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Let \(X=\) the number of nonzero digits in a randomly selected 4-digit PIN that has no restriction on the digits. What are the possible values of \(X\) ? Give three possible outcomes and their associated \(X\) values.

Short Answer

Expert verified
The possible values of \(X\) are 0, 1, 2, 3, and 4.

Step by step solution

01

Understanding the Problem

We need to determine the number of nonzero digits in a randomly selected 4-digit PIN. Each PIN has four positions, and each position can be filled with a digit ranging from 0 to 9.
02

Define the Range of X

Since each digit in the PIN can be either zero or a non-zero digit, the value of \(X\), which represents the count of nonzero digits, can range from 0 (if all are zeros) to 4 (if all are nonzero digits). Thus, the possible values for \(X\) are 0, 1, 2, 3, and 4.
03

Explore Possible Outcomes

Let's consider three possible 4-digit PIN outcomes and their corresponding values of \(X\): 1. PIN: 0000 - Here, all digits are zero, so \(X = 0\).2. PIN: 1230 - Only the last digit is zero, so \(X = 3\).3. PIN: 5678 - All digits are non-zero, so \(X = 4\).
04

Explain Possible Values

Based on the problem, any 4-digit PIN can have from 0 to 4 non-zero digits, which are represented by the possible values for \(X\). The total PINs such as 0000, 1230, and 5678 illustrate the variation in non-zero digits.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arranging, and analyzing combinations of elements within a given set. In this context, we apply combinatorics to explore the possible arrangements of digits in a 4-digit PIN. Each digit can be a number from 0 to 9, offering 10 choices per position in the PIN. Because there are four positions, the number of possible PIN combinations is determined by multiplying these possibilities together: - Total combinations = 10 choices (for the first digit) x 10 choices (for the second digit) x 10 choices (for the third digit) x 10 choices (for the fourth digit) = 10,000. This gives us 10,000 unique PINs. Combinatorics helps us understand how many different configurations we can potentially achieve from this four-digit selection, highlighting the importance of counting methods in probability scenarios.
Random Selection
Random selection in probability refers to the process of making a choice randomly from a set of possibilities, ensuring that each choice is equally likely to be selected. In the case of randomly selecting a 4-digit PIN, the assumption is that every digit from 0 to 9 has an equal chance of appearing in each of the four positions.
  • If a PIN is generated randomly, it guarantees that every specific PIN (such as 1234 or 5678) is equally likely to occur as any other.
  • This means the probability of any specific PIN combination is 1 out of 10,000, or 0.01%.
Random selection ensures fairness and lack of bias, but it also opens up the need to understand probabilities related to specific desires, such as calculating how often a PIN will contain a specific number of nonzero digits.
Digit Analysis
Digit analysis is a method used to evaluate and understand the composition and characteristics of numerical sequences, such as PIN codes. If we consider a 4-digit PIN, digit analysis involves examining each individual digit to ascertain specific details of interest, in this case, the number of nonzero digits present. When analyzing a PIN like 1230: - The first three digits (1, 2, 3) are non-zero, making the count of non-zero digits three. - Digit analysis helps determine configurations such as whether the digits form patterns, repeat, or meet specific requirements, such as zero versus non-zero occurrence. Understanding digit analysis is vital as it allows us to explore the diversity of possibilities within a seemingly constrained set, providing deeper insights particularly in setting security codes or other related fields.

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Most popular questions from this chapter

Show that \(E(X)=n p\) when \(X\) is a binomial random variable. [Hint: First express \(E(X)\) as a sum with lower limit \(x=1\). Then factor out \(n p\), let \(y=x-1\) so that the sum is from \(y=0\) to \(y=n-1\), and show that the sum equals 1.]

Compute the following binomial probabilities directly from the formula for \(b(x ; n, p)\) : a. \(b(3 ; 8,-35)\) b. \(b(5 ; 8, .6)\) c. \(P(3 \leq X \leq 5)\) when \(n=7\) and \(p=.6\) d. \(P(1 \leq X)\) when \(n=9\) and \(p=.1\)

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An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30 , decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a. What is the probability that exactly 10 of these are from the second section? b. What is the probability that at least 10 of these are from the second section? c. What is the probability that at least 10 of these are from the same section? d. What are the mean value and standard deviation of the number among these 15 that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a certain chemical, which it sells to customers in 5 -lb batches. Let \(X=\) the number of batches ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X_{\text {.] }}\)

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