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Chapter 12: Problem 55

Playing Pick 4. The Pick 4 games in many state lotteries announce a four-digit winning number each day. Each of the 10,000 possible numbers 0000 to 9999 has the same chance of winning. You win if your choice matches the winning digits. Suppose your chosen number is \(5974 .\) a. What is the probability that the winning number matches your number exactly? b. What is the probability that the winning number has the same digits as your number in any order?

Short Answer

Expert verified
a. Probability of exact match is \(\frac{1}{10,000}\). b. Probability of same digits in any order is \(\frac{3}{1,250}\).

Step by step solution

01

Understand the Total Number of Outcomes

The Pick 4 game consists of four-digit numbers ranging from 0000 to 9999. This gives a total number of outcomes: 10,000 (since each digit can be anything from 0 to 9).
02

Calculate Exact Matching Probability

For your chosen number to match the winning number exactly, only one combination is possible (5974). The probability is thus \[P(\text{exact match}) = \frac{1}{10,000}\]This means there is only one favorable outcome out of 10,000 possible outcomes.
03

Understand Permutations for Same Digits in Any Order

When considering the same digits in any order (5974), we need to determine how many different permutations of the digits 5, 9, 7, and 4 exist. Since all digits are unique, the total number of permutations is calculated using factorials: \[4! = 24\]Thus, there are 24 different ways to arrange these four digits.
04

Calculate Probability for Same Digits in Any Order

The probability that the winning number contains the digits 5, 9, 7, and 4 in any order is given by the number of those permutations over the total number of outcomes:\[P(\text{same digits, any order}) = \frac{24}{10,000}\]This simplifies to:\[P(\text{same digits, any order}) = \frac{3}{1,250}\]

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Key Concepts

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

Permutations
When we talk about permutations, we're looking at how many different ways we can arrange a set of items. In this case, we are interested in the four unique digits: 5, 9, 7, and 4. Imagine you had these digits written on separate cards. Every time you rearrange them into a new order, that's a different permutation.

Permutations are calculated using factorials, which are simply products of an integer and all the positive integers below it. Since we have four unique digits, the formula to find the permutations is:
  • Take the number 4.
  • Multiply it by the numbers lower than it: 4 脳 3 脳 2 脳 1
  • This gives us a total of 24 different permutations for arranging the digits in any order.


Permutations help us understand all the possible arrangements of a set of items. They are used to calculate the probability of specific ordered outcomes, which is vital in games like Pick 4.
Combinatorics
Combinatorics is all about counting combinations and permutations of objects. It explores fundamental counting principles that help us find probabilities and outcomes in various scenarios.

In the Pick 4 game, combinatorics simplifies understanding how numbers are arranged and reordered. In typical lottery questions, it involves calculating the number of possible outcomes where order may or may not matter. For this exercise, we used combinatorics to figure out how many ways the digits of the number 5974 can be rearranged.

Since all four digits are unique, we didn't have to deal with repeated elements. This makes our task straightforward using permutations as discussed earlier. Combinatorics provides the tools to study and solve problems involving these arrangements, and is fundamental in deriving probabilities for events in games, mathematics, and daily life.
Lottery probabilities
Lottery probabilities involve calculating the chances of a specific event occurring within a lottery game. Understanding these probabilities helps participants gauge their chances effectively.

For Part (a) of the Pick 4 exercise, the probability that your selected number (5974) matches the winning number exactly is calculated by considering that there is only one winning number among 10,000. Therefore, the probability is \[\frac{1}{10,000}\].

In Part (b), we explore the probability of the winning number containing the digits 5, 9, 7, and 4 in any order. Every permutation of these four digits represents a possible win. So, the probability is calculated by dividing the 24 permutations by the total possible numbers, resulting in the probability \[\frac{3}{1,250}\].

Knowing lottery probabilities enables strategic decision-making by making it clear how likely a specific event is within the context of what seems like a game of pure chance.

