Why is the probability to have classes every weekday not ${6^5}{25 \choose 2}\left/{30 \choose 7}\right.$?

Question

This is Blitzstein's Introduction to Probability (2019 2 ed) Ch 1, Exercise 54, p 51.

Alice attends a small college in which each class meets only once a week. She is deciding between 30 non-overlapping classes. There are 6 classes to choose from for each day of the week, Monday through Friday. Trusting in the benevolence of randomness, Alice decides to register for 7 randomly selected classes out of the 30, with all choices equally likely. What is the probability that she will have classes every day, Monday through Friday? (This problem can be done either directly using the naive deﬁnition of probability, or using inclusion-exclusion.)

My wrong answer using naive definition of probability

Total choices are $ ^{30}{C}_7 $. There are buckets with 6 classes each for each of the five days. So we will need to choose one from these buckets. Then from the 25 choices left, choose random 2. Prob: $\dfrac{(6^5 \cdot ^{25}{C}_2)}{({30}{C}_7)}$ This is greater than 1, so something is obviously wrong. But what exactly?

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An error in your approach is that you have counted some sets of 7 choices at least twice. If I make 5 choices followed by 2 more choices, and you do the same, my set of 5 initial choices may differ from yours, but we may finish with the same set of 7 choices. Your approach would count these as different sets of 7. — DanielWainfleet, Aug 21 '18 at 15:36
Consider a much smaller problem where you have only one bucket and it has two balls in it, you take two balls. You ask what the probability that every bucket (all one of them) were taken from. According to your method, "pick which appears in the first bucket" $2$ choices, then "pick what else appears in the bucket", $1$ choices. and this is out of the $\binom{2}{2}=1$ ways of picking things, giving a supposed probability of $\dfrac{2\cdot 1}{1}$. Notice, in the numerator it supposedly mattered what order they were picked when in reality it shouldn't have mattered. — JMoravitz, Aug 21 '18 at 15:36
If $a, b, c, d , e $ are the days and $a_k$ are the subjects available each day ----$a_1, b_1, c_1, d_1, e_1$ and $a_2$ from from the $^{25}C_2$ were considered different from $a_2, b_1, c_1, d_1, e_1 $ and $a_1$ from from the $^{25}C_2$ in my approach. I assumed order mattered in the numerator as you have mentioned. Thanks for the input. T — gooner, Aug 21 '18 at 15:43
Does this answer your question? Why isn't the probability that Alice will have classes every weekday $\dfrac{{6^5}{25 \choose 2}}{30 \choose 7}$?? — , Jul 21 '21 at 08:25

score 1 · Accepted Answer · answered Aug 21 '18 at 15:28

The problem with your approach is the typical one;: you are counting the same thing more than once. Let $k\in\{1,2,3,4,5\}$ and suppose that the contents of the $k$^th bucket consists of $\{a_k,b_k,c_k,d_k,e_k\}$. Then Alice could choose the $a$'s from all buckets, together with $b_1$ and $c_1$. But this single possibility is counted as $3$ different possibilities using your method of counting, since you could group $b_1$ and $c_1$ together, you could group $a_1$ and $c_1$ together, and you could group $a_1$ and $b_1$ together.

Oh, essentially, $a_1, b_1, c_1, d_1, e_1 $ and $a_2$ from from the $^{25}C_2$ were considered different from $a_2, b_1, c_1, d_1, e_1 $ and $a_1$ from the $^{25}C_2$ in my approach. Hence the double counting right. I got the point. Much appreciated — gooner, Aug 21 '18 at 15:41

score 0 · Answer 2 · answered Aug 21 '18 at 15:28

Ok, suppose first that each class is taught twice a day: once in the morning and once in the afternoon. Then your method for computing the part above the division bar (is it called enumerator in English? I believe so) would be something like this: count all ways she can have one class in the morning each day and, at some point in the week, two classes in the afternoon. (in more detail: the 6^5 are the morning classes and the 25C2 the afternoon classes)

Now we get some info on what is wrong. Suppose that Math and English are both taught at Mondays. You count the event of English on Monday morning and Math in on Monday afternoon and the event of Math on Monday morning and English on Monday afternoon as two separate options. But of course in reality Alice doesn't have that much choice and there is only one way she can have both Math and English.

Why is the probability to have classes every weekday not ${6^5}{25 \choose 2}\left/{30 \choose 7}\right.$?

My wrong answer using naive definition of probability

2 Answers2