Question Video: Selecting Appropriate Probability Distribution Function for a Discrete Random Variable Mathematics

Let 𝑋 denote a discrete random variable which can take the values 3, 5, and 6. Which of the following functions could represent the probability function of 𝑋? [A] 𝑓(π‘₯) = (4π‘₯ + 5)/2 [B] 𝑓(π‘₯) = 4/(5π‘₯ + 2) [C] 𝑓(π‘₯) = (π‘₯Β² + 3)/10 [D] 𝑓(π‘₯) = (π‘₯ βˆ’ 2)/8

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Video Transcript

Let 𝑋 denote a discrete random variable which can take the values three, five, and six. Which of the following functions could represent the probability function of 𝑋? (a) 𝑓 of π‘₯ equals four π‘₯ plus five over two. (b) 𝑓 of π‘₯ equals four over five π‘₯ plus two. (c) 𝑓 of π‘₯ equals π‘₯ squared plus three over 10. Or (d) 𝑓 of π‘₯ equals π‘₯ minus two over eight.

We recall first that the probability function of a discrete random variable 𝑋 is the set of all values the variable can take together with their associated probabilities. In this question, we’re told that 𝑋 can take the values three, five, and six. So a table of its probability function would have these three values in the first row and then their corresponding probabilities in the second.

If 𝑓 of π‘₯ represents the probability function of a discrete random variable, then there are two properties that it must satisfy. Firstly, 𝑓 of π‘₯ must be between zero and one for every value in the range of the discrete random variable. And secondly, the sum of 𝑓 of π‘₯ must be equal to one. That is, the sum of the probabilities of all possible π‘₯-values must be equal to one.

To determine whether each of these functions could be the probability function of 𝑋 then, we need to evaluate each function for the three π‘₯-values three, five, and six and then consider whether the values we’ve found satisfy the properties of a probability distribution function. Let’s begin with option (a). Substituting π‘₯ equals three first, we find that 𝑓 of three is equal to four multiplied by three plus five all over two. That’s 12 plus five over two, which is 17 over two, or 8.5.

Now, actually, we can stop here because this value is greater than one. And so it doesn’t satisfy the first property of a probability function, where we said that the value of 𝑓 of π‘₯ for each π‘₯-value must be between zero and one. So we can rule out option (a).

Now, let’s consider option (b). Evaluating 𝑓 of three, we have four over five multiplied by three plus two. That’s four over 15 plus two, which is four over 17. This is fine so far as it is between zero and one. 𝑓 of five then is four over five multiplied by five plus two. That’s four over 27. This value is also between zero and one, so we’ll keep going. 𝑓 of six, lastly, is four over five multiplied by six plus two, which is four over 32. This value is between zero and one. So the first property of a probability function is satisfied.

Next, we need to check whether the sum of these three values is equal to one. So we have four over 17 plus four over 27 plus four over 32. This isn’t a very nice sum, but it’s equal to 1867 over 3672. It certainly isn’t equal to one.

Another way to see this is to consider that each of the individual fractions that we’re summing is less than one-third and so their sum must be less than one. We found then that whilst the first property of probability functions is satisfied, the second isn’t. And so (b) cannot be the probability function of 𝑋.

Next, we’ll consider option (c). Substituting π‘₯ equals three, first, we have 𝑓 of three is equal to three squared plus three over 10. That’s nine plus three over 10, so 12 over 10, which is equal to 1.2. We can stop here because this value is greater than one. And so we can see that the first property of probability functions is not satisfied for option (c).

We expect then that option (d) is going to be our answer as it’s the only option left, but we must check this. Substituting π‘₯ equals three, we have 𝑓 of three is equal to three minus two over eight, which simplifies to one-eighth. 𝑓 of five is five minus two over eight, which simplifies to three-eighths. 𝑓 of six is six minus two over eight, which simplifies to four-eighths. Now that can of course be canceled to one-half, but we’ll keep it with the denominator of eight for now.

Each of these fractions is indeed between zero and one. So the first property of probability functions is satisfied. Summing the three values together, we have one-eighth plus three-eighths plus four-eighths, which is equal to eight-eighths. And of course that simplifies to one. So the second property, the sum of the probabilities must be equal to one, is also satisfied.

Our answer then is option (d), 𝑓 of π‘₯ equals π‘₯ minus two over eight, as it is the only one of the four functions which satisfies the two necessary properties of probability functions.

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