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91Ó°ÊÓ

Define the simple events for the experiments in Exercises \(16-20 .\) Three children are randomly selected and their gender is recorded.

Short Answer

Expert verified
Answer: The simple events in this experiment are: - Simple Event 1 (E1): {MMM} - Simple Event 2 (E2): {MMF} - Simple Event 3 (E3): {MFM} - Simple Event 4 (E4): {MFF} - Simple Event 5 (E5): {FMF} - Simple Event 6 (E6): {FMM} - Simple Event 7 (E7): {FMМ} - Simple Event 8 (E8): {FFF}

Step by step solution

01

Determine possible gender combinations

In this experiment, there are two possible outcomes for each child's gender: Male or Female. Let's denote Male as "M" and Female as "F". Since we are selecting three children, we can create possible combinations of their genders.
02

List all possible combinations

With two possible outcomes for each child and three children, we can calculate the total number of combinations using 2^(3) = 8 combinations. Now, we'll list out all those combinations: 1. MMM 2. MMF 3. MFM 4. MFF 5. FMF 6. FMМ 7. FMM 8. FFF
03

Define the simple events

After determining all possible gender combinations for the three children, we can define the simple events for this experiment. A simple event in this context is the possible outcome of selecting three children and recording their genders: - Simple Event 1 (E1): {MMM} - Simple Event 2 (E2): {MMF} - Simple Event 3 (E3): {MFM} - Simple Event 4 (E4): {MFF} - Simple Event 5 (E5): {FMF} - Simple Event 6 (E6): {FMM} - Simple Event 7 (E7): {FMМ} - Simple Event 8 (E8): {FFF} These simple events can be used to further analyze the probabilities and outcomes of this experiment.

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Key Concepts

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

Probability Combinations
Understanding probability combinations is fundamental when we're predicting or analyzing real-life situations. Let's consider a situation where we are trying to determine all the possible gender combinations of three randomly selected children. This scenario treats each selection as independent, meaning the gender of one child does not influence the gender of the others. Since there are two potential outcomes for gender (Male or Female), and each child can be either, we inherently work with a concept known as permutations and combinations, a pillar of probability theory.

To enumerate these, one handy method is to use the powers of the number of choices per child. Specifically, with two possible outcomes (M or F) and three children, the total number of combinations is given by the mathematical expression: \(2^3 = 8\). This formula implies that there are eight possible ways these genders can be combined. You can visualize or list them out systematically, ensuring you cover all the possibilities like so:
  • MMM (all males)
  • MMF
  • MFM
  • MFF
  • FMF
  • FMM
  • FFM
  • FFF (all females)
Each of these gender combinations represents a unique scenario, showcasing the power of probability combinations in quantifying potential outcomes in simple events.
Gender Probability
Gender probability typically deals with the likelihood of the occurrence of a particular gender in a given scenario. In our example of selecting three children, we want to determine the probability of getting different gender combinations. Let's assume that the probability of selecting a male or female child is equally likely, seeing as there’s no information provided to suggest otherwise.

Counting Individual Scenarios

For instance, to calculate the probability of getting the combination of two males and one female ('MMF', 'MFM', or 'FMM'), we would consider each of these a favorable outcome. Since there are three such combinations out of eight total possibilities, we would say the probability is \(\frac{3}{8}\) for that specific event.

However, we must also remember that real-life gender ratios may not always be equally distributed, and probabilities can shift when additional information or assumptions alter the likelihood of selecting either gender. When studying probability, it’s crucial to recognize these nuances and understand how they influence outcomes.
Probability Outcomes
Probability outcomes refer to all the possible results of a random process. In the context of our exercise, they tie back to the notion of simple events. So when we're looking at a family with three children, we've already established the eight possible outcomes in terms of their gender. However, delving deeper, each of these outcomes has an associated probability.

If we assume each gender has an equal chance of being chosen, each child represents a coin flip, with 'heads' perhaps being male and 'tails' female. Under this assumption, each of the eight simple events - 'MMM', 'MMF', 'MFM', 'MFF', 'FMF', 'FMM', 'FFM', and 'FFF' — would have an identical probability of \(\frac{1}{8}\) because all outcomes are equally likely in this scenario.

Visualizing with a Tree Diagram

To further aid understanding, a tree diagram is a fantastic visualization tool used in probability. It branches out the possible outcomes at each stage, allowing students to follow the paths to the end results—our simple events. By clearly mapping these out, learners can better grasp how the combinations come together and the probability associated with each outcome, enhancing their overall comprehension of basic probability concepts.

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Most popular questions from this chapter

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. A bowl contains five candies-red, brown, yellow, blue, and orange. Draw two candies at random, one for you to eat, and one for a friend. What is the probability that you get the orange candy and your friend does not get the red one?

Suppose that \(P(A)=.4\) and \(P(B)=.2 .\) If events \(A\) and \(B\) are independent, find these probabilities: a. \(P(A \cap B)\) b. \(P(A \cup B)\)

A large number of adults are classified according to whether they were judged to need eyeglasses for reading and whether they actually used eyeglasses when reading. The proportions falling into the four categories are shown in the table. A single adult is selected from this group. Find the probabilities given here. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Used Eyeglasses } \\ \text { for Reading } \end{array} & \\ \hline \text { Judged to Need Eyeglasses } & \text { Yes } & \text { No } \\ \hline \text { Yes } & .44 & .14 \\ \text { No } & .02 & .40 \end{array} $$ a. The adult is judged to need eyeglasses. b. The adult needs eyeglasses for reading but does not use them. c. The adult uses eyeglasses for reading whether he or she needs them or not. d. An adult used glasses when they didn't need them.

City crime records show that \(20 \%\) of all crimes are violent and \(80 \%\) are nonviolent, involving theft, forgery, and so on. Ninety percent of violent crimes are reported versus \(70 \%\) of nonviolent crimes. a. What is the overall reporting rate for crimes in the city? b. If a crime in progress is reported to the police, what is the probability that the crime is violent? What is the probability that it is nonviolent? c. Refer to part b. If a crime in progress is reported to the police, why is it more likely that it is a nonviolent crime? Wouldn't violent crimes be more likely to be reported? Can you explain these results?

A certain virus afflicted the families in three adjacent houses in a row of 12 houses. If houses were randomly chosen from a row of 12 houses, what is the probability that the three houses would be adjacent? Is there reason to believe that this virus is contagious?

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