### Video Transcript

In this video, we will practice
identifying discrete random variables and using probability distribution functions
and tables for them.

Letβs begin by considering what the
term discrete random variable means. Variable means that this is a
quantity which can take different values. Random means that the value taken
is determined by the outcome of a random event or experiment. Or you could say it depends on an
element of chance. The key part here is the word
discrete, which means that only a finite set of certain values can be taken. We call this set of values the
range of a discrete random variable. For example, the number of heads
observed when flipping two coins would be an example of a discrete random
variable. Its range would be the set of
values zero, one, two as we could have zero, one, or two heads. This is a finite or countable set
of values, and so π the number of heads when flipping two coins is a discrete
random variable.

Letβs now consider what we mean by
the probability distribution function of a discrete random variable, and suppose we
have a a discrete random variable π whose range is the set of values π₯ one, π₯
two, up to π₯ π. Now, itβs important to point out
here the convention that we follow. We use capital or uppercase letters
to represent the variable itself. And we use lowercase letters to
represent the values in its range. Now suppose we also have a function
π such that π of π₯ π is equal to the probability that our random variable π
takes the value π₯ π for π equals one, two, all the way up to π. Those for all values π₯ π in the
range of our discrete random variable capital π. Then, this function π is what we
call the probability distribution function of the discrete random variable π.

Itβs very common to represent a
probability distribution function in a table with the values in the range given in
the first row and the values of the probability distribution function π of π₯ given
in the second. More informally, we can think of
the probability distribution function of a discrete random variable simply as the
set of all values the variable can take together with their corresponding
probabilities. One important feature of a
probability distribution is that because the discrete random variable can only take
the values in its range and no other values, the sum of all the probabilities in the
probability distribution function must be equal to one.

As an example of a probability
distribution function, suppose weβre considering this spinner. We can introduce π to be the
discrete random variable represented by the number obtained when the spinner is
spun. This spinner has eight equal
sections, three of them labeled with the number one, one labeled with the number
two, two labeled with the number three, and the final two are labeled with the
number four. This means that the range of our
discreet random variable π is the set of values one, two, three, four as these are
the only values π can take.

We can write down the probability
distribution function for this discrete random variable which, remember, has the
values in the range in its first row and the corresponding probabilities in the
second. The values in the range as weβve
already said are one, two, three, and four. There were eight equal sections in
our spinner, so all of our probabilities will be fractions with denominators of
eight. Three sections have the number one
on them, one section has the number two, and two sections each have the numbers
three and four. The probabilities for three and
four of two-eighths could be simplified to one-quarter if we wish.

And so weβve written down the
probability distribution function for our discrete random variable π. We have the set of all values in
its range in the top row and the corresponding probabilities in the second. Notice that the sum of all the
probabilities in our probability distribution function is indeed equal to one. This key feature of probability
distribution functions is useful when identifying whether or not a given function
could be a probability distribution function as weβll see in our next example.

Can the function in the given table
be a probability distribution function?

Now, what we have here certainly
looks like it could be a probability distribution function. In the top row of the table, we
have a countable set of values, which could be the range of a discrete random
variable. And in the bottom row, we have some
decimal values, which could be the associated probabilities. In order to check whether this
could indeed represent a probability distribution function, we need to look at the
probabilities more closely. And we should recall that the sum
of all the probabilities in a probability distribution function must be equal to
one.

Looking at the second row of the
table and, in fact, looking just in the middle to values first of all, we can see
that the sum of these values of 0.43 and 0.69 is a value greater than one. Of course, if we then sum all four
values, we get an even larger value of 1.65. This means that this cannot be a
probability distribution function, as if a discrete random variable could take each
the values zero, one, four, and five with these associated probabilities would have
a total probability greater than one, which is impossible. So, our answer is no.

Letβs now consider an example in
which we need to calculate a missing probability in a probability distribution.

