/*! 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 31 A homeowner plants 6 bulbs selec... [FREE SOLUTION] | 91Ó°ÊÓ

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A homeowner plants 6 bulbs selected at. random from a box containing 5 tulip bulbs and 4 daffodil bulbs. What is the probability that he planted 2 daffodil bulbs and 4 tulip bulbs?

Short Answer

Expert verified
The probability that the homeowner planted 2 daffodil bulbs and 4 tulip bulbs is approximately 0.35714 or 35.714%

Step by step solution

01

Identify total number of bulbs and their types

There are 9 bulbs in total, 5 being tulip bulbs and 4 being daffodil bulbs. The homeowner plants a total of 6 bulbs.
02

Determine the number of ways to select daffodil bulbs

The homeowner plants 2 daffodil bulbs. Hence, the number of ways to select 2 daffodil bulbs from 4 is given by the combination formula C(4, 2) = 4! / [2!(4-2)!] = 6 ways.
03

Determine the number of ways to select tulip bulbs

The homeowner plants 4 tulip bulbs. Hence, the number of ways to select 4 tulip bulbs from 5 is given by the combination formula C(5, 4) = 5! / [4!(5-4)!] =5 ways.
04

Calculate total ways to select 6 bulbs

The total ways to select 6 bulbs from 9 (5 tulips and 4 daffodils) is given by C(9, 6) = 9! / [6!(9-6)!] = 84 ways.
05

Calculate the probability

The probability that homeowner planted 2 daffodil bulbs and 4 tulip bulbs is calculated as the product of the number of ways to select daffodil bulbs and tulip bulbs divided by the total number ways to select 6 bulbs. This is calculated as (6 * 5 ) / 84 = 30 / 84 = 0.35714.

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Key Concepts

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

Combination Formula
When solving problems related to probability, it's often necessary to calculate how many different ways a certain event can occur. This is where the combination formula comes into play. It's used when the order of outcomes doesn't matter and we need to find the number of possible groups that can be formed from a larger set.

Mathematically, it's represented as \( C(n, k) = \frac{n!}{k! \times (n-k)!} \), where \( n \) is the total number of items, \( k \) is the number of items to choose, and \(!\) denotes a factorial. The factorial function is crucial here because it provides the number of ways to arrange \( n \) items in order. The combination formula simplifies the complexity of counting without having to list out all the possibilities.

For example, if you have 5 different books and you want to know how many ways you can choose 2 to take on a trip, you would use the combination formula. By plugging the numbers into the formula, \( C(5, 2) = \frac{5!}{2! \times (5-2)!}\), it shows there are 10 possible pairs of books you could end up choosing. Understanding this concept is imperative when solving probability problems, as it directly relates to how likely certain outcomes are based on the combination of choices made.
Factorial
The concept of a factorial is quite simple, yet it's a powerful tool in mathematics, particularly in combinatorics and probability calculations. The factorial of a non-negative integer \( n \) is the product of all positive integers less than or equal to \( n \), and it's represented by the symbol \( n! \).

So, if you were to calculate \( 5! \), you're essentially finding the product of all positive integers from 1 up to 5: \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \), which gives you 120. Factorials grow very rapidly with larger values of \( n \), illustrating the exponential increase in the number of ways to arrange items as their quantity grows.

This concept is particularly relevant when working with combination problems because it helps determine the total number of possible arrangements and selections. A clear understanding of factorials is therefore critical for students tackling probability and permutation problems.
Discrete Probability
When it comes to the type of probability most commonly calculated in textbooks, we're usually dealing with discrete probability. This is about finding the likelihood of occurrence of finite, countable events. It can be thought of as the chance that a particular outcome will occur out of a set of possible outcomes. Discrete probability is often presented in the form of a fraction or a percentage.

The essential formula for discrete probability is \( P(E) = \frac{\text{Number of ways event E can occur}}{\text{Total number of possible outcomes}} \). It's fundamental when approaching problems where each event is distinct and separate from the others.

In our example involving the planting of bulbs, we calculated the probability of a specific event occurring by understanding all the distinct possible outcomes and then identifying how many of those outcomes corresponded to the event of interest. Mastering the principles of discrete probability enables students to accurately predict outcomes and assess the likelihood of various events in a given context.

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