### Video Transcript

Given that π of π₯ equals one over
15 π₯, for π₯ belonging to the set of values one, two, three up to π, is the
probability distribution of a discrete random variable π, calculate the following:
the value of π and the probability that π is greater than or equal to two and less
than four.

To answer the first part of this
question, we recall that if π of π₯ is the probability distribution of a discrete
random variable, then it has two properties. Firstly, π of π₯ must be between
zero and one for each π₯-value in the range of the discrete random variable. And secondly, the sum of all of the
π of π₯ values, that is, the sum of all of the probabilities in the probability
distribution, must be equal to one. We can find expressions for π of
π₯ for each of the values in the range of this discrete random variable.

π of one, first of all, is one
over 15 multiplied by one, which is simply one over 15. π of two is one over 15 multiplied
by two, which is two over 15. And π of three is one over 15
multiplied by three, which is three over 15. If we find the sum of these three
probabilities, we get six over 15. And as this is not equal to one, we
know that there must be other values in the range of this discrete random
variable. π of four then is one over 15
multiplied by four, which is four over 15. And adding this value to our sum
gives 10 over 15, which is still not equal to one.

The next value of π₯ is five. π of five is one over 15
multiplied by five, which is five over 15. And adding this value, we have 15
over 15, which is equal to one, so we can stop. This means that the value of π,
which is the largest value of π in the range of this discrete random variable, is
equal to five. And so weβve answered the first
part of the question.

Now, in the second part of the
question, weβre asked to calculate the probability that this discrete random
variable π is greater than or equal to two and less than four. In terms of a number line then,
weβre looking for the probability that π falls in this interval here. But π is a discrete random
variable, so it can only take the values in its range. π₯ can only be equal to one, two,
three, four, or five. It canβt, for example, be equal to
2.5. So the probability that π is
greater than or equal to two but strictly less than four is the sum of the
probabilities that π is equal to any values in its range which lie in this
interval. Itβs the probability that π is
equal to two plus the probability that π is equal to three.

Now, we know both of these
probabilities βcause we wrote them down in the first part of the question. The probability that π equals two,
or π of two, is two fifteenths. And the probability that π₯ equals
three is three fifteenths. So we have two fifteenths plus
three fifteenths, which is five fifteenths. And of course this can be
simplified by dividing both the numerator and denominator by five to one-third. So weβve completed the problem. The value of this unknown π in the
probability distribution of π is five. And the probability that π is
greater than or equal to two and less than four is one-third.