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Chapter 4: Problem 13

Express all probabilities as fractions. The author owns a safe in which he stores all of his great ideas for the next edition of this book. The safe "combination" consists of four numbers between 0 and 99 , and the safe is designed so that numbers can be repeated. If another author breaks in and tries to steal these ideas, what is the probability that he or she will get the correct combination on the first attempt? Assume that the numbers are randomly selected. Given the number of possibilities, does it seem feasible to try opening the safe by making random guesses for the combination?

Short Answer

Expert verified
The probability is \( \frac{1}{10^8} \). It is not feasible to guess the combination by random attempts.

Step by step solution

01

Understand the Total Number of Possibilities

Each number in the combination can be any number between 0 and 99. This gives 100 possible choices for each of the 4 numbers. Since the numbers can be repeated, the total number of different combinations is found by multiplying the number of choices for each number.
02

Calculate the Total Number of Combinations

Since there are 100 choices for each of the 4 positions, the total number of possible combinations is calculated as: \[ 100 \times 100 \times 100 \times 100 = 100^4 \]
03

Simplify the Calculation

Simplify the expression \(100^4\): \[ 100^4 = 100 \times 100 \times 100 \times 100 = 10^2 \times 10^2 \times 10^2 \times 10^2 = 10^8 \]
04

Calculate the Probability of One Specific Combination

Since there is only one correct combination out of the \(10^8\) possible combinations, the probability of guessing the correct one on the first attempt is: \[ \text{Probability} = \frac{1}{10^8} \]
05

Assessing Feasibility

Given the extremely low probability of \(\frac{1}{10^8}\) for guessing the correct combination on the first try, it does not seem feasible to attempt opening the safe by making random guesses.

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Key Concepts

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

Combinations
A combination is a selection of items from a larger pool where order does not matter. But in the context of our safe problem, we actually deal with something called a permutation, as the order of the digits matters.
In this scenario, we have a four-digit safe code, each digit ranging from 0 to 99. Since repeats are allowed, we use the multiplication principle to find the total number of permutations, represented by:
  • The number of choices for each digit, multiplied together.
This results in 100 choices per digit, giving us the total number of combinations as: \[ 100 \times 100 \times 100 \times 100 = 100^4 \]
Probability Calculation
Probability is a measure of the likelihood of a specific event occurring. It is calculated as the ratio of favorable outcomes to the total number of possible outcomes.
To find the probability of guessing the correct four-digit combination on the first try, we start with our previous calculation of possible combinations, which is: \[ 100^4 = 10^8 \]
Here, the probability of guessing the correct combination in one random guess is the number of successful outcomes (which is 1) over the total number of possible combinations (which is 10^8): \[ \text{Probability} = \frac{1}{10^8} \]
Random Guessing
Random guessing involves making an attempt to find the solution without any logical process or past information. In our scenario, if an author tries to guess the correct safe combination without any hints, each attempt is independent of the others.
The chances of guessing correctly on the first try are astronomically low. Given the probability of guessing correctly is \[\frac{1}{10^8} \], it demonstrates how impractical it is to rely on random guessing. Each new guess does not improve the odds, illustrating the challenge and futility of this approach.
Exponentiation
Exponentiation is a mathematical operation involving two numbers, the base and the exponent. The base is multiplied by itself as many times as the value of the exponent indicates.
In our case, we calculated the total number of combinations using exponentiation: \[ 100^4 \] This means 100 is multiplied by itself four times:
\[ 100 \times 100 \times 100 \times 100 = 10^8 \]
This simplified use of exponentiation helps us compute the total number of combinations quickly and effectively. Understanding exponentiation allows us to handle large numbers more conveniently and play a critical role in fields like probability theory.

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Most popular questions from this chapter

Find the probability and answer the questions. MicroSort's YSORT gender selection technique is designed to increase the likelihood that a baby will be a boy. At one point before clinical trials of the YSORT gender selection technique were discontinued, 291 births consisted of 239 baby boys and 52 baby girls (based on data from the Genetics \& IVF Institute). Based on these results, what is the probability of a boy born to a couple using MicroSort's YSORT method? Does it appear that the technique is effective in increasing the likelihood that a baby will be a boy?

Find the probability and answer the questions. Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective \(X\) chromosome that occurs without a paired \(X\) chromosome that is not defective. In the following, represent a defective \(X\) chromosome with lowercase \(x\), so a child with the \(x Y\) or Yx pair of chromosomes will have the disease and a child with \(X X\) or \(X Y\) or \(\mathrm{YX}\) or \(\mathrm{xX}\) or \(\mathrm{Xx}\) will not have the disease. Each parent contributes one of the chromosomes to the child. a. If a father has the defective \(x\) chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease? b. If a father has the defective \(x\) chromosome and the mother has good XX chromosomes, what is the probability that a daughter will inherit the disease? C. If a mother has one defective \(x\) chromosome and one good \(X\) chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease? d. If a mother has one defective \(x\) chromosome and one good \(X\) chromosome and the father has good XY chromosomes, what is the probability that a daughter will inherit the disease?

Express the indicated degree of likelihood as a probability value between 0 and \(1 .\) Based on a Harris poll, if you randomly select a traveler, there is a \(43 \%\) chance that his or her luggage is black.

A study addressed the issue of whether pregnant women can correctly predict the gender of their baby. Among 104 pregnant women, 57 correctly predicted the gender of their baby (based on data from "Are Women Carrying 'Basketballs'...," by Perry, DiPietro, Constigan, Birth, Vol. 26, No. 3). If pregnant women have no such ability, there is a \(0.327\) probability of getting such sample results by chance. What do you conclude?

Let \(A=\) the event of getting at least 1 defective iPhone when 3 iPhones are randomly selected with replacement from a batch. If \(5 \%\) of the iPhones in a batch are defective and the other \(95 \%\) are all good, which of the following are correct? a. \(P(\bar{A})=(0.95)(0.95)(0.95)=0.857\) b. \(P(A)=1-(0.95)(0.95)(0.95)=0.143\) c. \(P(A)=(0.05)(0.05)(0.05)=0.000125\)
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