Problem 32
Question
A fraternity votes on whether to accept each of 5 pledges. How many different outcomes are possible for the vote?
Step-by-Step Solution
Verified Answer
There are 32 different outcomes possible for the vote.
1Step 1: Understand the Scenario
Consider that each pledge can either be accepted or rejected. This means there are two possible outcomes for each pledge.
2Step 2: Identify the number of options per pledge
Each of the 5 pledges can be accepted or rejected, giving each pledge 2 choices.
3Step 3: Use the Counting Principle
According to the Counting Principle, if each of the 5 pledges has 2 possible outcomes, we multiply the number of outcomes per pledge together. This results in: \[ 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \]
4Step 4: Calculate the Total Number of Outcomes
Calculate the expression \(2^5\) to find the total number of outcomes. \[ 2^5 = 32 \]
Key Concepts
permutations and combinationsbinomial outcomesexponential calculation
permutations and combinations
Permutations and combinations are two important concepts in counting. They help us determine different ways to arrange or select objects.
In permutations, the order matters. For example, arranging 3 books (A, B, C) in different ways results in ABC, ACB, BAC, BCA, CAB, and CBA. That's 6 different ways.
In combinations, the order doesn't matter. For instance, selecting 2 books out of 3 (A, B, C) means choosing AB, AC, or BC. There are only 3 ways.
For the fraternity vote example, where the order of accepting or rejecting each pledge doesn't matter, we use combinations. Each pledge's outcome affects only the final count, not the order.
In permutations, the order matters. For example, arranging 3 books (A, B, C) in different ways results in ABC, ACB, BAC, BCA, CAB, and CBA. That's 6 different ways.
In combinations, the order doesn't matter. For instance, selecting 2 books out of 3 (A, B, C) means choosing AB, AC, or BC. There are only 3 ways.
For the fraternity vote example, where the order of accepting or rejecting each pledge doesn't matter, we use combinations. Each pledge's outcome affects only the final count, not the order.
binomial outcomes
A binomial outcome is when there are exactly two possible results for an event. Examples include flipping a coin (heads or tails) or deciding on a vote (accept or reject).
In our fraternity votes on 5 pledges, each pledge has a binomial outcome: either accepted or rejected. This creates a total of 2 distinct outcomes for each pledge.
Binomial outcomes are crucial when using the Counting Principle, as they simplify the calculation by assigning only 2 choices per event. Thus, with 5 pledges, the total number of binomial outcomes is given by multiplying the outcomes of each step together.
In our fraternity votes on 5 pledges, each pledge has a binomial outcome: either accepted or rejected. This creates a total of 2 distinct outcomes for each pledge.
Binomial outcomes are crucial when using the Counting Principle, as they simplify the calculation by assigning only 2 choices per event. Thus, with 5 pledges, the total number of binomial outcomes is given by multiplying the outcomes of each step together.
exponential calculation
Exponential calculation is a method to determine the total number of outcomes when each event in a sequence has the same number of choices.
Here’s how it works:
For each pledge in our example, we have 2 outcomes (accept or reject). There are 5 pledges, so we raise the number of outcomes to the power of the number of pledges: equation: \[ 2^5 = 32 \] .
You can see this as multiplying 2 by itself 5 times: \[ 2 \times 2 \times 2 \times 2 \times 2 = 32 \].
This exponentiation makes calculating large numbers of outcomes much easier and faster than listing all possibilities.
Here’s how it works:
For each pledge in our example, we have 2 outcomes (accept or reject). There are 5 pledges, so we raise the number of outcomes to the power of the number of pledges: equation: \[ 2^5 = 32 \] .
You can see this as multiplying 2 by itself 5 times: \[ 2 \times 2 \times 2 \times 2 \times 2 = 32 \].
This exponentiation makes calculating large numbers of outcomes much easier and faster than listing all possibilities.
Other exercises in this chapter
Problem 31
How many ways are there to mark the answers to a test that consists of 10 true-false questions followed by 10 multiple-choice questions with 5 options each?
View solution Problem 31
If the probability that the stock market goes up tomorrow is \(\frac{3}{5},\) then what are the odds a) in favor of the stock market going up tomorrow? b) again
View solution Problem 32
If the probability of a coal miners' strike this year is \(\frac{9}{10}\) then what are the odds a) in favor of a strike? b) against a strike?
View solution Problem 33
Make a list of all of the ways to arrange the letters in the word MILK. How many arrangements should be in your list?
View solution