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Chapter 11: Problem 77

Fill in the \(?\) to make \(p(x)\) a probability function. If not possible, say so. $$ \begin{array}{lccl} \hline x & 1 & 2 & 3 \\ \hline p(x) & 0.5 & 0.6 & ? \\ \hline \end{array} $$

Short Answer

Expert verified
Since the sum of known probabilities already exceeds 1, it is not possible to find a value for ? that can make \(p(x)\) a probability function.

Step by step solution

01

Identify Known Probabilities

List all known probabilities. According to the table, these are \(p(1)=0.5\) and \(p(2)=0.6\).
02

Sum Known Probabilities

Add all the known probabilities. When doing this, \(0.5 + 0.6 = 1.1\).
03

Determine Unknown Probability

To make the function a valid probability function, all probabilities must sum up to 1. As known values already exceed 1, we can conclude it's not possible to find a value for ? that can turn \(p(x)\) into a probability function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution
Probability distribution is a mathematical concept used to describe the probabilities of all possible outcomes of a random variable. Imagine you have a list of events, each with a chance of occurring. This is essentially what a probability distribution does: it provides a comprehensive view of possible outcomes and their likelihoods.
  • Each event has a probability assigned.
  • The sum of all probabilities in a distribution must equal 1, ensuring that one of the outcomes must indeed occur.
Think of probability distribution as a complete picture of all the assumptions and chances regarding the random variable. In our given exercise, each value of \(x\) has a probability \(p(x)\) associated with it. All known and unknown probabilities should adhere to the causality defined by the probability distribution.
Sum of Probabilities
The sum of probabilities is a fundamental rule when dealing with probability distributions. For a distribution to be considered complete and valid, the sum of all the individual probabilities must equal exactly 1.
  • Why 1? This is because 1 represents the certainty that one of the possible outcomes will definitely happen.
  • If the sum exceeds 1, it indicates that the probabilities have been overestimated and that the function is not valid.
In the problem at hand, when the known probabilities \(0.5\) and \(0.6\) are summed, the result is \(1.1\). This is already greater than 1, signaling the impossibility of making the function valid with the current setup. Students often encounter issues with determining unknown probabilities, and this rule acts as a core guideline.
Valid Probability Function
A valid probability function is essential in probability theory. For any function to be termed as a valid probability function:
  • Each individual probability must be non-negative, as it reflects reality where a negative chance is not possible.
  • The sum of all probabilities should strictly be 1, ensuring all potential outcomes are covered.
  • No single probability should be greater than 1, which would imply certainty.
In the context of the exercise, despite having two probabilities already known (0.5 and 0.6), their total exceeds the allowed sum of 1. This violation of the sum rule means it is impossible for the given setup to be a valid probability function, as there is no room left to fit the unknown \(?\) without breaking these critical rules. Understanding this helps students grasp why certain setups, like this one, cannot be reconciled into a valid probability framework.

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Most popular questions from this chapter

As we see in Exercise 11.38 on page \(653,\) during the \(2010-11\) NBA season, Ray Allen of the Boston Celtics had a free throw shooting percentage of \(0.881 .\) Assume that the probability Ray Allen makes any given free throw is fixed at 0.881 , and that free throws are independent. Let \(X\) be the number of free throws Ray Allen makes in two attempts. (a) What is the probability distribution of \(X ?\) (b) What is the mean of \(X ?\)

Worldwide, the proportion of babies who are boys is about 0.51 . A couple hopes to have three children and we assume that the sex of each child is independent of the others. Let the random variable \(X\) represent the number of girls in the three children, so \(X\) might be \(0,1,2,\) or 3 . Give the probability function for each value of \(X\).

A friend makes three pancakes for breakfast. One of the pancakes is burned on both sides, one is burned on only one side, and the other is not burned on either side. You are served one of the pancakes at random, and the side facing you is burned. What is the probability that the other side is burned? (Hint: Use conditional probability.)

State whether the process described is a discrete random variable, is a continuous random variable, or is not a random variable. Observe the average weight, in pounds, of everything you catch during a day of fishing.

In a certain board game participants roll a standard six-sided die and need to hit a particular value to get to the finish line exactly. For example, if Carol is three spots from the finish, only a roll of 3 will let her win; anything else and she must wait another turn to roll again. The chance of getting the number she wants on any roll is \(p=1 / 6\) and the rolls are independent of each other. We let a random variable \(X\) count the number of turns until a player gets the number needed to win. The possible values of \(X\) are \(1,2,3, \ldots\) and the probability function for any particular count is given by the formula $$P(X=k)=p(1-p)^{k-1}$$ (a) Find the probability a player finishes on the third turn. (b) Find the probability a player takes more than three turns to finish.
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