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Most popular questions from this chapter

Grades in a Business Course. Indiana University posts the grade distributions for its courses online. 11 Students in Business 100 in the spring 2019 semester received these grades: \(9 \%\) \(\mathrm{A}+, 15 \% \mathrm{~A}, 13 \% \mathrm{~A}-, 10 \% \mathrm{~B}+, 13 \% \mathrm{~B}, 8 \% \mathrm{~B}-7 \% \mathrm{C}+, 11 \% \mathrm{C}, 0 \% \mathrm{C}-, 2 \% \mathrm{D}+, 4 \% \mathrm{D}, 0 \% \mathrm{D}=\), and \(8 \%\) F. Choose a Business 100 student at random. To "choose at random" means to give every student the same chance to be chosen. The student's grade on a four-point scale (with A \(+=4.3\), \(\mathrm{A}=4, \mathrm{~A}-=3.7, \mathrm{~B}+=3.3, \mathrm{~B}=3.0, \mathrm{~B}-=2.7, \mathrm{C}+=2.3, \mathrm{C}=2.0, \mathrm{C}-=1.7, \mathrm{D}+=1.3, \mathrm{D}=1.0\), \(\mathrm{D}=0.7\), and \(\mathrm{F}=0.0\) ) is a random variable \(X\) with this probability distribution: \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline Value of \(X\) & \(0.0\) & \(0.7\) & \(1.0\) & \(1.3\) & \(1.7\) & \(2.0\) & \(2.3\) & \(2.7\) & \(3.0\) & \(3.3\) & \(3.7\) & \(4.0\) & \(4.3\) \\ \hline Probability & \(0.08\) & \(0.00\) & \(0.04\) & \(0.02\) & \(0.00\) & \(0.11\) & \(0.07\) & \(0.08\) & \(0.13\) & \(0.10\) & \(0.13\) & \(0.15\) & \(0.09\) \\ \hline \end{tabular} a. Is \(X\) a finite or continuous random variable? Explain your answer. b. Say in words what the meaning of \(P(X \geq 3.0)\) is. What is this probability? c. Write the event "the st udent got a grade poorer than \(\mathrm{B}\)-" in terms of values of the random variable \(X\). What is the probability of this event?

Education Among Young A dults. Choose a young adult (aged 25-29) at random. The probability is \(0.07\) that the person chosen did not complete high school, \(0.46\) that the person has a high school diploma but no further education, and \(0.37\) that the person has at least a bachelor's degree. a. What must be the probability that a randomly chosen young adult has some education beyond high school but does not have a bachelor's degree? b. What is the probability that a randomly chosen young adult has at least a high school education?

You read in a book on poker that the probability of being dealt a straight in a five-card poker hand is \(1 / 255\). This means that a. if you deal millions of poker hands, the fraction of them that contain a straight will be very close to \(1 / 255\). b. if you deal 255 poker hands, exactly one of them will contain a straight. c. if you deal 25,500 poker hands, exactly 100 of them will contain a straight.

Sample Space. In each of the following situations, describe a sample space \(S\) for the random phenomenon. a. A basket ball player shoots four free throws. You record the sequence of hits and misses. b. A basket ball player shoots four free throws. You record the number of baskets she makes.

Probability Models? In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate- that is, satisfies the rules of probability. Remember, a legitimate model need not be a practically reasonable model. If the assignment of probabilities is not legitimate, give specific reasons for your answer. a. Roll a six-sided die and record the count of spots on the upface: $$ \begin{array}{lll} P(1)=0 & P(2)=1 / 6 & P(3)=1 / 3 \\ P(4)=1 / 3 & P(5)=1 / 6 & P(6)=0 \end{array} $$ b. Deal a card from a shuffled deck: $$ \begin{array}{rlrl} P(\text { clubs }) & =12 / 52 & P(\text { diamonds }) & =12 / 52 \\ P(\text { hearts }) & =12 / 52 & P(\text { spades }) & =16 / 52 \end{array} $$ c. Choose a college student at random and record sex and enrollment status: $$ \begin{array}{rlrl} P(\text { female full-time }) & =0.56 & P(\text { male full-time }) & =0.44 \\\ P(\text { female part-time }) & =0.24 & P(\text { male part-time }) & =0.17 \end{array} $$
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