Question

Use the probability distribution for the random variable \(x\) to answer the questions in Exercises 12-16. $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ $$ \text { Find } p(4) $$

Short answer

Answer: \(p(4) = 0.05\)

Step-by-step solution

Understand the properties of a probability distribution

Recall that in a probability distribution, the sum of the probabilities for all possible values of the random variable must equal 1. In this case, we have a discrete distribution with values of \(x = 0,1,2,3,4,5\). We are given the probabilities for all values except \(x=4\), and we need to find \(p(4)\).

Calculate the sum of given probabilities

First, we need to add up the probabilities given in the table for \(x=0,1,2,3,5\). This helps us in finding how much probability is still left for \(x=4\). So, we have: $$ p(0)+p(1)+p(2)+p(3)+p(5) = 0.1 + 0.3 + 0.4 + 0.1 + 0.05 $$

Compute the sum

Now, compute the sum: $$ 0.1 + 0.3 + 0.4 + 0.1 + 0.05 = 0.95 $$ So, the sum of the given probabilities is \(0.95\).

Find the missing probability \(p(4)\)

We need the sum of all probabilities to be equal to 1. Since we have found the sum of the given probabilities to be 0.95, we can find the missing probability by subtracting the sum from 1: $$ p(4) = 1 - (p(0)+p(1)+p(2)+p(3)+p(5)) $$ Substitute the sum which we computed earlier: $$ p(4) = 1 - 0.95 $$

Calculate the final result

Calculate the value of \(p(4)\): $$ p(4) = 0.05 $$ So, the missing probability \(p(4)\) is \(0.05\).

Key concepts

Discrete Distribution

When we talk about probability, a discrete distribution comes into play when we are dealing with a countable number of possible outcomes. Think of it as a list where you can count each outcome on your fingers, like the number of heads in a series of coin flips or the roll of a die.

In this context, each possible outcome is assigned a specific probability, which is a number between 0 and 1, with the total probability summing up to 1. This is akin to saying that something will definitely happen, but each individual outcome has its own chance of occurring.

For instance, let's imagine you have a bag with differently colored marbles. If you have a finite number of colored marbles and you pick one at random, the probability distribution of picking each color is discrete—you can list out all the possibilities and their associated probabilities.

Random Variable

When dealing with probability distributions, the term random variable is essential. Think of it as a placeholder or a label for the outcomes of a random phenomenon. For example, if you're tossing a coin, you could have a random variable, say 'X', which represents the outcome, such as 'X = 0' for tails and 'X = 1' for heads.

A random variable can take on different values with certain probabilities, and in our context, it's what we call 'discrete.' That's because the possible values of 'X' form a finite or countable set, just like in the textbook exercise where the random variable 'x' could be one of the countable numbers from 0 to 5, representing something that we're measuring, counting, or keeping track of.

Probability Calculation

When you want to find out the likelihood of something happening, you engage in a probability calculation. It involves math, but not just any math—the kind that deals with chance and uncertainty. You look at all the possible outcomes, determine the chances of each happening, and then add up those chances to figure out the overall probability of an event.

In our exercise scenario, the probability distribution table already gives you some of the probabilities. Your job is to use those numbers to find the missing piece of the puzzle. By adding up the probabilities you know and subtracting from 1 (which represents absolute certainty), you can find the probability of the one missing outcome. So, if the sum of the known probabilities is less than 1, the difference gives you the probability of the missing outcome—that's the part of the pie that's not yet accounted for!