Problem 3 - Algebra 2 and Trigonometry

Question

In \(3-5,\) find exact probabilities showing all required computation. Five fair coins are tossed. Find the probability of the coins showing: \(\begin{array}{ll}{\text { a. at least four heads }} & {\text { b. at least three heads }} \\ {\text { c. at least two heads }} & {\text { d. at least one head }}\end{array}\)

Step-by-Step Solution

Answer

a) \(\frac{3}{16}\), b) \(\frac{1}{2}\), c) \(\frac{13}{16}\), d) \(\frac{31}{32}\).

Step 1: Define the Problem

We are examining the outcomes when five fair coins are tossed. Each coin has two possible outcomes: heads (H) or tails (T). The goal is to find probabilities of getting at least a certain number of heads.

Step 2: Understand the Total Number of Outcomes

Since each coin has 2 outcomes, and we have 5 coins, the total number of possible outcomes is calculated as \(2^5\). Therefore, there are 32 possible outcomes when five coins are tossed.

Step 3: Calculate Probability of Exactly k Heads

The probability of getting exactly \(k\) heads follows a binomial distribution, given by \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n\) is the number of trials (5 coins), \(p\) is the probability of heads (0.5), and \(k\) is the desired number of heads.

Step 4: Probability of At Least Four Heads

We need to find \(P(X \geq 4)\), which is the sum of the probabilities of getting exactly 4 heads and 5 heads: \[P(X=4) = \binom{5}{4} (0.5)^4 (0.5)^1 = 5 \cdot \frac{1}{32} = \frac{5}{32}\] \[P(X=5) = \binom{5}{5} (0.5)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32}\]Adding these gives \( P(X \geq 4) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}\).

Step 5: Probability of At Least Three Heads

We need to calculate \(P(X \geq 3)\), which includes 3, 4, or 5 heads: \[P(X=3) = \binom{5}{3} (0.5)^3 (0.5)^2 = 10 \cdot \frac{1}{32} = \frac{10}{32}\]Previously calculated \(P(X \geq 4) = \frac{6}{32}\)Adding these gives \(P(X \geq 3) = \frac{10}{32} + \frac{6}{32} = \frac{16}{32} = \frac{1}{2}\).

Step 6: Probability of At Least Two Heads

Find \(P(X \geq 2)\), which includes getting at least 2 heads: \[P(X=2) = \binom{5}{2} (0.5)^2 (0.5)^3 = 10 \cdot \frac{1}{32} = \frac{10}{32}\]Adding with \(P(X \geq 3) = \frac{16}{32}\) from Step 5\(P(X \geq 2) = \frac{10}{32} + \frac{16}{32} = \frac{26}{32} = \frac{13}{16}\).

Step 7: Probability of At Least One Head

Find \(P(X \geq 1)\), which includes all outcomes except the case with zero heads:\[P(X=0) = \binom{5}{0} (0.5)^0 (0.5)^5 = 1 \cdot \frac{1}{32} = \frac{1}{32}\]Thus, \(P(X \geq 1) = 1 - P(X=0) = 1 - \frac{1}{32} = \frac{31}{32}\).

Key Concepts

Binomial Distribution

The concept of the "Binomial Distribution" is central to this exercise. When dealing with multiple trials of a binary event, like coin tosses, the binomial distribution provides a way to calculate probabilities. It applies to situations where each trial is independent and has the same probability of success.

In our problem, tossing a fair coin gives each coin the probability of 0.5 (or 50%) to land on heads. The binomial distribution formula used is: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]

• \(n\) refers to the total number of coin tosses.
• \(k\) represents the exact number of heads you want to find the probability for.
• \(p\) is the probability of getting heads in a single toss.

This formula allows us to find the probability of any number of successes (heads) in a defined number of trials (coin tosses). Understanding this distribution is the key to solving problems like this efficiently.

Combinatorics

"Combinatorics" plays a vital role, providing the groundwork for counting ways to achieve different outcomes in the binomial setting. In our context, combinatorics is about finding how many ways we can choose \(k\) heads from \(n\) tosses.

The function \(\binom{n}{k}\) in the binomial formula is a combinatorial operator called a 'binomial coefficient'. It represents the number of ways to choose \(k\) successes (heads) out of \(n\) trials (tosses), calculated as: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

• Factorial, designated as \(!\), suggests the product of all positive integers up to that number.
• This counting is crucial for determining all possible ways to arrange heads and tails in sequences.

Mastering these counting techniques aids in evaluating probabilities accurately by counting the possible successes based on different configurations of heads and tails.

Coin Toss

A "Coin Toss" is the classic example used in probability theory to explain binary outcomes, where each trial has two possible results: heads or tails.

In the given problem, five fair coins are tossed, each flipping independently of the others. This independence and fairness ensure that the probability of heads on any single toss is 0.5. Therefore, with each experiment involving a toss sequence, understanding the mechanics of a single toss simplifies evaluating collective outcomes.

• The number of possible outcomes of n coins is \(2^n\).
• For 5 coins, this results in \(2^5 = 32\) possible sequences of heads and tails.

This simple mechanism of coin tossing underpins many foundational principles in probability and statistics. Becoming familiar with it helps learners extend these principles to a wide range of similar probability queries.

Probability Calculation

"Probability Calculation" is the final step that combines binomial distribution and combinatorics to find the occurrence likelihood of different head-count outcomes.

The goal is to calculate the probability of getting 'at least' a certain number of heads. The phrase "at least" implies combining probabilities over multiple values, rather than just one. For example:

• To find the probability of at least 4 heads, we combine probabilities for exactly 4 and 5 heads: \(P(X \geq 4) = P(X=4) + P(X=5)\).

Step by step, we use the binomial formula to calculate individual probabilities, while combining them yields the overall results for the desired conditions. Remember also that often the sum of probabilities for all possible outcomes equals 1, helping verify accuracy by considering complements when solving these problems.