Box Models - Extra Practice

Question 1

In the notes, one of the examples is about dice, and sets up a box model for the process of rolling a pair of fair six sided dice and summing the spots. In the notes we make 2 draws at random with replacement from the following box, and sum the draws:

\[ \boxed{ \fbox{1}\; \fbox{2} \;\fbox{3} \;\fbox{4}\; \fbox{5} \; \fbox{6}} \]

Why not make one draw from the following box instead? (The tickets are the possible sums when we roll a pair of six sided dice.)

\[ \boxed{ \fbox{2}\; \fbox{3}\; \fbox{4} \; \fbox{5} \; \fbox{6} \; \fbox{7} \; \fbox{8} \; \fbox{9} \; \fbox{10}\; \fbox{11} \;\fbox{12}} \]

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Because the sums are not equally likely and so this box does not correctly represent the probabilities of each sum. For example, the chance of the sum of spots being \(2\) is \(\frac{1}{36}\), but the chance of the sum being \(7\) is \(\frac{6}{36}\), since there is only one way to get a sum of \(2\) ( + ), but \(6\) ways to get a sum of \(7\) (for example: + , + , + and so on).

If we did indeed want to model the result of two die rolls with a box from which we would only draw once, what would the box be? (List all the tickets in the box, which numbers, and how many of each.)

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The box would be:

\[ \boxed{ \begin{array}{c} \fbox{2}\; \fbox{3}\; \fbox{3}\; \fbox{4}\; \fbox{4}\; \fbox{4}\; \fbox{5}\; \fbox{5}\; \fbox{5}\; \fbox{5} \\ \fbox{6}\; \fbox{6}\; \fbox{6}\; \fbox{6}\; \fbox{6}\; \fbox{7}\; \fbox{7}\; \fbox{7}\; \fbox{7}\; \fbox{7}\; \fbox{7} \\ \fbox{8}\; \fbox{8}\; \fbox{8}\; \fbox{8}\; \fbox{8}\; \fbox{9}\; \fbox{9}\; \fbox{9}\; \fbox{9} \\ \fbox{10}\; \fbox{10}\; \fbox{10}\; \fbox{11}\; \fbox{11}\; \fbox{12} \end{array} } \]

Question 2

If we are setting up a box for modeling the number of heads in three tosses of a fair coin, would either of the boxes below work? If not, why not?

  1. \(\boxed{\fbox{0}\; \fbox{1}}\): Draw three times at random with replacement, and sum the draws.
  2. \(\boxed{\fbox{0} \;\fbox{1}\; \fbox{2} \;\fbox{3}}\): Draw once, the result is the number of heads.
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Only (a) would work, and the reasons are similar to the previous question. For (b) to be correct, the box would need to be: \[ \boxed{\fbox{0}\; \fbox{1}\; \fbox{1}\; \fbox{1}\; \fbox{2}\; \fbox{2}\; \fbox{2}\; \fbox{3}} \]

Question 3

A standard deck of cards has 52 cards, consisting of 4 “suits” (hearts , diamonds , spades \(\spadesuit\), and clubs \(\clubsuit\)). Each suit has 13 cards (Ace, King, Queen, Jack, 2, 3, 4, 5, 6, 7, 8, 9, and 10). Half of the cards are red (hearts and diamonds) and half of the cards are black (spades and clubs). I would like to shuffle the cards, draw five cards with replacement, and count the number of hearts in these five cards. What would be the box model for this set up?

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The box would contain \(52\) tickets, of which \(13\) would be marked with a \(1\) to represent the hearts, and the remaining \(39\) with \(0\) to represent the non-hearts. We could then draw \(5\) tickets with replacement and sum the values. This would give us the number of hearts in the drawn tickets.
Now, since we are drawing with replacement, all that matters is the probability of drawing a heart, which is \(1\) in \(4\), and so we could draw \(5\) tickets with replacement from the equivalent box \(\boxed{\fbox{0}\; \fbox{0}\; \fbox{0}\; \fbox{1}}\).

Question 4

One ticket will be drawn at random from each of the two boxes below:

\(A: \boxed{\fbox{1}\; \fbox{2} \;\fbox{3}}\)

\(B: \boxed{ \fbox{1} \;\fbox{2} \;\fbox{3} \;\fbox{4}}\)

  1. What is the chance the number drawn from \(A\) is greater than the one drawn from \(B\)?
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Box \(A\) has 3 tickets and box \(B\) has 4, so there are \(3 \times 4 = 12\) possible pairs of draws, and they are all equally likely. Here is the grid of all 12 pairs, with the ticket from \(A\) first:

\[ \begin{array}{c|cccc} & \fbox{1} & \fbox{2} & \fbox{3} & \fbox{4} \\ \hline \fbox{1} & (\fbox{1}, \fbox{1}) & (\fbox{1}, \fbox{2}) & (\fbox{1}, \fbox{3}) & (\fbox{1}, \fbox{4}) \\ \fbox{2} & (\fbox{2}, \fbox{1}) & (\fbox{2}, \fbox{2}) & (\fbox{2}, \fbox{3}) & (\fbox{2}, \fbox{4}) \\ \fbox{3} & (\fbox{3}, \fbox{1}) & (\fbox{3}, \fbox{2}) & (\fbox{3}, \fbox{3}) & (\fbox{3}, \fbox{4}) \end{array} \]

The rows are the ticket drawn from \(A\) and the columns are the ticket drawn from \(B\).