Let π denote a discrete random
variable which can take the values two, six, seven, and eight. Given that the probability π
equals two is equal to the probability π equals six, which is three over 22, and
the probability π equals seven is four elevenths, find the probability π equals
eight. Give your answer as a fraction.

Letβs begin by representing the
information weβve been given in a slightly different format. We can use a table to display this
probability distribution function. In the top row, weβll have the four
values in the range of this discrete random variable, which are two, six, seven, and
eight. In the second row, weβll fill in
the probabilities weβve been given: three over 22 for both two and six and four over
11 for seven. Weβre missing one of the
probabilities: the probability that π equals eight. And this is the value weβre asked
to find.

To do so, we need to recall that
the sum of all the probabilities in a probability distribution function must be
equal to one because the discrete random variable can only take values within its
range. We can therefore form an equation
and then substitute the three probabilities weβve been given in the question. By thinking of four over 11 as the
equivalent fraction eight over 22, we then have the equation 14 over 22 plus the
probability π equals eight is equal to one. Subtracting 14 over 22 from each
side and then simplifying 14 over 22 to seven over 11, we have that the probability
π equals eight is equal to one minus seven over 11, which is four over 11.

So, by using the fact that the sum
of all the probabilities in a probability distribution function must be equal to
one, we found the missing probability. The probability that π is equal to
eight is four elevenths.

Letβs now consider a similar but
slightly more complex example in which all of the probabilities in our probability
distribution have been expressed algebraically.

The function in the given table is
a probability function of a discrete random variable π. Find the value of π.

Notice that each probability in
this probability distribution function has been expressed in terms of the variable
π whose value weβve been asked to find. In order to do this, we need to
recall the key fact that the sum of all the probabilities in a probability
distribution function must be equal to one. We can therefore form an equation
using the four values in the second row of our table. Three π plus eight π squared plus
four π squared plus eight π is equal to one. This simplifies to 12π squared
plus 11π equals one. And then subtracting one from each
side, we see that we have the quadratic equation 12π squared plus 11π minus one is
equal to zero.

This equation can be solved in a
variety of ways. But the easiest method for this
particular quadratic is going to be to solve by factoring. With a little bit of trial and
error or perhaps using factoring by grouping, we see that this quadratic factors as
12π minus one multiplied by π plus one. We then follow the usual method for
solving a quadratic equation by factoring. We set each factor in turn equal to
zero and solve the resulting linear equation, giving two values of π: π equals one
twelfth and π equals negative one.

So, we have two possibilities for
the value of π, both correct roots of this quadratic equation. But only one value makes sense in
the context of this problem. If we look back at our table, we
see, for example, that the probability π equals zero is three π. If we use the value one twelfth,
then this becomes three twelfths or one-quarter. But if we use the value negative
one, this gives negative three. Remember that probabilities must
always be between zero and one, so we canβt have a probability of negative
three. This means that the value π equals
negative one, whilst being a correct solution to the quadratic equation, is not
correct in the context of this problem as a value of π.

We can check whether this value of
one twelfth is correct by calculating all of the probabilities. Three π gives three twelfths,
which is equivalent to 36 over 144. And in the same way, we can find
the probabilities as fractions with a denominator of 144 for eight π squared, four
π squared, and eight π. When we sum these four values
together, we do indeed get 144 over 144 or one, and so this confirms that our value
of π is correct. The value of π then is one
twelfth.

In this example, we highlighted
another key feature of probability distribution functions, which is that all of the
individual probabilities must be between zero and one. So, weβll include this in our
definition of a probability distribution function when we return to it later. Next, though, letβs consider how we
can write a probability distribution function ourselves from a worded description of
a discrete random variable.

Two boys and two girls are ranked
according to their scores on an exam. Assume that no two scores are alike
and that all possible rankings are equally likely. Find the probability distribution
of the random variable π expressing the highest ranking achieved by a girl, for
example, π equals two if the top-ranked student is a boy and the second-ranked
student is a girl.

