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What is the probability of exactly three heads when you flip a coin five times? What is the probability of three or more heads when you flip a coin five times?

Short Answer

Expert verified
The probability of exactly three heads is 0.3125 and for three or more heads is 0.5.

Step by step solution

01

Define the Problem

We need to calculate two probabilities: 1) the probability of getting exactly three heads in five coin flips, and 2) the probability of getting three or more heads in five coin flips. The coin is fair, so the probability of heads (H) or tails (T) is 50% or 0.5 for each flip.
02

Use the Binomial Probability Formula

The probability of getting exactly k successes (heads) in n trials (flips) is given by the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient, \( p \) is the probability of success (0.5 for heads), \( n = 5 \) (total flips), and \( k = 3 \) (desired heads).
03

Calculate for Exactly Three Heads

Substitute the values into the formula: \[ P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{2} \] Calculate the binomial coefficient and powers: \[ \binom{5}{3} = \frac{5!}{3! (5-3)!} = 10 \] \[ P(X = 3) = 10 \times (0.5)^5 = 10 \times 0.03125 = 0.3125 \] Thus, the probability of exactly three heads is 0.3125.
04

Calculate for Three or More Heads

We need the probability of three, four, or five heads. Thus:\[ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) \]. We already calculated \( P(X=3) = 0.3125 \). Now calculate \( P(X=4) \) and \( P(X=5) \).
05

Calculate Probability for Four Heads

Use the formula for four heads:\[ P(X = 4) = \binom{5}{4} (0.5)^4 (0.5)^{1} \]\[ \binom{5}{4} = 5 \]\[ P(X = 4) = 5 \times (0.5)^5 = 5 \times 0.03125 = 0.15625 \].
06

Calculate Probability for Five Heads

Use the formula for five heads:\[ P(X = 5) = \binom{5}{5} (0.5)^5 (0.5)^{0} \]\[ \binom{5}{5} = 1 \]\[ P(X = 5) = 1 \times (0.5)^5 = 1 \times 0.03125 = 0.03125 \].
07

Sum Probabilities for Three or More Heads

Add the probabilities:\[ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) = 0.3125 + 0.15625 + 0.03125 = 0.5 \].The probability of getting three or more heads is 0.5.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics that deals with the analysis of random events. It provides a framework to quantify the likelihood of various outcomes in an experiment.
Probability theory is foundational to many studies involving stats and uncertainties. It defines how likely an event is to happen. For instance, when flipping a coin, we often want to determine probabilities such as getting a certain number of heads.
  • Each possible outcome in a probability model is assigned a probability value between 0 and 1. A probability of 0 means an event is impossible, while a probability of 1 indicates certainty.
  • All probabilities in a given situation must add up to 1. This means that every possible outcome has been considered. For a single coin flip, the probability of either heads or tails together must equal 1.
  • The probability of getting heads or tails with a fair coin is equally 0.5, assuming that the coin isn't biased.
Understanding probability theory helps in analyzing more complex events, such as determining the probability of several coin flips resulting in a specific number of heads.
Binomial Coefficient
The binomial coefficient is a crucial component in calculating probabilities in binomial scenarios. But what exactly is it?
The binomial coefficient, denoted as \( \binom{n}{k} \), is used in the binomial probability formula. It represents the number of ways to select \( k \) successes (like heads in a coin flip) out of \( n \) trials.
The formula for the binomial coefficient is:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
  • This results in a whole number which counts different arrangements of successes in trials.
  • For example, \( \binom{5}{3} \), which is 10, indicates there are 10 different ways to obtain exactly 3 heads in 5 flips.
By using binomial coefficients, we can easily solve probability problems with large sample sizes, without manually listing every possible outcome. This is why it's essential in probability theory, particularly in analyzing sequences of events such as our coin flip example.
Discrete Mathematics
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. This includes concepts like integers, graphs, and statements in logic.
In discrete math, problems often involve countable numbers, such as determining the number of successful coin flips in several trials.
  • One major area of discrete math is combinatorics, the study of counting. It involves counting different combinations (like our problem of coin flips).
  • Discrete probability, a part of discrete mathematics, focuses on systems with distinct outcomes seen in events like rolling dice or flipping coins.
  • These methods help simplify complex problems that would otherwise be difficult to analyze.
Discrete mathematics gives tools to understand and solve probability problems, like deciding the probability of achieving a specified number of heads in multiple coin flips, with precision and logical reasoning.

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Most popular questions from this chapter

A candy machine in a school has d different kinds of candy. Assume (for simplicity) that all these kinds of candy are equally popular and there is a large supply of each. Suppose that c children come to the machine, and each child purchases one package of candy. One of the kinds of candy is a Snackers bar. a. What is the probability that any given child purchases a Snackers bar? b. Let Yi be the number of Snackers bars that Child i purchases—Yi is either 0 or 1. What is the expected value of Yi? c. Let Y be the random variable Y1 + Y2 +···+ Yc. What is the expected value of Y ? d. What is the expected number of Snackers bars that are purchased? e. Does the same result apply to any of the varieties of candy?

In three flips of a coin, is the event that two flips in a row are heads independent of the event that there is an even number of heads?

Given an array \(A\) of length \(n\) (chosen from some set that has an underlying ordering), you can select the largest element of the array by first setting \(L=A[1]\) and then comparing \(L\) to the remaining elements of the array, one at a time, replacing \(L\) with \(A[i]\) if \(A[i]\) is larger than \(L\). Assume that the elements of \(A\) are randomly chosen. For \(i>1\), let \(X_{i}=1\) if an element \(i\) of \(A\) is larger than any element of \(A[1: i-1]\). Let \(X_{1}=1 .\) What does \(X_{1}+X_{2}+\cdots+X_{n}\) have to do with the number of times you assign a value to \(L ?\) What is the expected number of times you assign a value to \(L\) ?

There is a retired professor who used to love to go into a probability class of 30 or more students and announce, "I will give even money odds that there are two people in this classroom with the same birthday." With 30 students in the room, what is the probability that all have different birthdays? What is the minimum number of students that must be in the room so that the professor has probability at least \(1 / 2\) of winning the bet? What is the probability that he wins his bet if there are 50 students in the room? Does this probability make sense to you? (There is no wrong answer to this last question!) Explain why or why not. (A programmable calculator, spreadsheet, computer program, or computer algebra system will be helpful in this problem.)

If you roll eight dice, what is the probability that each of the numbers 1 through 6 appear on top at least once? What about with nine dice?

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