Problem 25 - College Algebra

Question

For the following exercises, four coins are tossed. Find the probability of tossing either two heads or three heads.

Step-by-Step Solution

Answer

The probability is \( \frac{5}{8} \).

Step 1: Identify the Sample Space

When four coins are tossed, each coin has 2 possibilities: heads (H) or tails (T). The total number of outcomes is calculated as \(2^4 = 16\). Hence, the sample space consists of 16 possible outcomes: HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, TTHH, HTTT, THTT, TTHT, TTTH, TTTT, TTTH.

Step 2: Define the Desired Events

We are looking for the probability of getting either 2 heads or 3 heads. - For 2 heads (2H), the combinations are: HHTT, HTHT, HTTH, THHT, TTHH, THTH. - For 3 heads (3H), the combinations are: HHHT, HHTH, HTHH, THHH. Count these combinations: 6 combinations for 2 heads and 4 combinations for 3 heads.

Step 3: Calculate Probability of Each Event

The probability of getting each specific event can be calculated using the formula: \[ P(E) = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} \]Calculate the probability:- Probability of 2 heads: \( P( ext{2H}) = \frac{6}{16} = \frac{3}{8} \)- Probability of 3 heads: \( P( ext{3H}) = \frac{4}{16} = \frac{1}{4} = \frac{2}{8} \)

Step 4: Combine Probabilities of Events

Since the events of getting 2 heads or 3 heads are mutually exclusive (they cannot happen at the same time), their probabilities can be added together: \[ P( ext{2H or 3H}) = P( ext{2H}) + P( ext{3H}) = \frac{3}{8} + \frac{2}{8} = \frac{5}{8} \]

Step 5: State the Final Probability

The final probability of tossing either two heads or three heads when four coins are tossed is \( \frac{5}{8} \).

Key Concepts

Sample Space

In probability, the concept of a **sample space** is crucial for any experiment. It represents all the possible outcomes that an experiment can have. For instance, when you toss four coins, each coin can result in either heads (H) or tails (T). This means there are two possible outcomes per coin. To find the total number of outcomes for four coins, we use the equation \(2^4\), which equals 16. Thus, the sample space contains the following outcomes: HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, TTHH, HTTT, THTT, TTHT, TTTH, TTTT, and TTTH.
Sample space helps us understand the entirety of possible results before delving into specific probabilities. Knowing all possible outcomes allows us to accurately calculate the likelihood of any particular event occurring within the experiment.

Mutually Exclusive Events

**Mutually exclusive events** are events that cannot occur simultaneously. In simpler terms, if one event happens, the other cannot. In our exercise of tossing four coins, consider the events of tossing exactly two heads or tossing exactly three heads. You cannot simultaneously get exactly two heads and three heads in a single coin toss. Thus, these are mutually exclusive.
Understanding this concept is key because mutually exclusive events allow us to add their separate probabilities together to determine the probability of either event occurring. This simplifies the calculation and understanding of such probabilities in complex problems.

Combinatorics

**Combinatorics** is a branch of mathematics that deals with counting, arranging, and analyzing combinations of objects. In the context of probability, it's used to count how many ways specific outcomes can occur from a set of possibilities.
For example, when calculating the number of ways to get two heads or three heads in four coin tosses, combinatorics clarifies how many combinations fit each scenario. For two heads (2H), the combinations are HHTT, HTHT, HTTH, THHT, TTHH, and THTH, resulting in 6 ways. For three heads (3H), the combinations are HHHT, HHTH, HTHH, and THHH, resulting in 4 ways. Combinatorics helps in systematically identifying these combinations, ensuring no possibilities are missed.

Probability Formula

The **probability formula** is a fundamental part of calculating the chances of an event happening. This formula can be expressed as \( P(E) = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} \). It essentially compares the number of ways an event can happen to the total number of outcomes.
Using our exercise as a case study, we find that for tossing exactly two heads (2H), the probability is \( \frac{6}{16} \), since there are 6 successful outcomes for this event out of a sample space of 16. Similarly, for three heads (3H), the probability is \( \frac{4}{16} \).

• The probability of getting two heads \( P(\text{2H}) = \frac{3}{8} \)
• The probability of getting three heads \( P(\text{3H}) = \frac{1}{4} = \frac{2}{8} \)

Since these events are mutually exclusive, we add their probabilities to get a final result of \( \frac{5}{8} \) for either two or three heads. This formula equips us to systematically determine event probabilities and helps demystify complex probability problems.