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Compute the probability that in three flips of a coin, the coin comes up heads on the first flip or on the last flip.

Short Answer

Expert verified
The probability is \( \frac{3}{4} \).

Step by step solution

01

Understanding the Event

We need to find the probability that a coin flipped three times results in heads on either the first flip or the last flip. This can be expressed as the probability of the union of two events: getting heads on the first flip (event A) and getting heads on the last flip (event B).
02

Identify Individual Probabilities

The probability of getting heads on any single flip is \( P(H) = \frac{1}{2} \). Therefore, the probabilities of events A and B are both \( \frac{1}{2} \).
03

Understand Combined Events

Calculate the probability of both events A and B occurring simultaneously (i.e., getting heads on both the first and last flips), which is \( P(A \cap B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \).
04

Apply the Addition Rule

Using the addition rule for probability: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). Substitute the known values: \( P(A \cup B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{4} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4} \).
05

Conclusion

Therefore, the probability of getting heads on either the first flip or the last flip in three coin flips is \( \frac{3}{4} \).

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Key Concepts

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

Coin Flipping
Coin flipping is one of the most fundamental examples used to understand basic probability concepts. Imagine you have a coin with two sides: heads and tails. When you flip this coin, it's equally likely to land on either side, making the probability of heads or tails each \( \frac{1}{2} \). When involved in multiple coin flips, each flip is independent of the others. This means the outcome of one flip doesn't affect the others. For example, flipping a coin three times doesn’t change the fundamental probability of each flip (still \( \frac{1}{2} \) for heads). The challenge here is to figure out the probability of specific outcomes occurring over multiple flips.
Addition Rule
The addition rule is crucial for calculating probabilities involving the union of events. Essentially, it helps us find the probability of either of two events happening. In mathematical terms, if you have two events, A and B, that could overlap, the rule states:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
In simpler terms, you add the probability of each event but subtract the probability of them both happening, as this overlap would be counted twice.For instance, if you need heads on either the first flip or the last flip of a coin, consider each possibility's individual probability (such as heads on the first flip) and adjust for any overlaps (like heads on both). This way, you're accurately considering how these events interact.
Union of Events
The idea of the union of events in probability relates to combining multiple potential outcomes into one. When discussing the union, noted as \( A \cup B \), we mean any situation where at least one of the events (A or B) occurs.Understanding this concept allows us to examine more complex scenarios beyond a single event's probability. For example, concerning coin flipping, if event A represents heads on the first flip and event B represents heads on the last flip, both events together form \( A \cup B \). We then want to calculate the probability that either, or both, of these events occur.Utilizing the union of events, and by applying the addition rule, students can tackle a variety of probability problems that involve multiple independent events.

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Most popular questions from this chapter

Which is more likely, or are both equally likely? a. Drawing an ace and a king when you draw two cards from among the 13 spades, or drawing an ace and a king when you draw two cards from an ordinary deck of 52 playing cards? b. Drawing an ace and a king of the same suit when you draw two cards from a deck, or drawing an ace and a king when you draw two cards from among the 13 spades?

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 \(L\) ?

The eight kings and queens are removed from a deck of 52 cards, and then two of these cards are selected. What is the probability that the king or queen of spades is among the cards selected?

Evaluate the sum $$ \sum_{i=0}^{10} i\left(\begin{array}{c} 10 \\ i \end{array}\right)(.9)^{i}(.1)^{10-i} $$ which arose in computing the expected number of right answers a person would have on a 10 -question test with probability \(.9\) of answering each question correctly. First, use the binomial theorem and calculus to show that $$ 10(.1+x)^{9}=\sum_{i=0}^{10} i\left(\begin{array}{c} 10 \\ i \end{array}\right)(.1)^{10-i} x^{i-1} $$ Substituting \(x=.9\) almost gives the sum you want on the right side of the equation, except that in every term of the sum, the power on 9 is one too small. Use some simple algebra to fix this and then explain why the expected number of right answers is 9 .

Give an example of two random variables \(X\) and \(Y\) such that \(E(X Y) \neq E(X) E(Y)\). Here \(X Y\) is the random variable with \((X Y)(s)=X(s) Y(s)\)

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