/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 157 A pizza parlor has 12 different ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A pizza parlor has 12 different toppings available for its pizzas, and 2 of these toppings are pepperoni and anchovies. If a customer picks 2 toppings at random, find the probability that a. neither topping is anchovies b. pepperoni is one of the toppings

Short Answer

Expert verified
The probability that neither topping is anchovies is \(5/6\) and the probability that pepperoni is one of the toppings is \(1/6\).

Step by step solution

01

Calculate total possible combinations

Firstly, calculate how many different pairs of toppings a customer could choose. Use the formula for combinations \[C(n,k) = \frac{n!}{k!(n - k)!}\] Here, n equals 12 since we have 12 different toppings and k equals 2 since the customer picks two toppings. Plugging into the formula gives a total of \(C(12, 2) = 66\) possible pairs.
02

Calculate neither topping is anchovies

The task is to find the number of pairs not including anchovies. If anchovies are excluded, we only consider 11 possible toppings. Thus, into our formula, n is now 11 yet we still choose 2 toppings. Calculating gives 55 pairs. The probability is then found by dividing the number of favorable outcomes (pairs not including anchovies) by the total possible outcomes. Thus, \[P(\text{no anchovies}) = \frac{C(11, 2)}{C(12, 2)} = \frac{55}{66} = \frac{5}{6}\].
03

Calculate pepperoni is one of the toppings

Now we need to find the number of pairs that include pepperoni. If one of the toppings is guaranteed to be pepperoni, then the second topping is chosen from the remaining 11. This is a combination of 11 things taken 1 at a time, which gives us 11 pairs. Using the probability formula from above, the probability will be \[P(\text{pepperoni included}) = \frac{C(11, 1)}{C(12, 2)} = \frac{11}{66} = \frac{1}{6}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. In this context, it helps us determine the number of ways to choose or arrange items from a larger set. In our pizza problem, we are interested in finding out how many different pairs of pizza toppings a person can choose from a set of 12 toppings.

The basic formula used in these calculations is the combination formula, represented as \(C(n, k) = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items to choose from, \(k\) is the number of items to choose, and \(n!\) (read as 'n factorial') is the product of all positive integers up to \(n\).

Combinatorics is useful when order does not matter, which is typical in cases like our pizza example, where choosing pepperoni and olives is the same as choosing olives and pepperoni.
Probability Calculation
Probability calculation helps us quantify the likelihood of certain outcomes. In the context of our pizza toppings problem, we determine the probability of specific events occurring when picking two toppings at random.

  • The total number of possible outcomes for our problem is based on all possible combinations of two toppings from the available 12, calculated as \(C(12, 2) = 66\).
  • To find the probability of an event, divide the number of favorable outcomes by the total possible outcomes.
  • For instance, the probability of picking two toppings that do not include anchovies is derived by calculating \(C(11, 2) = 55\), and then dividing by the total, resulting in \(\frac{55}{66} = \frac{5}{6}\).
  • Similarly, the probability that one of the chosen toppings is pepperoni is \(\frac{11}{66} = \frac{1}{6}\).
These calculations allow us to make informed interpretations about the likelihood of these specific preferences.
Mathematical Combinations
Mathematical combinations are a core part of both combinatorics and probability theory. They provide a way to figure out how many ways we can select items from a group, without considering the order.

When dealing with the pizza problem, combinations were used to calculate the different sets of two toppings that could be chosen from the 12 available. The process involves plugging values into the combination formula \(C(n, k)\) to find the possible selection outcomes.

  • The calculation \(C(12, 2) = 66\) tells us there are 66 ways to choose 2 toppings from 12.
  • In personalized cases, such as excluding certain toppings like anchovies, the combinations adapt to consider only the non-excluded toppings, reducing the count accordingly, as seen in \(C(11, 2) = 55\).
  • Understanding these concepts of combinations helps simplify the task of calculating complex probabilities and possible outcomes.
Comprehending combinations effectively allows one to navigate numerous mathematical applications beyond just this simple pizza selecting example.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Five hundred employees were selected from a city's large private companies and asked whether or not they have any retirement benefits provided by their companies. Based on this information, the following two-way classification table was prepared. $$ \begin{array}{lcc} &{\text { Have Retirement Benefits }} \\ \hline { 2 - 3 } & \text { Yes } & \text { No } \\ \hline \text { Men } & 225 & 75 \\ \text { Women } & 150 & 50 \\ \hline \end{array} $$ a. Suppose one employee is selected at random from these 500 employees. Find the following probabilities. i. Probability of the intersection of events "woman" and "yes" ii. Probability of the intersection of events "no" and "man" b. Mention what other joint probabilities you can calculate for this table and then find them. You may draw a tree diagram to find these probabilities.

A random sample of 80 lawyers was taken, and they were asked if they are in favor of or against capital punishment. The following table gives the two-way classification of their responses. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Favors Capital } \\ \text { Punishment } \end{array} & \begin{array}{c} \text { Opposes Capital } \\ \text { Punishment } \end{array} \\ \hline \text { Male } & 32 & 24 \\ \text { Female } & 13 & 11 \\ \hline \end{array} $$ a. If one lawyer is randomly selected from this group, find the probability that this lawyer i. favors capital punishment ii. is a female iii. opposes capital punishment given that the lawyer is a female iv. is a male given that he favors capital punishment \(\mathrm{v}\). is a female and favors capital punishment vi. opposes capital punishment or is a male b. Are the events "female" and "opposes capital punishment" independent? Are they mutually exclusive? Explain why or why not.

Given that \(P(A \mid B)=.44\) and \(P(A\) and \(B)=.33\), find \(P(B)\).

Suppose that \(20 \%\) of all adults in a small town live alone, and \(8 \%\) of the adults live alone and have at least one pet. What is the probability that a randomly selected adult from this town has at least one pet given that this adult lives alone?

A veterinarian assigned to a racetrack has received a tip that one or more of the 12 horses in the third race have been doped. She has time to test only 3 horses. How many ways are there to randomly select 3 horses from these 12 horses? How many permutations are possible?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.