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(a) Draw a tree diagram to display all the possible head-tail sequences that can occur when you flip a coin three times. (b) How many sequences contain exactly two heads? (c) Probability extension: Assuming the sequences are all equally likely, what is the probability that you will get exactly two heads when you toss a coin three times?

Short Answer

Expert verified
There are 3 sequences with exactly two heads and the probability is \( \frac{3}{8} \).

Step by step solution

01

Understanding the Problem

We need to determine all possible outcomes for flipping a coin three times. For each flip, there are two possible outcomes: Heads (H) or Tails (T). We will represent these outcomes in a structured tree diagram.
02

Creating the Tree Diagram

Starting with the first coin flip, we have two branches: H and T. Each of these branches further divides into H and T for the second coin flip. For the third flip, each of these branches divides once more into H and T. This tree represents all possible sequences of outcomes for three flips.
03

Analyzing the Tree Diagram

The full tree shows all sequences: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. There are eight sequences in total, representing all possible outcomes of three coin flips.
04

Counting Sequences with Exactly Two Heads

From the tree diagram, identify sequences with exactly two H's: HHT, HTH, and THH. Count these sequences to find that there are three such sequences.
05

Calculating the Probability

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. There are three sequences with exactly two heads, and eight total sequences, so the probability is \( \frac{3}{8} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tree Diagram
A tree diagram is an excellent visual tool for systematically displaying all possible outcomes of a probabilistic event. In the scenario of flipping a coin three times, the diagram helps visualize every possible sequence that can occur. Each level of the tree represents a stage of the process, in this case, each coin flip.

- **First Coin Flip:** At the initial stage, the two possible outcomes are Heads (H) and Tails (T). We draw two branches from a starting point to represent these possibilities. - **Second Coin Flip:** From each outcome of the first flip (H and T), we draw two new branches for the second flip. Again, each can result in either H or T. - **Third Coin Flip:** Similarly, each outcome of the second flip splits further into two branches representing the third flip results, H or T.

By organizing the flips as branches, the tree diagram clarifies all eight possible sequences (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT), thus laying out the entire sample space.
Coin Flip Sequences
When considering sequences of coin flips, it's vital to understand the concept of permutations. For a sequence with three flips, each flip can result in one of two outcomes: H for heads or T for tails.

Each stage of the flipping introduces additional combinations: - **For one flip:** There are 2 outcomes: H and T. - **For two flips:** Each initial outcome branches into two more, resulting in HH, HT, TH, TT. - **For three flips:** Each of the four results can again be either H or T, leading to the eight sequences outlined in the tree diagram.

Each sequence represents a unique outcome of flipping a coin three times, affirming the combinatorial nature of such probabilistic experiments.
Outcomes Calculation
Calculating outcomes and their probabilities in coin flips involves a straightforward process. In this exercise, we determine the probability of getting exactly two heads in three coin flips.

Here's how we do it:- **Identifying Favorable Outcomes:** First, refer to the tree diagram to find sequences with exactly two heads: HHT, HTH, and THH.- **Count Favorable Outcomes:** Three sequences meet the criteria of having exactly two heads.- **Calculate Probability:** The probability formula is the number of favorable outcomes divided by the total number of possible outcomes. Hence, the probability is \( \frac{3}{8} \), since there are three favorable sequences out of eight possible.

This method of calculating probability allows for clear, quantifiable understanding of the likelihood of different outcomes in coin flip events, an important aspect in the study of probability.

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Most popular questions from this chapter

During the Computer Daze special promotion, a customer purchasing a computer and printer is given a choice of three free software packages. There are 10 different software packages from which to select. How many different groups of software packages can be selected?

USA Today gave the information shown in the table about ages of children receiving toys. The percentages represent all toys sold. What is the probability that a toy is purchased for someone (a) 6 years old or older? (b) 12 years old or younger? (c) between 6 and 12 years old? (d) between 3 and 9 years old? A child between 10 and 12 years old looks at this probability distribution and asks, "Why are people more likely to buy toys for kids older than I am (13 and over) than for kids in my age group (10-12)?" How would you respond? $$\begin{array}{l|c} \hline \text { Age (years) } & \text { Percentage of Toys } \\ \hline 2 \text { and under } & 15 \% \\ 3-5 & 22 \% \\ 6-9 & 270 / 6 \\ 10-12 & 14 \% \\ 13 \text { and over } & 22 \% \end{array}$$

You toss a pair of dice. (a) Determine the number of possible pairs of outcomes. (Recall that there are six possible outcomes for each die.) (b) There are three even numbers on each die. How many outcomes are possible with even numbers appearing on each die? (c) Probability extension: What is the probability that both dice will show an even number?

Sometimes probability statements are expressed in terms of odds. The odds in favor of an event \(A\) is the ratio \(\frac{P(A)}{P(n o t A)}=\frac{P(A)}{P\left(A^{c}\right)}\). For instance, if \(P(A)=0.60\), then \(P\left(A^{c}\right)=0.40\) and the odds in favor of \(A\) are \(\frac{0.60}{0.40}=\frac{6}{4}=\frac{3}{2}\), written as 3 to 2 or \(3: 2 .\) (a) Show that if we are given the odds in favor of event \(A\) as \(n: m\), the probability of event \(A\) is given by \(P(A)=\frac{n}{n+m} \cdot\) Hint \(:\) Solve the equation \(\frac{n}{m}=\frac{P(A)}{1-P(A)}\) for \(P(\bar{A})\). (b) A telemarketing supervisor tells a new worker that the odds of making a sale on a single call are 2 to \(15 .\) What is the probability of a successful call? (c) A sports announcer says that the odds a basketball player will make a free throw shot are 3 to \(5 .\) What is the probability the player will make the shot?

John runs a computer software store. Yesterday he counted 127 people who walked by his store, 58 of whom came into the store. Of the 58 , only 25 bought something in the store. (a) Estimate the probability that a person who walks by the store will enter the store. (b) Estimate the probability that a person who walks into the store will buy something. (c) Estimate the probability that a person who walks by the store will come in and buy something. (d) Estimate the probability that a person who comes into the store will buy nothing.

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