B)= \dfrac{n(A \text{ or } B)}{n(S)} = \dfrac{12}{15}\).
Are the two events mutually exclusive?
To test whether two events are mutually exclusive, we can use the addition rule. For two mutually exclusive events, \[P(A \text{ and } B) \text{ is an empty set, therefore } P(A \text{ or } B) = P(A) + P(B)\] \[\text{Since } P(A \text{ or } B) = \frac{12}{15} \text{ and } P(A) + P(B) = \frac{6}{15} + \frac{7}{15} = \frac{13}{15}\] \[P(A \text{ or } B) \ne P(A) + P(B)\] Therefore the intersection of \(A\) and \(B\) is nonzero. This means that the events \(A\) and \(B\) are not mutually exclusive.
Worked example 5: The addition rule
The probability that a person drinks tea is \(\text{0,5}\). The probability that a person drinks coffee is \(\text{0,4}\). The probability that a person drinks tea, coffee or both is \(\text{0,8}\). Determine the probability that a person drinks tea and coffee.
Determine if the events are mutually exclusive
Let the probability that a person drinks tea \(= P(T)\) and the probability that a person drinks coffee \(= P(C)\).
From the information provided in the question, we know that:
 \(P(T) = \text{0,5}\)
 \(P(C) = \text{0,4}\)
 \(P(T \text{ or } C) = \text{0,8}\)
 \(P(T) + P(C) = \text{0,5} + \text{0,4} = \text{0,9}\)
 \(\text{Therefore } P(T \text{ or } C) \ne P(T) + P(C)\)
Therefore the events are not mutually exclusive.
Compute the probability that a person drinks tea and coffee
Using the addition rule, we know that:
\[\begin{array}{rll} P(A \text{ or } B) & = P(A) + P(B)  P(A \text{ and } B) \\ \therefore P(T \text{ or } C) & = P(T) + P(C)  P(T \text{ and } C) &\\ \text{0,8}& = \text{0,4} + \text{0,5}  P(T \text{ and } C) &\\ \therefore P(T \text{ and } C) & = \text{0,4} + \text{0,5}  \text{0,8} = \text{0,1} & \end{array}\]Therefore the probability that a person drinks tea and coffee is \(\text{0,1}\).
Worked example 6: The complementary rule
Joe wants to open a tuckshop at his school but is not sure which cool drinks to stock. Before opening, he interviewed a sample of learners to determine what types of cool drinks they like. From his research, he determined that the probability that a learner drinks cola is \(\text{0,3}\), the probability that a learner drinks lemonade is \(\text{0,6}\) and the probability that a learner drinks neither is \(\text{0,2}\). Determine:
 the probability that a learner drinks cola and lemonade.
 the probability that a learner drinks only cola or only lemonade.
Determine the probability that a learner drinks cola or lemonade
Let the probability that a learner drinks cola \(= P(C)\) and the probability that a learner drinks lemonade \(= P(L)\).
From the information provided in the question, we know that:
 \(P(C) = \text{0,3}\)
 \(P(L) = \text{0,6}\)
 \(P(\text{not } (C \text{ or } L)) = \text{0,2}\)
Using the complementary rule:
\begin{align*} P(\text{not } (C \text{ or } L)) &= \text{1}  P(C \text{ or } L) \\ \therefore P(C \text{ or } L) &= \text{1}  P(\text{not } (C \text{ or } L)) \\ & = \text{1}  \text{0,2} \\ & = \text{0,8} \end{align*}Calculate the probability that a learner drinks cola and lemonade
Using the addition rule:
\begin{align*} P(C \text{ or } L) & = P(C) + P(L)  P(C \text{ and } L) \\ \therefore P(C \text{ and } L) & = P(C) + P(L)  P(C \text{ or } L) \\ & = \text{0,3} + \text{0,6}  \text{0,8} \\ & = \text{0,1} \end{align*}The probability that a learner drinks both cola and lemonade is \(\text{0,1}\).
Determine the probability that a learner drinks only cola or only lemonade
This question requires us to calculate the probability that a learner likes lemonade or cola but not both of them. We can write this as: \[P(\text{only } C \text{ or only } L) = P(C \text{ or } L)  P(C \text{ and } L)\] since a learner can like either cola or lemonade but not both.
We already know \(P(C \text{ or } L) = \text{0,8}\) and \(P(C \text{ and } L) = \text{0,1}\), therefore the probability of a learner drinking only cola or only lemonade is \(\text{0,7}\).
The product and addition rules
Determine whether the following events are dependent or independent and give a reason for your answer:
The two events are independent because the outcome of the first throw has no effect on the outcome of the second throw.
The two events are independent because the set of cards in the deck is unchanged each time Celia chooses one randomly.
The two events are independent because there are the same collection of beads each time Thandi chooses one.
