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A coin is tossed twice. Let \(Z\) denote the number of heads on the first toss and \(W\) the total number of heads on the 2 tosses. If the coin is unbalanced and a head has a \(40 \%\) chance of occurring, find (a) the joint probability distribution of \(W\) and \(Z\); (b) the marginal distribution of \(W\); (c) the marginal distribution of \(Z\); (d) the probability that at least 1 head occurs.

Short Answer

Expert verified
The joint probability distribution of \(W\) and \(Z\) would be a \(3x2\) table with probability values for each combination of \(Z\) and \(W\). The marginal distribution of \(W\) would include probabilities for \(W=0\), \(W=1\), \(W=2\) obtained by summing probabilities from the joint distribution. The same applies for the marginal distribution of \(Z\) where probabilities for \(Z=0\) and \(Z=1\) are found from the joint distribution. The probability of obtaining at least one head would be the sum of probabilities for outcomes HT, TH, HH or equivalently \(1 - P_{TT}\).

Step by step solution

01

Outline Possible Outcomes and Probabilities

The possible outcomes for each coin flip are Head (H) or Tail (T) which means for two flips we have four possible outcomes: HH, HT, TH, TT. In this case, the probability of obtaining a head \(P(H) = 0.4\), and the probability obtaining a tail \(P(T) = 0.6\). These will be used to establish the sample space and corresponding probabilities. The variables \(Z\) and \(W\) represent number of heads on the first toss and total number of heads respectively.
02

Define Joint Probability Distribution of \(W\) and \(Z\)

Here, we establish the selectable outcomes for \(W\) (0, 1, 2 heads in total) and \(Z\) (0, 1 head on first flip) and the corresponding probabilities. It is important to remember that each coin flip is an independent event.
03

Determine the Marginal Distributions of \(W\) and \(Z\)

The marginal distribution of \(W\) and \(Z\) can be obtained by summing up the joint probabilities. For \(W\), add the probabilities for each possible number of total heads (0, 1, 2). For \(Z\), sum the probabilities for each possible number of heads on the first flip (0, 1 ).
04

Calculate the Probability of at Least One Head

This can be found by summing the probabilities of the outcomes that contain at least one head (HH, HT, TH) from the joint distribution table. Alternatively, one can also find this probability by subtracting the probability of not getting any heads (TT) from 1.

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Key Concepts

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

Probability and Statistics
Understanding the foundations of probability and statistics is crucial for analyzing real-life scenarios where outcomes are uncertain. Probability refers to the likelihood of an event’s occurrence and is expressed as a number between 0 and 1, where 0 indicates impossibility, and 1 represents certainty. The field of statistics often deals with the collection, analysis, interpretation, and presentation of data. One of the fundamental aspects covered in this domain is the probability distribution, which is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.

For instance, when a coin is tossed, there are two possible outcomes: heads or tails. If the coin in our situation has a 40% chance of landing on heads (unbalanced), this alters the typical 50-50 split found in a balanced coin and influences the calculation of probabilities. Dealing with real data often involves randomness and variability; hence, understanding both probability and statistics is essential to interpret the results effectively. Let's dive deeper into this with the concept of marginal distribution, which can help us analyze the outcomes more thoroughly.
Marginal Distribution
The term marginal distribution refers to the probability distribution of a subset of a collection of random variables. It is derived from the joint probability distribution of a set of variables, which shows the probability that each combination of outcomes occurs. By summing up the probabilities in the joint distribution that correspond to each outcome of a single variable, we obtain the marginal distribution for that variable.

In the exercise at hand, we calculated the marginal distributions of the variables \(W\) and \(Z\). \(W\) represents the total number of heads in two tosses, while \(Z\) denotes the number of heads on the first toss. To find the marginal distribution for each variable, we sum the probabilities across the rows or columns of the joint distribution table for \(W\) and \(Z\) respectively. This method simplifies the joint distribution into individual distributions, enabling us to analyze each variable separately, which can be particularly insightful when dealing with large data sets or complex probabilities.
Independent Events
The concept of independent events is pivotal in probability theory and plays a significant role in calculating the joint probability distribution of two or more random processes. Two events are said to be independent if the occurrence of one event has no effect on the probability of occurrence of the other event. In simpler terms, knowing the outcome of one event does not provide any information about the outcome of another if they are independent.

In the context of our exercise, each coin toss is an independent event. The probability of getting heads or tails on the first toss has no bearing on the outcome of the second toss. This assumption is essential when we define the joint probability distribution for \(W\) and \(Z\), as it allows us to compute the probabilities of combined outcomes by multiplying the probabilities of individual outcomes. If the tosses were not independent, the calculations would become more complex, as we would need to know the conditional probabilities. Having a clear understanding of independent events is, therefore, crucial in assessing probabilities in many random processes.

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Most popular questions from this chapter

A tobacco company produces blends of tobacco with each blend containing various proportions of Turkish, domestic, and other tobaccos. The proportions of Turkish and domestic in a blend are random variables with joint density function \((X=\) Turkish and \(Y=\) domestic \()\) $$ f(x, y)=\left\\{\begin{array}{ll} 24 x y, & 0 \leq x, y<_{-} 1: x+y \leq 1, \\ 0, & \text { elsewhere; } \end{array}\right. $$ (a) Find the probability that in a given box the Turkish tobacco accounts for over half the blend. (b) Find the marginal density function for the proportion of the domestic tobacco. (c) Find the probability that the proportion of Turkish tobacco is less than \(1 / 8\) if it is known that the blend contains \(3 / 4\) domestic tobacco.

The proportion of the budgets for a certain type of industrial company that is allotted to environmental and pollution control is coming under scrutiny. A data collection project determines that the distribution of these proportions is given by $$ f(y)=\left\\{\begin{array}{ll} 5(1-y)^{4}, & 0 \leq y \leq 1, \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Verify that the above is a valid density. (b) What is the probability that a company chosen at random expends less than \(10 \%\) of its budget on environmental and pollution controls? (c) What is the probability that a company selected at random spends more than \(50 \%\) on environmental and pollution control?

Let \(X\) denote the: number of heads and \(Y\) the number of heads minus the number of tails when 3 coins are tossed. Find the joint probability distribution of \(X\) and \(Y\).

A candy company distributes boxes of chocolates with a mixture of creams, toffees, and cordials. Suppose that the weight of each box is 1 kilogram, but the individual weights of the creams, toffees, and cordials vary from box to box. For a randomly selected box, let \(X\) and \(Y\) represent the weights of the creams and the toffees, respectively, and suppose that the joint density function of these variables is $$ f(x, y)=\left\\{\begin{array}{ll} 24 x y, & 0 \leq x \leq 1, \quad 0 \leq y \leq 1, \\ & x+y \leq 1 ,\\\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Find the probability that in a given box the cordials account for more than \(1 / 2\) of the weight. (b) Find the marginal density for the weight of the creams. (c) Find the probability that the weight of the toffees in a box is less than \(1 / 8\) of a kilogram if it is known that creams constitute \(3 / 4\) of the weight.

Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pound per square inch (psi). Let \(X\) denote the actual air pressure for the right tire and \(Y\) denote the actual air pressure for the left tire. Suppose that \(X\) and \(Y\) are random variables with the joint density $$ f(x, y)=\left\\{\begin{array}{ll} k\left(x^{2}+y^{2}\right), & 30 \leq x<50; \\ & 30 \leq y<50; \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Find \(k\). (b) Find \(\mathrm{P}(30 \leq X \leq 40\) and \(40 \leq Y<50)\) (c) Find the probability that both tires are underfilled.

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