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# If Alice must've have classes on at least 2 days, why do you need the intersection of 3 $A_i^C$'s? [closed]

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Closed as not constructive by Peter Taylorâ€­ on Aug 11, 2021 at 07:20

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Can someone please rectify my MathJax? Please see the red phrase below.

1. The question itself never touts or postulates outright that Alice "must have classes on at least 2 days", which feels like an esoteric deduction. So why must she have classes on at least 2 days?

2. If Alice must've classes on at least 2 days, then don't we need merely the first 2 summations $\sum\limits_i P(A_i^C) - \sum\limits_{i < j} P(A_i^C \cap A_j^C)$? Why do we need $\sum\limits_{i < j<k} P(A_i^C \cap A_j^C \cap A_k^C)$?

Blitzstein, 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 denition of probability, or using inclusion-exclusion.)

I modified the solution in the Selected Solutions PDF, p 8.

### Inclusion-Exclusion Method

We will use inclusion-exclusion to find the probability of the complement, which is the event that she has at least one day with no classes. Let $A_i$ be the event of having at least one class on the $i^{th}$ day of the week. So $A_1$ means not having any classes on Mondays. $\cup A_i^C$ means there's at least one day with no classes.

Then $P(\cup\limits_{i = 5} A_i^C) = \sum\limits_i P(A_i^C) - \sum\limits_{i < j} P(A_i^C \cap A_j^C) + \sum\limits_{i < j<k} P(A_i^C \cap A_j^C \cap A_k^C)$

$\color{red}{\text{(terms with the intersection of 4 or more$A_i^C$'s are not needed since Alice must have classes on at least 2 days)}}$. We have

$P(A_1^C) = \dfrac{\dbinom{24}{7}}{\dbinom{30}{7}}, P(A_1^C \cap A_2^C) = \dfrac{\dbinom{18}{7}}{\dbinom{30}{7}}, P(A_1^C \cap A_2^C \cap A_3^C) = \dfrac{\dbinom{12}{7}}{\dbinom{30}{7}}$

and similarly for the other intersections. So $P(\cup\limits_{i = 5} A_i^C) = 5\dfrac{\dbinom{24}{7}}{\dbinom{30}{7}} - \dbinom{5}{2} \dfrac{\dbinom{18}{7}}{\dbinom{30}{7} + \dbinom{5}{3} \dfrac{\dbinom{12}{7}}{\dbinom{30}{7} = \dfrac{263}{377}$.

Therefore, $P(\cap\limit_{i = 5} A_i^C) = \dfrac{114}{377}$.

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