The two events are dependent because Mark has fewer calculators to choose from when he picks again.
Given that \(P(A) = \text{0,7}\); \(P(B) = \text{0,4}\) and \(P(A \text{ and } B) = \text{0,28}\),
are events \(A\) and \(B\) mutually exclusive? Give a reason for your answer.
are the events \(A\) and \(B\) independent? Give a reason for your answer.
For events to be independent: \(P(A) × P(B) = P(A \text{ and } B)\). \(P(A) \times P(B) = \text{0,7} \times \text{0,4} = \text{0,28} = P(A \text{ and } B)\). Therefore the events are independent.
In the following examples, are \(A\) and \(B\) dependent or independent?
Therefore the events are dependent.
Therefore the events are independent.
\(n(A) = 5; n(B) = 4; n(S) = 20 \text{ and } n(A \text{ or } B) = 8\).
For \(A\) and \(B\) to be mutually exclusive: \(P(A) + P(B) = P(A \text{ or } B)\).
\[\frac{5}{20} + \frac{4}{20} = \frac{9}{20} \ne \frac{8}{20}\]The events are therefore not mutually exclusive.
For \(A\) and \(B\) to be independent, \(P(A) \times P(B) = P(A \text{ and } B)\)
\begin{align*} P(A \text{ or } B) &= P(A) + P(B)  P(A \text{ and } B) \\ \therefore P(A \text{ and } B) &= P(A) + P(B)  P(A \text{ or } B) \\ &= \frac{5}{20} + \frac{4}{20}  \frac{8}{20} \\ &=\frac{1}{20} \\ P(A) \times P(B) &= \frac{1}{4} \times \frac{1}{5} \\ & = \frac{1}{20} = P(A \text{ and } B) \end{align*} Therefore the events are independent.Simon rolls a die twice. What is the probability of getting:
two threes.
\[P(\text{two threes}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\]
a prime number then an even number.
There are 3 possible prime numbers on a die, namely, 2, 3, and 5, and there are 3 possible even numbers, namely, 2, 4, and 6.
\begin{align*} P(\text{prime number then even number}) &= P(\text{prime number}) \times P(\text{even number}) \\ &= \frac{3}{6} \times \frac{3}{6} = \frac{9}{36} = \frac{1}{4} \end{align*}no threes.
If no threes are rolled then, for each of the events, 5 possibilities remain.
\[P(\text{no threes}) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}\]only one three.
In the sample space there are 36 possible outcomes. There are two ways of getting only one three: either getting a 3 on the first throw and a number other than three on the second throw or a 3 on the second throw and a number other than 3 on the first throw. The outcomes containing only one 3 are: \((3; 1)\); \((3; 2)\); \((3; 4)\); \((3; 5)\); \((3; 6)\); \((1; 3)\); \((2; 3)\); \((4; 3)\); \((5; 3)\); \((6; 3)\).
\[P(\text{only one } 3) = \frac{10}{36} = \frac{5}{18}\]at least one three.
In the sample space there are 36 possible outcomes. The outcomes containing at least one 3 are: \((3; 1)\); \((3; 2)\); \((3; 3)\); \((3; 4)\); \((3; 5)\); \((3; 6)\); \((1; 3)\); \((2; 3)\); \((4; 3)\); \((5; 3)\); \((6; 3)\).
\[P(\text{at least one } 3) = \frac{11}{36}\]The Mandalay Secondary soccer team has to win both of their next two matches in order to qualify for the finals. The probability that Mandalay Secondary will win their first soccer match against Ihlumelo High is \(\frac{2}{5}\) and the probability of winning their second soccer match against Masiphumelele Secondary is \(\frac{3}{7}\). Assume each match is an independent event.
To calculate the probability of not winning a match use:
\[P(\text{not win}) = 1  P(\text{win})\] \begin{align*} \text{Therefore } P(\text{not win and not win}) &= \frac{3}{5} \times \frac{4}{7} \\ &= \frac{12}{35} \end{align*}This solution makes use of the complementary rule which students should be familiar with. We will revise the rule in more detail later.
There are two possible outcomes: winnot win or not winwin. Let win \(= W\).
\begin{align*} P(\text{(W;not W) or (not W;W)}) &= P(\text{W;not W}) + P(\text{not W;W})\\ & = P(\text{W}) \times P(\text{not W}) + P(\text{not W}) \times P(\text{W}) \\ &= \frac{2}{5} \times \frac{4}{7} + \frac{3}{5} \times \frac{3}{7}\\ &= \frac{8}{35} + \frac{9}{35} = \frac{17}{35} \end{align*}This is an openended question designed to get learners to think critically about the dependent or independent nature of real life events. Example answers could include injuries to or suspensions of players during the first match, team morale if they win or lose the first match, etc.