So we have four students and
theyβre going to be ranked according to their scores on an exam. We need to begin by listing all the
different ways we can place these two boys and two girls in the four positions. Now notice that we donβt care which
girl or which boy is in each position, only whether the person in each rank is a boy
or a girl.

Letβs begin by putting a girl in
the first position. We can then put the other girl in
position two, position three, or position four. Each time, the remaining two gaps
must both be filled by boys. If, instead, we were to have a boy
in the first position, then in the same way, the second boy could either be in the
second position, the third position, or the fourth position. And in each case, the remaining two
positions would both be filled by girls. So we see that we have six possible
ordering of two girls and two boys. Remember, each time weβre not
interested in which boy or which girl it is. The question tells us that all
possible rankings are equally likely, which means that the probability associated
with each of these orderings is one-sixth.

The discrete random variable π
weβre interested in is the highest ranking achieved by a girl. In each of the first three cases,
thereβs a girl in first place, so the value of π is one. In the fourth case, the girls are
in third and fourth positions, so the highest ranking achieved by a girl, the value
of π, is three. In the fifth case, the girls are in
second and fourth positions, so the highest ranking is two. And in the sixth case, the girls
were in second and third positions. The highest ranking is again
two. These values in the final column
give the range of our discrete random variable π. It can take the values one, two, or
three.

We then need to fill in the
associated probabilities. Remember, the value one appears
three times, so its total probability is three times one-sixth. Thatβs three-sixths, which can be
simplified to one-half. The value two appears twice. So its total probability is
two-sixths, which simplifies to one-third. And finally, the value three only
appears once, so its probability is one-sixth. And so we found the probability
distribution of the random variable π. The values in the range are one,
two, and three with corresponding probabilities one-half, one-third, and
one-sixth.

In our final example, weβll briefly
consider how to answer questions where weβre asked to find the probability that a
discrete random variable is greater than or less than a particular value.

Let π denote a discrete random
variable which can take the values negative two, two, four, and five. Given that the probability π
equals negative two is 0.15, the probability π equals two is 0.43, and the
probability π equals four is 0.25, find the probability π is greater than two.

In this question, weβre not asked
to determine the probability that our random variable is equal to a particular
value, but rather the probability that it is greater than a given value. This may look a little confusing at
first, but the key to remember is that a discrete random variable can only take the
values in its range and no others. Letβs begin by representing
information in the question using a table. We have the values in the range of
our discrete random variable in the top row and the corresponding probabilities in
the second row.

Now, we do have a missing
probability, the probability that π equals five, but weβll worry about that later
if necessary. We want to find the probability
that π is greater than two. Now remember that our discrete
random variable can only take the values in its range. For π to be strictly greater than
two, this means that it must either take the value four or the value five. There are no other values it can
take. The probability π is greater than
two then is equal to the sum of the probability π equals four and the probability
π equals five.

Now remember, we donβt know the
probability π equals five, although we can easily calculate it by remembering that
the sum of all probabilities in the distribution must be equal to one. Instead, a slightly easier way may
be to recall that the probability π is greater than two will be equal to one minus
the probability π is less than or equal to two. So, instead, we can just subtract
the probabilities for π equaling negative two and π equaling two from one, one
minus 0.15 and 0.43, which is equal to 0.42.

The key point to remember is that a
discrete random variable can only take the values in its range. So, if weβre asked for a cumulative
probability as we were here, we just need to consider which values in the range will
satisfy this.

Letβs now review the key points
covered in this video. A discrete random variable is a
quantity whose value is determined by chance, and it can only take a countable or
finite number of values, which we call its range. The probability distribution
function π of a discrete random variable is a list or table of all the values in
the range together with their corresponding probabilities. The probability distribution
function has two key properties. Firstly, as with all probabilities,
each individual probability must be between zero and one. And secondly, the sum or total of
all the probabilities in the probability distribution function must be equal to
one. Weβve seen through examples how we
can write down a probability distribution function and how we can use these
properties to calculate missing values or probabilities.