Question Video: Finding an Unknown in the Probability Distribution of a Discrete Random Variable Mathematics

Given that 𝑓(π‘₯) = 1/15 π‘₯, π‘₯ ∈ {1, 2, 3, ..., π‘˜} is a probability distribution of a discrete random variable 𝑋, calculate the following. The value of π‘˜. 𝑃(2 ≀ 𝑋 < 4)

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Video Transcript

Given that 𝑓 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 𝑋, calculate the following: the value of π‘˜ and the probability that 𝑋 is greater than or equal to two and less than four.

To answer the first part of this question, we recall that if 𝑓 of π‘₯ is the probability distribution of a discrete random variable, then it has two properties. Firstly, 𝑓 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 𝑓 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 𝑓 of π‘₯ for each of the values in the range of this discrete random variable.

𝑓 of one, first of all, is one over 15 multiplied by one, which is simply one over 15. 𝑓 of two is one over 15 multiplied by two, which is two over 15. And 𝑓 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. 𝑓 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. 𝑓 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 𝑋 in the range of this discrete random variable, is equal to five. And so we’ve answered the first part of the question.

Now, in the second part of the question, we’re asked to calculate the probability that this discrete random variable 𝑋 is greater than or equal to two and less than four. In terms of a number line then, we’re looking for the probability that 𝑋 falls in this interval here. But 𝑋 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 can’t, for example, be equal to 2.5. So the probability that 𝑋 is greater than or equal to two but strictly less than four is the sum of the probabilities that 𝑋 is equal to any values in its range which lie in this interval. It’s the probability that 𝑋 is equal to two plus the probability that 𝑋 is equal to three.

Now, we know both of these probabilities ’cause we wrote them down in the first part of the question. The probability that 𝑋 equals two, or 𝑓 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 we’ve completed the problem. The value of this unknown π‘˜ in the probability distribution of 𝑋 is five. And the probability that 𝑋 is greater than or equal to two and less than four is one-third.

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