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Chapter 5: Problem 21

The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails \((44\) female and 36 male) and 60 are orange swordtails \((36\) female and 24 male). A fish is randomly captured from the aquarium: a. What is the probability the selected fish is a green swordtail? b. What is the probability the selected fish is male? c. What is the probability the selected fish is a male green swordtail? d. What is the probability the selected fish is either a male or a green swordtail?

Short Answer

Expert verified
a) \(\frac{4}{7}\), b) \(\frac{3}{7}\), c) \(\frac{9}{35}\), d) \(\frac{26}{35}\).

Step by step solution

01

Calculate Total Probability for Green Swordtails

To find the probability that a randomly selected fish is a green swordtail, divide the number of green swordtails by the total number of fish in the aquarium. There are 80 green swordtails out of a total of 140 fish. Thus, the probability is:\[P(\text{Green Swordtail}) = \frac{80}{140} = \frac{4}{7}\]
02

Calculate Total Probability for Male Fish

To find the probability that a randomly selected fish is male, sum the number of male fish (both green and orange swordtails), then divide by the total number of fish. There are 36 male green swordtails and 24 male orange swordtails, totalling 60 male fish. Thus, the probability is:\[P(\text{Male}) = \frac{36 + 24}{140} = \frac{60}{140} = \frac{3}{7}\]
03

Calculate Probability for Male Green Swordtails

Calculate the probability of selecting a male green swordtail. There are 36 male green swordtails. Divide this by the total number of fish to find the probability:\[P(\text{Male and Green Swordtail}) = \frac{36}{140} = \frac{9}{35}\]
04

Calculate Probability for Either Male or Green Swordtail

The probability of selecting either a male fish or a green swordtail should be calculated by adding their individual probabilities and subtracting the overlap (which is male green swordtails counted in both categories):\[P(\text{Male or Green Swordtail}) = P(\text{Male}) + P(\text{Green Swordtail}) - P(\text{Male and Green Swordtail})\]Substituting the values we have:\[P(\text{Male or Green Swordtail}) = \frac{3}{7} + \frac{4}{7} - \frac{9}{35} = \frac{7}{7} - \frac{9}{35} = 1 - \frac{9}{35} = \frac{26}{35}\]

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Key Concepts

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

Random Selection
Random selection is a fundamental principle in probability that refers to the process of selecting an item from a set without any bias or predetermined outcome. In our exercise, we randomly capture a fish from an aquarium. This implies that each fish has an equal chance of being selected. Understanding this concept is key to calculating probabilities, such as the probability of selecting a green swordtail among all the fish. Here鈥檚 how:
  • There are 140 fish in total, making this the entire set.
  • Randomly selecting means each fish, regardless of color or gender, has a \(\frac{1}{140}\) chance of being picked.
  • Calculating probabilities, like for green swordtails, involves dividing the number of favorable outcomes (e.g., 80 green swordtails) by the total number of possible outcomes (140 fish).
Understanding random selection helps in establishing the baseline probability of events independently before considering other factors like intersection or conditioning.
Conditional Probability
Conditional probability refers to the likelihood of an event occurring given that another event has already occurred. This concept is crucial when we want to determine probabilities that are not independent, as seen in some of our calculations. For example:
  • Finding the probability of selecting a green swordtail, given that the fish is male, involves examining only the subset of male fish.
  • In the exercise, once we know a fish is male, our sample space reduces from 140 fish to 60 male fish.
  • The conditional probability is then calculated by focusing on the male green swordtails (36) within this reduced space.
The formula used is \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), where \(| \cap |\) indicates an intersection of events. Understanding this helps you predict outcomes considering given conditions.
Intersection of Events
The intersection of events refers to the probability that two events will happen at the same time. We often use the symbol "\(\cap\)" to denote intersection. In our exercise, an example of this is determining the probability of a fish being both male and a green swordtail:
  • The intersection here are the 36 male green swordtails.
  • This probability isn鈥檛 independent; it's derived from intersecting two groups: males and green swordtails.
  • We can calculate the probability of this intersection by dividing the number of male green swordtails (36) by the total number of fish (140).
The formula \(P(A \cap B)\) aids in understanding events that can happen simultaneously, helping us organize how to compare and contrast overlapping outcomes.

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Most popular questions from this chapter

A large company must hire a new president. The board of directors prepares a list of five candidates, all of whom are equally qualified. Two of these candidates are members of a minority group. To avoid bias in the selection of the candidate, the company decides to select the president by lottery. a. What is the probability one of the minority candidates is hired? b. Which concept of probability did you use to make this estimate?

Berdine's Chicken Factory has several stores in the Hilton Head, South Carolina, area. When interviewing applicants for server positions, the owner would like to include information on the amount of tip a server can expect to earn per check (or bill). $$ \begin{array}{|ccc|} \hline \text { Amount of Tip } & \text { Number } \\ \hline \$ 0 \text { up to } \$ 20 & 200 \\ 20 \text { up to } & 50 & 100 \\ 50 \text { up to } & 100 & 75 \\ 100 \text { up to } & 200 & 75 \\ 200 \text { or more } & \frac{50}{500} \\ \text { Total } & & 500 \\ \hline \end{array} $$ a. What is the probability of a tip of \(\$ 200\) or more? b. Are the categories "\$0 up to \$20," "\$20 up to \$50," and so on considered mutually exclusive? c. If the probabilities associated with each outcome were totaled, what would that total be? d. What is the probability of a tip of up to \(\$ 50 ?\) e. What is the probability of a tip of less than \(\$ 200 ?\)

The events \(A\) and \(B\) are mutually exclusive. Suppose \(P(A)=.30\) and \(P(B)=.20\). What is the probability of either \(A\) or \(B\) occurring? What is the probability that neither A nor \(B\) will happen?

Althoff and Roll, an investment firm in Augusta, Georgia, advertises extensively in the Augusta Morning Gazette, the newspaper serving the region. The Gazette marketing staff estimates that \(60 \%\) of Althoff and Roll's potential market read the newspaper. It is further estimated that \(85 \%\) of those who read the Gazette remember the Althoff and Roll advertisement. a. What percent of the investment firm's potential market sees and remembers the advertisement? b. What percent of the investment firm's potential market sees, but does not remember, the advertisement?

There are 100 employees at Kiddie Carts International. Fifty-seven of the employees are hourly workers, 40 are supervisors, 2 are secretaries, and the remaining employee is the president. Suppose an employee is selected: a. What is the probability the selected employee is an hourly worker? b. What is the probability the selected employee is either an hourly worker or a supervisor? c. Refer to part (b). Are these events mutually exclusive? d. What is the probability the selected employee is neither an hourly worker nor a supervisor?
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