/*! 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 32 Express all probabilities as fra... [FREE SOLUTION] | 91影视

91影视

Express all probabilities as fractions. Mendel conducted some his famous experiments with peas that were either smooth yellow plants or wrinkly green plants. If four peas are randomly selected from a batch consisting of four smooth yellow plants and four wrinkly green plants, find the probability that the four selected peas are of the same type.

Short Answer

Expert verified
The probability that the four selected peas are of the same type is \( \frac{1}{35} \).

Step by step solution

01

- Understand the Total Sample Space

First, determine the total number of ways to select 4 peas from the 8 available peas. This can be calculated using the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Here, \(n = 8\) and \(k = 4\). Calculate: \[ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]So, there are 70 possible ways to choose 4 peas out of 8.
02

- Calculate the Favorable Outcomes for Smooth Yellow Peas

Next, find the number of ways to select 4 smooth yellow peas from the 4 available smooth yellow peas. This is also a combination and can be calculated as: \[ \binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1 \]There is only 1 way to choose 4 smooth yellow peas from 4.
03

- Calculate the Favorable Outcomes for Wrinkly Green Peas

Now, find the number of ways to select 4 wrinkly green peas from the 4 available wrinkly green peas. Similar to the previous step, calculate: \[ \binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1 \]There is also only 1 way to choose 4 wrinkly green peas from 4.
04

- Calculate the Total Favorable Outcomes

Add the favorable outcomes from both smooth yellow peas and wrinkly green peas: \[ 1 + 1 = 2 \]So, there are 2 favorable ways out of the total 70 ways to choose 4 peas that are of the same type.
05

- Calculate the Probability

Finally, calculate the probability that the 4 selected peas are of the same type. Use the fraction of favorable outcomes over the total outcomes: \[ \frac{2}{70} = \frac{1}{35} \]Thus, the probability is \( \frac{1}{35} \).

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 91影视!

Key Concepts

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

Combination Formula
The combination formula is a fundamental concept in probability and statistics. It helps us determine how many ways we can choose a subset of items from a larger set without regard to the order of selection. The formula is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

Here:
  • \(n\) represents the total number of items.
  • \(k\) represents the number of items to choose.
  • \(!\) denotes factorial, which is the product of all positive integers up to that number. For example, \(!4 = 4 脳 3 脳 2 脳 1 = 24\).

In the example given, we have 8 peas (\(n = 8\)) and want to choose 4 of them (\(k = 4\)). Using the formula, we calculate:
\[ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{40320}{24 \times 24} = 70 \]
This shows there are 70 different ways to select 4 peas from a total of 8.
Sample Space
The sample space in probability refers to the set of all possible outcomes of an experiment. It's essential for determining probabilities because it gives us the context for calculating favorable outcomes.

For the problem at hand, the sample space is the total number of ways to select 4 peas from a batch of 8 (4 smooth yellow and 4 wrinkly green). We previously calculated that there are 70 possible ways to do this.

Understanding the sample space helps us see all possible scenarios. It enables us to identify and count the number of favorable outcomes among these possibilities.
Favorable Outcomes
Favorable outcomes are the specific outcomes that satisfy the condition of our probability question. In this problem, we want to know the probability that all 4 selected peas are of the same type.

We have two types of peas: smooth yellow and wrinkly green. There is only one way to choose 4 smooth yellow peas from the 4 available, calculated as follows:
\[ \binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{24}{24 \times 1} = 1 \]
Similarly, there is only one way to choose 4 wrinkly green peas from the 4 available:
\[ \binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{24}{24 \times 1} = 1 \]
Therefore, the total number of favorable outcomes is 1 (smooth yellow) + 1 (wrinkly green) = 2.
Elementary Statistics
Elementary statistics involves the basic concepts and techniques used to summarize and make inferences about data. A key component is understanding probabilities, which is the likelihood of an event occurring.

In this exercise, we used elementary statistics methods to calculate the probability of a specific event: selecting 4 peas of the same type from a batch. Here's a step-by-step summary:
  • First, calculate the total number of possible outcomes using the combination formula.
  • Next, determine the number of favorable outcomes (same type peas).
  • Finally, find the probability by dividing the number of favorable outcomes by the total number of possible outcomes.

Thus, we calculated that the probability is \(\frac{1}{35}\). Understanding these basic steps and concepts helps build a strong foundation in elementary statistics and probability.

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

Express all probabilities as fractions. If radio station call letters must begin with either \(\mathrm{K}\) or \(\mathrm{W}\) and must include either two or three additional letters, how many different possibilities are there?

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 \(\mathrm{X}\) chromosome that occurs without a paired \(\mathrm{X}\) 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 XY or YX or \(x\) X or \(X x\) 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 \(X X\) 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?

Find the probability.At Least One. In Exercises \(5-12,\) find the probability. It has been reported that \(20 \%\) of iPhones manufactured by Foxconn for a product launch did not meet Apple's quality standards. An engineer needs at least one defective iPhone so she can try to identify the problem(s). If she randomly selects 15 iPhones from a very large batch, what is the probability that she will get at least 1 that is defective? Is that probability high enough so that she can be reasonably sure of getting a defect for her work?

a. Five 鈥渕athletes鈥 celebrate after solving a particularly challenging problem during competition. If each mathlete high fives each other mathlete exactly once, what is the total number of high fives? b. If n mathletes shake hands with each other exactly once, what is the total number of handshakes? c. How many different ways can five mathletes be seated at a round table? (Assume that if everyone moves to the right, the seating arrangement is the same.) d. How many different ways can n mathletes be seated at a round table?

Describe the simulation procedure. (For example, to simulate 10 births, use a random number generator to generate 10 integers between 0 and 1 inclusive, and consider 0 to be a male and 1 to be a female.) Lefties Ten percent of people are left-handed. In a study of dexterity, 15 people are randomly selected. Describe a procedure for using software or a TI- \(83 / 84\) Plus calculator to simulate the random selection of 15 people. Each of the 15 outcomes should be an indication of one of two results: (1) Subject is left-handed; (2) subject is not left-handed.

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.