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Chapter 6: Problem 7

Distribution of Two Thumbtacks When a certain type of thumbtack is flipped, the probability of its landing tip up (U) is \(0.60\) and the probability of its landing tip down (D) is \(0.40\). Now suppose we flip two such thumbtacks: one red, one blue. Make a list of all the possible arrangements using \(\mathrm{U}\) for up and \(\mathrm{D}\) for down, listing the red one first; include both UD and DU. Find the probabilities of each possible outcome, and record the result in table form. Be sure the total of all the probabilities is \(1 .\)

Short Answer

Expert verified
The possible outcomes are UU, UD, DU, DD. The probability for each outcome respectively is: 0.36, 0.24, 0.24, 0.16 and they do sum up to 1.

Step by step solution

01

Enumeration of all possible outcomes

When two thumbtacks are flipped, the outcomes for each can be either up (U) or down (D). For two tacks, one red and one blue, with red listed first, the possible outcomes, when considering order, are: 'Red Up and Blue Up', 'Red Up and Blue Down', 'Red Down and Blue Up', 'Red Down and Blue Down'. In shorthand, this would be: RU BU, RU BD, RD BU, RD BD. Or in terms of U and D only: UU, UD, DU, DD.
02

Calculation of the probabilities for each outcome

The probability of each outcome is equal to the product of the probabilities of each event. Therefore: \n \[ P(RU BD) = P(RU) * P(BD) = 0.60 * 0.40 = 0.24 \] \n \[ P(RU BU) = P(RU) * P(BU) = 0.60 * 0.60 = 0.36 \]\n \[ P(RD BU) = P(RD) * P(BU) = 0.40 * 0.60 = 0.24 \]\n \[ P(RD BD) = P(RD) * P(BD) = 0.40 * 0.40 = 0.16 \]
03

Verification of Probability Sum

The sum of probabilities for all possible outcomes should be 1. By adding the calculated probabilities: 0.24 (for RU BD) + 0.36 (for RU BU) + 0.24 (for RD BU) + 0.16 (for RD BD), the total is indeed 1.

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Key Concepts

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

Outcome Enumeration
When dealing with probability distributions involving multiple objects, like thumbtacks, the first step is to list all possible outcomes. This is known as Outcome Enumeration. In our exercise, we have two thumbtacks: a red one and a blue one. Each thumbtack can land either tip up (U) or tip down (D).
To capture all possibilities, we arrange the outcomes in pairs, always listing the red tack's outcome first. This gives us four possible outcomes:
  • Red Up, Blue Up (UU)
  • Red Up, Blue Down (UD)
  • Red Down, Blue Up (DU)
  • Red Down, Blue Down (DD)
By systematically listing each scenario, we ensure that no possibilities are missed. This process is crucial in probability, as an incomplete list can lead to incorrect probability calculations.
Probability Calculation
Once all outcomes are enumerated, we calculate the probability for each. Probability Calculation is done using the multiplication rule for independent events. Each thumbtack is an independent event; thus, the probability of a combined outcome is the product of the individual probabilities of each event.
In this case:
  • The probability of the red thumbtack landing up is 0.60, and down is 0.40.
  • The same probabilities apply to the blue thumbtack, with 0.60 for up and 0.40 for down.
By multiplying these values, we find:
  • \[ P(UU) = 0.60 \times 0.60 = 0.36 \]
  • \[ P(UD) = 0.60 \times 0.40 = 0.24 \]
  • \[ P(DU) = 0.40 \times 0.60 = 0.24 \]
  • \[ P(DD) = 0.40 \times 0.40 = 0.16 \]
These values represent the chance of each specific outcome occurring. Correct calculation of these probabilities is key to understanding the likelihood of events in a probability distribution.
Event Probability
After calculating the probabilities of each individual outcome, we can assess the Event Probability. An event in probability theory is defined as one or more outcomes of an experiment. Here, the experiment is flipping the thumbtacks.
In practical terms, checking that the total probability equals 1 verifies that the enumeration and calculations include all possible results. For our thumbtacks:
  • Aggregate probabilities add up: 0.36 (UU) + 0.24 (UD) + 0.24 (DU) + 0.16 (DD) = 1.
Ensuring total probability sums to 1 confirms the event encompasses every foreseeable outcome. This completeness is fundamental for applying probability distributions in various real-world scenarios, as it guarantees that all potential outcomes have been considered and accounted for.

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Most popular questions from this chapter

A study of human body temperatures using healthy men showed a mean of \(98.1^{\circ} \mathrm{F}\) and a standard deviation of \(0.70^{\circ} \mathrm{F}\). Assume the temperatures are approximately Normally distributed. a. Find the percentage of healthy men with temperatures below \(98.6^{\circ} \mathrm{F}\) (that temperature was considered typical for many decades). b. What temperature does a healthy man have if his temperature is at the 76 th percentile?

The distribution of the math portion of SAT scores has a mean of 500 and a standard deviation of 100 , and the scores are approximately Normally distributed. a. What is the probability that one randomly selected person will have an SAT score of 550 or more? b. What is the probability that four randomly selected people will all have SAT scores of 550 or more? c. For 800 randomly selected people, what is the probability that 250 or more will have scores of 550 or more? d. For 800 randomly selected people, on average how many should have scores of 550 or more? Round to the nearest whole number. e. Find the standard deviation for part d. Round to the nearest whole number. f. Report the range of people out of 800 who should have scores of 550 or more from two standard deviations below the mean to two standard deviations above the mean. Use your rounded answers to part \(\mathrm{d}\) and \(\mathrm{e}\). g. If 400 out of 800 randomly selected people had scores of 550 or more, would you be surprised? Explain.

The undergraduate admission rate at Harvard University is about \(6 \%\) a. Assuming the admission rate is still \(6 \%\), in a sample of 100 applicants to Harvard, what is the probability that exactly 5 will be admitted? Assume that decisions to admit are independent. b. What is the probability that exactly 95 out of 100 applicants will be rejected?

Distribution of Two Dices When two dices are thrown, the probability of getting a multiple of 3 (M) is \(0.33\) and the probability of not getting a multiple of \(3(\mathrm{~N})\) is \(0.67\). Make a list of all possible arrangements for getting a multiple of 3 , using \(\mathrm{M}\) for multiples and \(\mathrm{N}\) for numbers that are not. Find the probabilities of each arrangement, and record the results in table form. Be sure the total of all the probabilities is \(1 .\)

College women have a mean height of 65 inches and a standard deviation of \(2.5\) inches. The distribution of heights for this group is Normal. Choose the correct StatCrunch output for finding the percentage of college women with heights of less than 63 inches, and report the correct percentage.
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