Problem 13 - Algebra 2

Question

Find each probability if a coin is tossed 5 times. \(P(\text { at least } 4 \text { tails) }\)

Step-by-Step Solution

Answer

\(P(\text{at least 4 tails}) = \frac{3}{16}\)

Step 1: Understand the Problem

We need to find the probability of getting at least 4 tails when a coin is tossed 5 times. This means we need to consider the scenarios where we get exactly 4 tails and exactly 5 tails.

Step 2: Determine the Sample Space

Each coin toss has two outcomes: heads (H) or tails (T). For 5 tosses, the sample space has \(2^5 = 32\) outcomes.

Step 3: Define the Events

Let's define the events: - Event A: getting exactly 4 tails.- Event B: getting exactly 5 tails. We need to find \(P(A \text{ or } B)\).

Step 4: Calculate \(P(A)\)

To calculate \(P(A)\), determine the number of ways to get exactly 4 tails. This is given by choosing 1 position out of 5 for the head, which is done in \( \binom{5}{4} = 5 \) ways.

Step 5: Calculate \(P(B)\)

To calculate \(P(B)\), determine the number of ways to get exactly 5 tails. There is only 1 way to have all 5 outcomes be tails, so \(\binom{5}{5} = 1\).

Step 6: Calculate \(P(\text{at least 4 tails})\)

Now we use the equation \(P(A \text{ or } B) = P(A) + P(B)\). Calculate each probability:- \(P(A) = \frac{5}{32}\)- \(P(B) = \frac{1}{32}\)So, \(P(\text{at least 4 tails}) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}\).

Key Concepts

Binomial Distribution

A binomial distribution is a specific probability distribution. It applies to scenarios where there are two possible outcomes, often referred to as a "success" or "failure." For the exercise at hand, each coin toss results in either heads or tails.
The key features of a binomial distribution are:

• There are a fixed number of trials, like the 5 tosses in this problem.
• Each trial is independent, meaning the result of one toss does not impact another.
• The probability of success (getting tails in this case) remains the same for each trial.

For our problem, we describe the distribution of tails in the sequence of 5 coin tosses. We are interested in two specific events: getting exactly 4 and 5 tails respectively. Knowing these characteristics helps us use binomial distribution techniques to compute probabilities.

Sample Space

A sample space encompasses all potential outcomes of an experiment. Here, the experiment involves tossing a coin 5 times, leading to numerous possible results.
In mathematical terms, if each coin toss has two results (heads, tails), the sample space for 5 tosses is computed as:

• Number of outcomes = \(2^5 = 32\) outcomes.

Each possible sequence from these 32 reflects a distinct outcome from our experiment, like HHHHH, THTTT, or TTTTT. Understanding the sample space is crucial because it serves as the foundation for calculating probabilities for each configuration within these 32 possible results.

Combination

Combinations play a vital role when calculating specific probabilities. A combination helps find the number of ways to choose items where order does not matter.
For our problem, we use combinations to determine the number of sequences that result in exactly 4 or 5 tails.

• Choosing 4 tails from 5 tosses:\[\binom{5}{4} = 5\]This tells us there are 5 scenarios of having exactly 4 tails.
• Choosing 5 tails is straightforward:\[\binom{5}{5} = 1\]This reveals that there is only one scenario for getting 5 tails.

Using combinations allows us to distinguish and count the occurrence of events we're interested in, such as getting a certain number of tails in a set sequence of tosses.

Event Probability

Event probability measures the likelihood of an event happening relative to the sample space.
In calculating the probability of events "at least 4 tails" when tossing a coin 5 times, we first find individual event probabilities and combine them.

• Probability of exactly 4 tails (\(P(A)\)):\[P(A) = \frac{5}{32}\]
• Probability of exactly 5 tails (\(P(B)\)):\[P(B) = \frac{1}{32}\]
• Combining these gives us the overall probability of getting at least 4 tails:\[P(A \text{ or } B) = \frac{5}{32} + \frac{1}{32} = \frac{6}{32} = \frac{3}{16}\]

By calculating the individual probabilities and summing them up, we derive meaningful insights about the entire event and its likelihood.