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10.1 Revision
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10.3 More Venn diagrams
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Sometimes the presence or absence of one event tells us something about other events. We call events dependent if knowing whether one of them happened tells us something about whether the others happened. Independent events give us no information about one another; the probability of one event occurring does not affect the probability of the other events occurring.
Two events, \(A\) and \(B\) are independent if and only if \[P(A \text{ and } B) = P(A) \times P(B)\]
At first it might not be clear why we should call events that satisfy the equation above independent. We will explore this further using a number of examples.
Roll a single \(\text{6}\)-sided die and consider the following two events:
Now answer the following questions:
Are the events \(E\) and \(T\) dependent or independent according to the definition? (Hint: compute the probabilities in the definition of independence.)
So, why do we call it independence when \(P(A\text{ and }B) = P(A) \times P(B)\)? For two events, \(A\) and \(B\), independence means that knowing the outcome of \(B\) does not affect the probability of \(A\).
Consider the following Venn diagram.
The probability of \(A\) is the ratio between the number of outcomes in \(A\) and the number of outcomes in the sample space, \(S\). \[P(A) = \frac{n(A)}{n(S)}\]
Now, let's say that we know that event \(B\) happened. How does this affect the probability of \(A\)? Here is how the Venn diagram changes:
A lot of the possible outcomes (all of the outcomes outside \(B\)) are now out of the picture, because we know that they did not happen. Now the probability of \(A\) happening, given that we know that \(B\) happened, is the ratio between the size of the region where \(A\) is present (\(A\text{ and }B\)) and the size of all possible events (\(B\)). \[P(A\text{ if we know }B) = \frac{n(A\text{ and }B)}{n(B)}\] If \(P(A) = P(A\text{ if we know }B)\) we call them independent, because knowing \(B\) does not change the probability of \(A\).
With some algebra, we can prove that this statement of independence is the same as the definition of independence that we saw at the beginning of this section. For independent events \[P(A\text{ and }B) = P(A) \times P(B)\] This is equivalent to \begin{align*} P(A) &= P(A\text{ and }B) \div P(B) \\ &= \frac{n(A\text{ and }B)}{n(S)} \div \frac{n(B)}{n(S)} \\ &= \frac{n(A\text{ and }B)}{n(B)} \\ &= P(A\text{ if we know }B) \end{align*} That is why we call events independent!
(For enrichment only):
The ratio \[\frac{P(A \text{ and } B)}{P(B)}\] is called a conditional probability and written using the notation \(P(A\;|\;B)\). This notation is read as “the probability of \(A\) given \(B\).”
If (and only if) \(A\) and \(B\) are independent: \(P(A\;|\;B) = P(A)\) and \(P(B\;|\;A) = P(B)\). Try to prove this using the definition of independence.
A bag contains \(\text{5}\) red and \(\text{5}\) blue balls. We remove a random ball from the bag, record its colour and put it back into the bag. We then remove another random ball from the bag and record its colour.
Since there are a total of \(\text{10}\) balls, of which \(\text{5}\) are red, the probability of getting a red ball is \[P(\text{first ball red}) = \frac{5}{10} = \frac{1}{2}\]
The problem states that the first ball is placed back into the bag before we take the second ball. This means that when we draw the second ball, there are again a total of \(\text{10}\) balls in the bag, of which \(\text{5}\) are blue. Therefore the probability of drawing a blue ball is \[P(\text{second ball blue}) = \frac{5}{10} = \frac{1}{2}\]
When drawing two balls from the bag, there are \(\text{4}\) possibilities. We can get
We want to know the probability of the second outcome, where we have to get a red ball first. Since there are \(\text{5}\) red balls and \(\text{10}\) balls in total, there are \(\frac{5}{10}\) ways to get a red ball first. Now we put the first ball back, so there are again \(\text{5}\) red balls and \(\text{5}\) blue balls in the bag. Therefore there are \(\frac{5}{10}\) ways to get a blue ball second if the first ball was red. This means that there are \[\frac{5}{10}\times\frac{5}{10}=\frac{25}{\text{100}}\] ways to get a red ball first and a blue ball second. So, the probability of getting a red ball first and a blue ball second is \(\frac{1}{4}\).
According to the definition, events are independent if and only if \[P(A\text{ and }B) = P(A) \times P(B)\]
In this problem:
In the previous example, we picked a random ball and put it back into the bag before continuing. This is called sampling with replacement. In this example, we will follow the same process, except that we will not put the first ball back into the bag. This is called sampling without replacement.
So, from a bag with \(\text{5}\) red and \(\text{5}\) blue balls, we remove a random ball and record its colour. Then, without putting back the first ball, we remove another random ball from the bag and record its colour.
We will look directly at the number of possible ways in which we can get the \(\text{4}\) possible outcomes when removing \(\text{2}\) balls. In the previous example, we saw that the \(\text{4}\) possible outcomes are
For the first outcome, we have to get a red ball first. Since there are \(\text{5}\) red balls and \(\text{10}\) balls in total, there are \(\frac{5}{10}\) ways to get a red ball first. After we have taken out a red ball, there are now \(\text{4}\) red balls and \(\text{5}\) blue balls left. Therefore there are \(\frac{4}{9}\) ways to get a red ball second if the first ball was also red. This means that there are \[\frac{5}{10}\times\frac{4}{9}=\frac{20}{90}\] ways to get a red ball first and a red ball second. The probability of the first outcome is \(\frac{2}{9}\).
