Question

The following questions are from ARTIST (reproduced with permission) A bowl has 100 wrapped hard candies in it. 20 are yellow, 50 are red, and 30 are blue. They are well mixed up in the bowl. Jenny pulls out a handful of 10 candies, counts the number of reds, and tells her teacher. The teacher writes the number of red candies on a list. Then, Jenny puts the candies back into the bowl, and mixes them all up again. Four of Jenny's classmates, Jack, Julie, Jason, and Jerry do the same thing. They each pick ten candies, count the reds, and the teacher writes down the number of reds. Then they put the candies back and mix them up again each time. The teacher's list for the number of reds is most likely to be (please select one): a. 8,9,7,10,9 b. 3,7,5,8,5 c. 5,5,5,5,5 d. 2,4,3,4,3 e. 3,0,9,2,8

Short answer

Option b: 3,7,5,8,5 is most likely.

Step-by-step solution

Calculate Probability of Drawing a Red Candy

First, to solve this exercise, calculate the probability of drawing a red candy from the bowl. There are 50 red candies out of 100 total candies, so the probability \( P(\text{red}) \) is given by: \[ P(\text{red}) = \frac{50}{100} = 0.5 \]

Expected Number of Red Candies

Since Jenny and her classmates each pick 10 candies, we find the expected number of red candies in a sample of 10.This is calculated using the probability from Step 1. \[ E(\text{reds in 10 candies}) = 10 \times 0.5 = 5 \]So, on average, we would expect 5 red candies in each pick of 10 candies.

Analyze Options Against Expected Outcome

Now, compare the given options with the expected average: - Option a: 8,9,7,10,9 (largely above 5) - Option b: 3,7,5,8,5 (sums around 5) - Option c: 5,5,5,5,5 (exactly 5 each time) - Option d: 2,4,3,4,3 (below 5) - Option e: 3,0,9,2,8 (too variable) Based on the expected value calculated, the most realistic scenario should have red counts averaging around 5.

Select the Most Likely List

Considering the calculations and the understanding that each student's draw is a random, independent event based on probabilities: Option b: 3,7,5,8,5, although showing variability, floats around the expected average of 5. This makes it the most realistic distribution of results because it reflects natural variation around the mean expectation.

Key concepts

Expected Value

The concept of expected value is an essential tool in probability. It helps us predict the average outcome of a random process. Expected value, often denoted as \( E(X) \), is calculated by multiplying each possible outcome by its probability and then summing all these products.
In the exercise, Jenny and her classmates each pick 10 candies. Calculating the expected number of red candies helps predict the average results over time. Since the probability of picking a red candy is 0.5, the expected number of red candies in 10 draws is:

• Expected value for reds = Total candies drawn \( \times \) Probability of red
• \( E(\text{reds}) = 10 \times 0.5 = 5 \)

This means, on average, we would expect to find 5 red candies each time 10 candies are selected.

Random Sampling

Random sampling involves selecting items from a population with every item having an equal chance of being chosen. This concept is highlighted in the exercise where Jenny and her classmates pull out 10 candies each time. By mixing the candies back into the bowl after each draw, they ensure each selection is an independent event where all candies are equally likely to be picked.
Random sampling is crucial for achieving results that are representative of the entire group. It prevents bias and helps generate reliable data that adhere to the expected distribution confirmed by statistical probabilities.

Probability Distribution

Probability distribution describes how probabilities are spread over different outcomes. In the case of Jenny's exercise, the probability distribution of drawing red candies shapes the list of observed red candy counts.
Options a, b, c, d, and e reflect different distributions. Option b aligns well with the expected distribution because although numbers vary, they average close to the expected value of 5. This distribution illustrates normal variability around a mean, reflecting real-world randomness rather than fixed outcomes or erratic data points.
Understanding probability distribution helps predict possible outcomes and analyze how observed data aligns with expected results.