Let 𝑥 denote a discrete random variable that can take the values zero, one, two, and three. Given that the probability 𝑥 equals zero is equal to one-ninth, the probability 𝑥 is equal to one is equal to four-ninths, the probability 𝑥 is equal to two is equal to 𝑎, and the probability 𝑥 is equal to three equals three 𝑎, find the value of 𝑎.
In order to answer this question, we will recall some of the key properties of discrete random variables. A discrete random variable has a countable number of possible values. The probability of each outcome can be described using a probability distribution function. If 𝑓 of 𝑥 tells us the probability of each event occurring, then the sum of all 𝑓 of 𝑥 values must be equal to one. And Each individual 𝑓 of 𝑥 value must be between zero and one inclusive. And in fact, we can use this first property to answer the problem.
We’re told that the probability 𝑥 is equal to zero is equal to one-ninth, the probability 𝑥 is equal to one is four-ninths, the probability 𝑥 is equal to two is 𝑎, and the probability 𝑥 equals three is equal to three 𝑎. And we’re also told that these are the only possible outcomes. So the probability that 𝑥 equals zero, one, two, and three, the sum of these is equal to one. We can replace each of these expressions with its corresponding probability value. So one-ninth plus four-ninths plus 𝑎 plus three 𝑎 equals one. One-ninths plus four-ninths is five-ninths. And 𝑎 plus three 𝑎 is four 𝑎.
We then solve this equation for 𝑎 by subtracting five-ninths from both sides, giving us four 𝑎 equals four-ninths. And finally, we’re going to divide through by four. Well, it follows that four-ninths divided by four is equal to one-ninths. And so given the information about our discrete random variable 𝑥, 𝑎 must be equal to one-ninth.