For the second outcome, we have to get a red ball first. As in the first outcome, there are \(\frac{5}{10}\) ways to get a red ball first; and there are now \(\text{4}\) red balls and \(\text{5}\) blue balls left. Therefore there are \(\frac{5}{9}\) ways to get a blue ball second if the first ball was red. This means that there are \[\frac{5}{10}\times\frac{5}{9}=\frac{25}{90}\] ways to get a red ball first and a blue ball second. The probability of the second outcome is \(\frac{5}{18}\).
We can compute the probabilities of the third and fourth outcomes in the same way as the first two, but there is an easier way. Notice that there are only \(\text{2}\) types of ball and that there are exactly equal numbers of them at the start. This means that the problem is completely symmetric in red and blue. We can use this symmetry to compute the probabilities of the other two outcomes.
In the third outcome, the first ball is blue and the second ball is red. Because of symmetry this outcome must have the same probability as the second outcome (when the first ball is red and the second ball is blue). Therefore the probability of the third outcome is \(\frac{5}{18}\).
In the fourth outcome, the first and second balls are both blue. From symmetry, this outcome must have the same probability as the first outcome (when both balls are red). Therefore the probability of the fourth outcome is \(\frac{2}{9}\).
To summarise, these are the possible outcomes and their probabilities:
To determine the probability of getting a red ball on the first draw, we look at all of the outcomes that contain a red ball first. These are
The probability of the first outcome is \(\frac{2}{9}\) and the probability of the second outcome is \(\frac{5}{18}\). By adding these two probabilities, we see that the probability of getting a red ball first is \[P(\text{first ball red}) = \frac{2}{9} + \frac{5}{18} = \frac{1}{2}\]
This is the same as in the previous exercise, which should not be too surprising since the probability of the first ball being red is not affected by whether or not we put it back into the bag before drawing the second ball.
To determine the probability of getting a blue ball on the second draw, we look at all of the outcomes that contain a blue ball second. These are
The probability of the first outcome is \(\frac{5}{18}\) and the probability of the second outcome is \(\frac{2}{9}\). By adding these two probabilities, we see that the probability of getting a blue ball second is \[P(\text{second ball blue}) = \frac{5}{18} + \frac{2}{9} = \frac{1}{2}\]
This is also the same as in the previous exercise! You might find it surprising that the probability of the second ball is not affected by whether or not we replace the first ball. The reason why this probability is still \(\frac{1}{2}\) is that we are computing the probability that the second ball is blue without knowing the colour of the first ball. Because there are only two equal possibilities for the second ball (red and blue) and because we don't know whether the first ball is red or blue, there is an equal chance that the second ball will be one colour or the other.
We have already calculated the probability that the first ball is red and the second ball is blue. It is \(\frac{5}{18}\).
According to the definition, events are independent if and only if \[P(A\text{ and }B) = P(A) \times P(B)\]
In this problem:
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.
Use the following Venn diagram to determine whether events \(X\) and \(Y\) are
The events are mutually exclusive if and only if \((X \text{ and } Y) = \emptyset\). From the Venn diagram we see that \((X \text{ and } Y)\) contains \(\text{3}\) elements. Therefore \(X\) and \(Y\) are not mutually exclusive.
The events are independent if and only if \(P(X \text{ and } Y) = P(X) \times P(Y)\). In this case, \begin{align*} P(X \text{ and } Y) &= \frac{3}{35} \\ P(X) &= \frac{14}{35} \\ P(Y) &= \frac{10}{35} \end{align*} So \(P(X) \times P(Y) = \frac{4}{35} \ne P(X \text{ and } Y)\). Therefore \(X\) and \(Y\) are dependent.
Of the \(\text{30}\) learners in a class \(\text{17}\) have black hair, \(\text{11}\) have brown hair and \(\text{2}\) have red hair. A learner is selected from the class at random.
\(P(M) = \text{0,45}\); \(P(N) = \text{0,3}\) and \(P(M \text{ or } N) = \text{0,615}\). Are the events \(M\) and \(N\) mutually exclusive, independent or neither mutually exclusive nor independent?
From the sum rule, \begin{align*} P(M \text{ and } N) &= P(M) + P(N) - P(M \text{ or } N) \\ &= \text{0,45} + \text{0,3} - \text{0,615} \\ &= \text{0,135} \end{align*} \(P(M \text{ and } N) \ne 0\), therefore the events are not mutually exclusive.
Therefore the events are independent since \[P(M) \times P(N) = \text{0,135} = P(M \text{ and } N)\]
(For enrichment)
Prove that if event \(A\) and event \(B\) are mutually exclusive with \(P(A) \ne 0\) and \(P(B) \ne 0\), then \(A\) and \(B\) are always dependent.
Proof by contradiction.
Assume that \(A\) and \(B\) are independent events. From the definition of independence, we then have \(P(A \text{ and } B) = P(A) \times P(B)\). Since \(A\) and \(B\) are mutually exclusive, we know that \(P(A \text{ and } B) = 0\). Therefore \(P(A) \times P(B) = 0\).
For the product of two numbers to be zero, at least one of the two numbers must be zero. But in the problem statement, we were told \(P(A) \ne 0\) and \(P(B) \ne 0\). Therefore we have a contradiction and our original assumption, that \(A\) and \(B\) are independent events, must be false.
Conclusion: \(A\) and \(B\) are dependent.
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10.1 Revision
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10.3 More Venn diagrams
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