Questions: Angela Barnett 10/17/24 6:07 PM work: Section 11.1 Empirical and etical Probabilities (Law of Large Question 5, 11.1.13 HW Score: 95.42%, 19.08 of 20 points Part 1 of 3 Points: 0.67 of 1 Save In a given week, a veterinarian treated the animals in the accompanying table. Use this information to complete parts (a) through (c). a) Determine the empirical probability that the next animal she treats is a rabbit. b) Determine the empirical probability that the next animal she treats is a bird. c) Determine the empirical probability that the next animal she treats is a cat. Animal Number Treated Dog 35 Cat 40 Bird 20 Rabbit 15 a) P(rabbit) = (Type an integer or a simplified fraction.)

Angela Barnett 10/17/24 6:07 PM work: Section 11.1 Empirical and etical Probabilities (Law of Large Question 5, 11.1.13 HW Score: 95.42%, 19.08 of 20 points Part 1 of 3 Points: 0.67 of 1 Save In a given week, a veterinarian treated the animals in the accompanying table. Use this information to complete parts (a) through (c). a) Determine the empirical probability that the next animal she treats is a rabbit. b) Determine the empirical probability that the next animal she treats is a bird. c) Determine the empirical probability that the next animal she treats is a cat. Animal Number Treated Dog 35 Cat 40 Bird 20 Rabbit 15 a) P(rabbit) = (Type an integer or a simplified fraction.)
Transcript text: Angela Barnett 10/17/24 6:07 PM work: Section 11.1 Empirical and etical Probabilities (Law of Large Question 5, 11.1.13 HW Score: $95.42 \%, 19.08$ of 20 points ers) Part 1 of 3 Points: 0.67 of 1 Save In a given week, a veterinarian treated the animals in the accompanying table. Use this information to complete parts (a) through (c). a) Determine the empirical probability that the next animal she treats is a rabbit. b) Determine the empirical probability that the next animal she treats is a bird. c) Determine the empirical probability that the next animal she treats is a cat. \begin{tabular}{|l|c|} \hline Animal & Number Treated \\ \hline Dog & 35 \\ Cat & 40 \\ Bird & 20 \\ Rabbit & 15 \\ \hline \end{tabular} on 2 $\qquad$ ion 3 a) $P($ rabbit $)=$ $\qquad$ (Type an integer or a simplified fraction.) tion 4 stion 5 Clear all Check answer 6:07 PM
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Solution

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Solution Steps

To solve the problem, we need to calculate the empirical probability for each animal type. The empirical probability is determined by dividing the number of times an event occurs by the total number of events. In this case, the event is treating a specific type of animal, and the total number of events is the total number of animals treated.

  1. Calculate the total number of animals treated by summing the numbers for all animal types.
  2. For each animal type (rabbit, bird, cat), divide the number treated by the total number of animals to find the empirical probability.
Step 1: Calculate Total Number of Animals Treated

The total number of animals treated is given by the sum of all animals: Total Animals=35+40+20+15=110 \text{Total Animals} = 35 + 40 + 20 + 15 = 110

Step 2: Calculate Empirical Probability for Rabbit

The empirical probability that the next animal treated is a rabbit is calculated as: P(Rabbit)=Number of RabbitsTotal Animals=15110≈0.1364 P(\text{Rabbit}) = \frac{\text{Number of Rabbits}}{\text{Total Animals}} = \frac{15}{110} \approx 0.1364

Step 3: Calculate Empirical Probability for Bird

The empirical probability that the next animal treated is a bird is calculated as: P(Bird)=Number of BirdsTotal Animals=20110≈0.1818 P(\text{Bird}) = \frac{\text{Number of Birds}}{\text{Total Animals}} = \frac{20}{110} \approx 0.1818

Step 4: Calculate Empirical Probability for Cat

The empirical probability that the next animal treated is a cat is calculated as: P(Cat)=Number of CatsTotal Animals=40110≈0.3636 P(\text{Cat}) = \frac{\text{Number of Cats}}{\text{Total Animals}} = \frac{40}{110} \approx 0.3636

Final Answer

The empirical probabilities are:

  • P(Rabbit)≈0.1364 P(\text{Rabbit}) \approx 0.1364
  • P(Bird)≈0.1818 P(\text{Bird}) \approx 0.1818
  • P(Cat)≈0.3636 P(\text{Cat}) \approx 0.3636

Thus, the final answers are: P(Rabbit)≈0.1364 \boxed{P(\text{Rabbit}) \approx 0.1364} P(Bird)≈0.1818 \boxed{P(\text{Bird}) \approx 0.1818} P(Cat)≈0.3636 \boxed{P(\text{Cat}) \approx 0.3636}

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