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Compute the variance Var(X) of the random variable X that counts the number of heads in four flips of a fair coin.

Short Answer

Expert verified
The variance Var(X) is 1.

Step by step solution

01

Define the Random Variable

The random variable X counts the number of heads in four flips of a fair coin. Each flip is a Bernoulli trial with p=0.5 (probability of getting heads). Thus, X follows a Binomial distribution with parameters n=4 and p=0.5.
02

Recall the Variance Formula for Binomial Distribution

For a binomially distributed random variable X, the variance Var(X) is given by the formula Var(X)=np(1p).
03

Substitute the Values

Substitute n=4 and p=0.5 into the variance formula: Var(X)=40.5(10.5).
04

Calculate the Result

Solve the expression 40.50.5 to find Var(X). This simplifies to 40.25=1.
05

Write Down the Conclusion

After computing the expression, we find that the variance Var(X)=1.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a fundamental concept in probability theory. It is used to model the number of successful outcomes in a fixed number of independent experiments, or trials, where each trial has the same probability of success. For example, if we flip a fair coin four times, we might be interested in counting how many times it lands on heads.

In our case, each coin flip represents a trial, and the probability of getting a head (success) is 0.5. The binomial distribution is characterized by two parameters: n, the number of trials, and p, the probability of success. For coin flips, n=4 and p=0.5.
  • "Success" is getting a head on a coin flip.
  • It has a defined number of trials, like our four flips.
  • Each trial is independent of others, meaning the outcome doesn't affect others.
The variance in a binomial distribution helps us understand the spread of our results around the expected number of successes. The formula for variance in a binomial distribution is Var(X)=np(1p). By plugging our coin parameters into this formula, we determined the variance to be 1. This implies that the number of heads we expect to see is on average evenly distributed around its expected value.
Bernoulli Trial
A Bernoulli trial is one simple kind of experiment where there are only two possible outcomes: success or failure. This is directly applied in our example of flipping a coin.

Each flip of a coin is a Bernoulli trial because it can result in either a head (success) or a tail (failure). In terms of the probability, the success (a head) has a fixed probability p, which is 0.5 in this scenario.
  • The probability of failure (a tail) is 1p, which would also be 0.5 here.
  • The assortment of successes and failures in a series of Bernoulli trials forms a binomial distribution.
  • These trials are foundational for understanding many types of probability models.
Understanding each flip as a Bernoulli trial helps in calculating probabilities over multiple trials and is an integral concept leading into more complex probability theory topics.
Probability Theory
Probability theory forms the backbone of statistics and is all about quantifying uncertainty. It encompasses a wide range of concepts used to analyze random events and outcomes.

In practical terms, it means understanding that various outcomes of an experiment are not deterministic but random. For the coin flip example, each flip is an event whose result can't be predicted with certainty, but with the help of probability theory, we can analyze and predict the likelihood of different outcomes.
  • Probability helps us determine the likelihood of events like getting exactly two heads in four coin flips.
  • It underpins statistical inference methods for making predictions and decisions.
  • Probability distributions, like the binomial distribution used here, serve as tools for modeling random variables.
Using probability theory, we can translate our real-world observations into mathematical models, allowing us to make assumptions and predictions about a wide range of phenomena. It's essential for fields like finance, medicine, and engineering.

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Most popular questions from this chapter

Wagga Wagga University has 15.000 students. Let X be the number of courses for which a randomly chosen student is registered. No student is registered for more than seven courses, and each student is registered for at least one course. The number of students registered for i courses where 1i7 is 150,450,1950,3750,5850,2550 , and 300 , respectively. Compute the expected value of the random variable X.

A television show features the following weekly game: A sports car is hidden behind one door, and a goat is hidden behind each of two other doors. The moderator of the show invites the contestant to pick a door at random. Then, by tradition, the moderator is obligated to open one of the two doors not chosen to reveal a goat (there are two goats, so there is always such a door to open). At this point, the contestant is given the opportunity to stand pat (do nothing) or to choose the remaining door. Suppose you are the contestant, and suppose you prefer the sports car over a goat as your prize. What do you do? (Hint: It may help to model this as a two-stage dependent trials process, but it may not be obvious how to do this). (a) Suppose you decide to stand with your original choice. What are your chances of winning the car? (b) Suppose you decide to switch to the remaining door. What are your chances of winning the car? (c) Suppose you decide to flip a fair coin. If it comes up heads, you change your choice; otherwise, you stand pat. What are your chances of winning the car?

Suppose that Ω is a sample space with a probability density function p, and suppose that AΩ. Let P(A) denote the probability of A. Assume that P(A)>0. Define a function p1 on A as follows: For ωA,p1(ω)=p(ω)/P(A). (a) Show that if ω1,ω2A and p(ω1),p(ω2)0, then p(ω1)p(ω2)=p1(ω1)p1(ω2) (b) Show that if B and C are nonempty subsets of A with elements that have positive probabilities, then P(B)P(C)=P1(B)P1(C) (c) Show that p1 is a probability density function on Ω1=A.

Compute the variance Var(X) of the random variable X that counts the number of heads in four flips of a coin that lands heads with a frequency of 1/3.

A fair die is rolled, and a fair coin is tossed. The sample space is taken to be Ω= Ω1×Ω2 where Ω1 is the six- clement sample space for the die and Ω2 is the twoelement sample space for the coin. Let AΩ1 be the event "a 5 is rolled." Let BΩ2 be the event "heads." Let CΩ be the event "at most two spots on the top face of the die (with heads or tails on the coin) or at least five spots on the top face of the die together with heads on the coin." Let D be the event "at least a 5 on the die (with heads or tails on the coin)." Which of the following sets of events are independent sets? Explain your answer. (a) A,B (b) A,B,C (c) \(\{B, C]\) (d) B,C,D

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