Problem 20 - Algebra for College Students

Question

Solve each problem.Six coins are tossed. Find the probability of getting at least two heads. \(\frac{57}{64}\)

Step-by-Step Solution

Answer

The probability of getting at least two heads is \( \frac{57}{64} \).

Step 1: Understanding the Problem

We need to find the probability of getting at least two heads when tossing six coins. This means we are interested in the events where the number of heads is 2 through 6.

Step 2: Total Possible Outcomes

When tossing six coins, each coin has 2 outcomes (heads or tails). Therefore, the total number of possible outcomes is calculated as:\[ 2^6 = 64 \]

Step 3: Calculate At Least Two Heads

To find the probability of getting at least two heads, we need to calculate the probability of not getting two heads and then subtract from 1. First, let's find the probability of getting zero or one head.

Step 4: Calculate Zero or One Head Probability

Find the probability of getting exactly zero heads (all tails) and exactly one head.

Step 5: Probability of Zero Heads

There is 1 outcome where all coins are tails. Thus, the probability of zero heads is:\[ P(0 \, \text{heads}) = \frac{1}{64} \]

Step 6: Probability of One Head

The number of ways to choose 1 head from 6 tosses is given by \( \binom{6}{1} \). This results in:\[ \binom{6}{1} = 6 \]So, the probability of exactly one head is:\[ P(1 \, \text{head}) = \frac{6}{64} \]

Step 7: Sum of Zero or One Head Probabilities

Add the probabilities of getting zero or one head:\[ P(0 \, \text{heads or} \, 1 \, \text{head}) = \frac{1}{64} + \frac{6}{64} = \frac{7}{64} \]

Step 8: Final Probability Calculation

The probability of getting at least two heads is the complement of getting zero or one head:\[ P(\text{at least 2 heads}) = 1 - P(0 \, \text{heads or} \, 1 \, \text{head}) = 1 - \frac{7}{64} = \frac{57}{64} \]

Step 9: Conclusion

The probability of getting at least two heads when tossing six coins is \( \frac{57}{64} \).

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics dealing with the study of countable, discrete structures. In simpler terms, it involves counting groups of items and understanding how these groups are formed.
In probability, combinatorics helps us determine the number of ways to achieve a certain result. For example, when we toss six coins, each coin can land as either heads or tails. To determine the total possible outcomes for these coin tosses, we use combinatorics.

Here are some key points about combinatorics in probability problems:

• It involves calculating combinations and permutations.
• Combinations are used when the order does not matter, permutations are used when the order does matter.
• The formula for combinations, denoted as \( \binom{n}{k} \), calculates the number of ways to choose \( k \) items from \( n \) without regard to order. For example, in this problem, \( \binom{6}{1} \) represents the number of ways to choose 1 head out of 6 tosses.

By using combinatorics, we can compute the probability of different outcomes efficiently, which is essential for solving complex probability problems.

Binomial Experiments

A binomial experiment is a specific type of probability experiment that meets all of the following conditions:

• The experiment consists of a fixed number of trials.
• Each trial is independent of the others.
• There are exactly two possible outcomes for each trial (often referred to as "success" and "failure").
• The probability of success is the same for each trial.

This coin-tossing problem is a classic example of a binomial experiment. Each coin flip is a trial with two possible outcomes—heads or tails—making it straightforward to apply binomial probability.For six coin tosses:

• Each toss is independent.
• The probability of getting heads (or tails) is constant, often assumed to be \( \frac{1}{2} \) for a fair coin.
• The number of trials (n) is 6.

These features classify it as a binomial experiment, allowing us to calculate the probability of getting a certain number of heads using methods such as combinatorics or probability formulas.

Complement Rule in Probability

The complement rule is a fundamental principle in probability that helps calculate the likelihood of an event not occurring. It states:

If \( P(A) \) is the probability of event A occurring, then the probability of event A not occurring is \( 1 - P(A) \).
This can be written as:

• \( P(A') = 1 - P(A) \)

This rule is particularly useful when calculating the probability of getting at least a certain number of successes in problems like our coin-tossing example.In the exercise:

• We are tasked with finding the probability of getting at least two heads. Instead of directly calculating all outcomes with two or more heads, we first find the probability of getting less than two heads (either zero or one head), and then subtract this from one.
• By using the complement rule, we simplify the problem significantly and arrive at the solution: \( \frac{57}{64} \).

Using the complement rule eases calculations in probability, making it an indispensable tool for solving probability-related problems efficiently.