The number from \(A\) is greater than the number from \(B\) in the 3 pairs below the diagonal: \((\fbox{2}, \fbox{1})\), \((\fbox{3}, \fbox{1})\) and \((\fbox{3}, \fbox{2})\). So \[ P(A > B) = \frac{3}{12} = \frac{1}{4}. \]

  1. What is the chance that the number drawn from \(A\) is equal to the one drawn from \(B\)?
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Using the grid from part (a), the two numbers are equal in the 3 pairs on the diagonal: \((\fbox{1}, \fbox{1})\), \((\fbox{2}, \fbox{2})\) and \((\fbox{3}, \fbox{3})\). So \[ P(A = B) = \frac{3}{12} = \frac{1}{4}. \]

  1. What is the chance the number drawn from \(A\) is smaller than the one drawn from \(B\)?
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Using the grid from part (a), the number from \(A\) is smaller in the 6 pairs above the diagonal: \((\fbox{1}, \fbox{2})\), \((\fbox{1}, \fbox{3})\), \((\fbox{1}, \fbox{4})\), \((\fbox{2}, \fbox{3})\), \((\fbox{2}, \fbox{4})\) and \((\fbox{3}, \fbox{4})\). So \[ P(A < B) = \frac{6}{12} = \frac{1}{2}. \]

As a check, every pair is in exactly one of the three cases, so the three answers add up to 1: \(\frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1\). (We could also have found this answer as \(1 - \frac{1}{4} - \frac{1}{4} = \frac{1}{2}\).)

Question 5

I want to estimate the proportion of people in Berkeley who speak at least two languages. I stand at the corner of University and Shattuck and ask each person who goes by how many languages they speak, and keep a count of how many speak at least two. Can I set this up as a box model? If so, how? If not, why not?

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This cannot be set up as a box model as described here. We can imagine that the box represents the people living in Berkeley (Berkeley residents), with one ticket per Berkeley resident, marked “1” if that person speaks at least two languages and “0” if not. It is a box with an unknown number of 0s and 1s. Later, when we study sampling, we will think about such boxes, when we want to estimate the proportions. The number of draws in that situation will be our sample size. The main problem here is that standing at the corner of University and Shattuck and asking the people who walk by is not the same as taking a random sample of Berkeley residents. So it is not a box model of what we want (an estimated proportion of Berkeley residents who speak at least 2 languages), since the sample we get is not a random draw from the box defined above. Every resident is not equally likely to be chosen - we get some arbitrary selection of people, who may not even be Berkeley residents.

Question 6

In this increasingly genAI reliant world of ours, one happy casualty is the “Homework help” company Chegg. It was an unashamedly unethical venture that made its money by facilitating cheating. In 2021, it had about 7.8 million subscribers (as of 2021). Suppose Chegg surveyed all its subscribers asking if they had ever cheated on an assignment using Chegg (trying to show that they did not), and suppose it got back 1.5 million responses of which 87% of the respondents say that they had never used Chegg to cheat on an assignment.

  1. Could this survey be set up as a box model? If so, what tickets would go in the box and how many would we draw? If not, why not?
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No, we cannot set it up as a box model. We could imagine a box with one ticket for each of the 7.8 million subscribers, marked “1” if that subscriber has used Chegg to cheat and “0” if not. The survey would then be 1.5 million draws from this box.

The problem is that the 1.5 million responses are not draws at random from the box representing the Chegg subscribers, the respondents volunteered to reply, and so they chose themselves.
Chegg asked all 7.8 million subscribers, and only the 1.5 million who decided to answer are in the sample. Subscribers who cheated may have been less likely to respond to a survey from Chegg about cheating. (This is called non-response bias, and we will see it in the chapter From Samples to Populations.)

So the 87% is not the result of drawing tickets at random from the box, and a box model doesn’t describe this survey.

  1. Would this sample of 1.5 million be a representative sample of high school and college students?
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No. Chegg’s subscribers are not a random sample of high school and college students. They are students (or their parents) who chose to pay for a homework-help service, so they could differ from other students in many ways, including how likely they are to cheat. On top of that, the 1.5 million who responded chose themselves out of the 7.8 million subscribers, as described in part (a).

A sample of 1.5 million is very large, but a large sample does not fix bias. If the way people end up in the sample is not random, collecting more of them just gives a more precise estimate of the wrong quantity.