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In Problem 5 , if the coin is assumed fair, for \(n=3\) what are the probabilities associated with the values that \(X\) can take on?

Short Answer

Expert verified
The probabilities associated with the values that \(X\) can take on when the coin is tossed \(n = 3\) times are as follows: - \(P(X = 0) = \frac{1}{8}\) - \(P(X = 1) = \frac{3}{8}\) - \(P(X = 2) = \frac{3}{8}\) - \(P(X = 3) = \frac{1}{8}\)

Step by step solution

01

List all possible outcomes

Since we are tossing the coin \(3\) times, we will have a total of \(2^3 = 8\) possible outcomes. Let's list them: 1. HHH 2. HHT 3. HTH 4. HTT 5. THH 6. THT 7. TTH 8. TTT
02

Define the random variable X

We need to define the random variable \(X\) for this problem. We are not given any specific definition for \(X\), so we will assume that \(X\) represents the number of Heads obtained in the \(3\) coin tosses.
03

Assign values of X for each outcome

Now, let's assign the corresponding values of \(X\) for each of the possible outcomes: 1. HHH -> \(X = 3\) 2. HHT -> \(X = 2\) 3. HTH -> \(X = 2\) 4. HTT -> \(X = 1\) 5. THH -> \(X = 2\) 6. THT -> \(X = 1\) 7. TTH -> \(X = 1\) 8. TTT -> \(X = 0\)
04

Calculate probabilities for each value of X

We can now calculate the probabilities for each value of \(X\). Considering that the coin is fair, the probability of each outcome is \(\frac{1}{8}\): - \(P(X = 0)\): There is only one outcome with \(X = 0\) (TTT), so the probability is \(\frac{1}{8}\). - \(P(X = 1)\): There are three outcomes with \(X = 1\) (HTT, THT, TTH), so the probability is \(\frac{3}{8}\). - \(P(X = 2)\): There are three outcomes with \(X = 2\) (HHT, HTH, THH), so the probability is \(\frac{3}{8}\). - \(P(X = 3)\): There is only one outcome with \(X = 3\) (HHH), so the probability is \(\frac{1}{8}\).
05

Present final probabilities

The probabilities associated with the values that \(X\) can take on when the coin is tossed \(n = 3\) times are as follows: - \(P(X = 0) = \frac{1}{8}\) - \(P(X = 1) = \frac{3}{8}\) - \(P(X = 2) = \frac{3}{8}\) - \(P(X = 3) = \frac{1}{8}\)

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Key Concepts

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

Random Variables
Random variables are fundamental in probability and statistics as they help us translate real-world phenomena into mathematical language. Think of a random variable as a function or a rule that assigns a real number to each outcome of a random process. In our example of tossing a coin three times, we defined the random variable \(X\) as the number of heads obtained. This means:
  • Each outcome from the coin toss (like HHT or TTT) is converted into a numerical value through \(X\).
  • The value \(X = 2\) for outcome HHT indicates that there are two heads in this sequence.
Random variables can be discrete or continuous. In this problem, \(X\) is a discrete random variable because it takes on a finite number of possible values (0, 1, 2, or 3). Understanding random variables is crucial as they allow us to quantify uncertainties and analyze outcomes logically.
Probability Distribution
Once a random variable \(X\) is defined, we are interested in understanding how likely it is to take on each of its possible values. A probability distribution gives us this likelihood. In simple terms, it’s a function that maps each possible value of \(X\) to a probability.

For our coin-tossing exercise:
  • We calculated probabilities for \(X = 0, 1, 2,\) and \(3\).
  • The sum of these probabilities always equals 1, emphasizing that one of the outcomes is certain to happen.
Distributions provide a comprehensive picture of the behavior of random variables. They allow us to predict outcomes and make informed decisions. By looking at the distribution, such as the probabilities we found in the coin toss example, we can understand which outcomes are more likely and which are less likely.
Binomial Probability
The problem of tossing a fair coin three times and counting the number of heads is a classic example of binomial probability. A scenario that fits the binomial probability model has several key characteristics:
  • The experiment is repeated a fixed number of times, \(n\). For us, \(n = 3\).
  • Each trial is independent, meaning the outcome of one toss does not affect another.
  • Each trial results in one of two outcomes often labeled as "success" or "failure"—here, heads (success) or tails (failure).
The probability distribution associated with binomial scenarios is known as the binomial distribution. It predicts the probability of obtaining a certain number of successes in \(n\) trials. The binomial formula is:\[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \(\binom{n}{k}\) is a binomial coefficient representing the number of ways to choose \(k\) successes out of \(n\) trials, \(p\) is the probability of success on a single trial, and \(k\) is the number of successes. It's a powerful tool that helps us calculate probabilities efficiently in situations similar to our exercise.

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Most popular questions from this chapter

There are \(N\) distinct types of coupons, and each time one is obtained it will, independently of past choices, be of type \(i\) with probability \(P_{i}, i=1, \ldots\), \(N\). Let \(T\) denote the number one need select to obtain at least one of each type. Compute \(P\\{T=n\\}\).

A salesman has scheduled two appointments to sell encyclopedias. His first appointment will lead to a sale with probability \(.3\), and his second will lead independently to a sale with probability .6. Any sale made is equally likely to be either for the deluxe model, which costs \(\$ 1000\), or the standard model, which costs \(\$ 500\). Determine the probability mass function of \(X\), the total dollar value of all sales.

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If \(n\) married couples are seated at random at a round table, approximately what is the probability that no wife sits next to her husband? When \(n=10\) compare your approximation with the exact answer given in Example \(5 \mathrm{n}\) of Chapter 2 .

A satellite system consists of \(n\) components and functions on any given day if at least \(k\) of the \(n\) components function on that day. On a rainy day each of the components independently functions with probability \(p_{1}\), whereas on a dry day they each independently function with probability \(p_{2}\). If the probability of rain tomorrow is \(\alpha\), what is the probability that the satellite system will function?

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