A pencil bag contains 2 red pens and 4 green pens. A pen is drawn from the bag and then replaced before a second pen is drawn. Calculate:
The events are independent so:
\[P(\text{red pen first}) = \frac{2}{6} = \frac{1}{3}\]The events are independent so:
\[P(\text{green pen second}) = \frac{4}{6} = \frac{2}{3}\]A lunch box contains \(\text{4}\) sandwiches and \(\text{2}\) apples. Vuyele chooses a food item randomly and eats it. He then chooses another food item randomly and eats that. Determine the following:
First item sandwich and second item apple (SA): \[\frac{4}{6} \times \frac{2}{5} = \frac{8}{30} = \frac{4}{15}\]
There are two possible outcomes of getting an apple second:
 first item sandwich and second item apple (SA): \[= \frac{4}{15} \text{ (from b)}\]
 first item apple and second item apple (AA): \[\frac{2}{6} \times \frac{1}{5} = \frac{2}{30}= \frac{1}{15}\]
Therefore the events are dependent.
Given that \(P(A) = \text{0,5}\); \(P(B) = \text{0,4}\) and \(P(A \text{ or } B) = \text{0,7}\), determine by calculation whether events \(A\) and \(B\) are:
Therefore, \(A\) and \(B\) are not mutually exclusive.
Therefore \(A\) and \(B\) are independent.
\(A\) and \(B\) are two events in a sample space where \(P(A) = \text{0,3}\); \(P(A \text{ or } B) = \text{0,8}\) and \(P(B) = k\). Determine the value of \(k\) if:
For \(A\) and \(B\) to be mutually exclusive: \(P(A) + P(B) = P(A \text{ or } B)\)
\begin{align*} \text{0,3} + k&= \text{0,8} \\ \therefore k&= \text{0,5} \end{align*}For \(A\) and \(B\) to be independent:
\begin{align*} P(A \text{ and } B) &= P(A) \times P(B) \\ \text{Therefore } P(A \text{ and } B) &= \text{0,3}k \end{align*} \begin{align*} P(A \text{ or } B) &= P(A) + P(B)  P(A \text{ and } B)\\ \text{0,8} &= \text{0,3} + k  \text{0,3}k \\ \text{0,8} &= \text{0,3} + \text{0,7}k \\ \therefore \text{0,7}k &= \text{0,5} \\ \therefore k &= \frac{5}{7} \end{align*}\(A\) and \(B\) are two events in sample space \(S\) where \(n(S) = 36\); \(n(A) = 9\); \(n(B) = 4\) and \(n(\text{not } (A \text{ or } B)) = 24\). Determine:
For independent events \(P(A) × P(B) = P(A \text{ and }B)\)
\[P(A) × P(B) = \frac{1}{4} \times \frac{1}{9} = \frac{1}{36} = P(A \text{ and } B)\]Therefore \(A\) and \(B\) are independent.
The probability that a Mathematics teacher is absent from school on a certain day is \(\text{0,2}\). The probability that the Science teacher will be absent that same day is \(\text{0,3}\).
Learner dependent. For example: No, there could a bug or illness spreading through the school, therefore the absence of both teachers may be dependent.
Let the probability that the Mathematics teacher is absent \(= P(M)\) and the probability that the Science teacher is absent \(= P(S)\).
\[P(M \text{ or } S) = P(M) + P(S)  P(M \text{ and } S)\] \begin{align*} \text{Assuming the events are independent:} \\ P(M \text{ and } S) &= \text{0,2} \times \text{0,3} = \text{0,06} \\ \text{Therefore } P(M \text{ or } S) &= \text{0,2} + \text{0,3}  \text{0,06} \\ &=\text{0,44} \end{align*}Langa Cricket Club plays two cricket matches against different clubs. The probability of winning the first match is \(\frac{3}{5}\) and the probability of winning the second match is \(\frac{4}{9}\). Assuming the results of the matches are independent, calculate the probability that Langa Cricket Club will:
Let \(P(M) =\) the probability of winning the first match and \(P(N) =\) the probability of winning the second match.
\begin{align*} P(M \text{ and } N) &= \frac{3}{5} \times \frac{4}{9} \\ &= \frac{12}{45} \\ &= \frac{4}{15} \end{align*}Let the probability that team A will finish \(= P(A)\) and the probability team B will finish \(= P(B)\). The teams are working separately, therefore the two events are independent.
\begin{align*} P(A \text{ and } B)&= P(A) \times P(B) \\ &=\text{0,4} \times \text{0,25} \\ &=\text{0,1} \text{ or } \text{10}\% \end{align*}Thabo and Julia were arguing about whether people prefer tea or coffee. Thabo suggested that they do a survey to settle the dispute. In total, they surveyed 24 people and found that 8 of them preferred to drink coffee and 12 of them preferred to drink tea. The number of people who drink tea, coffee or both is 16. Determine:
Therefore the events are independent.
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