Recitation 19 Problem 3. Suppose there are 100 people in a room. Assume that their birthdays are in dependent and uniformly distributed. As stated in lecture notes, with probability >99% there will be two that have the same birthday. Now suppose you find out the birthdays of all the people in the room except one-call her "Jane"and find all 99 dates to be different (a) What's wrong with the following argument With probability greater than 99%, some pair of people in the room have the same birthday. Since the 99 people we asked all had different birth days, it follows that with probability greater than 99%o Jane has the same birthday as some other person in the room Solution. Here's the problem with the argument: Let A be the event that some two people in the room have the same birthday. Let b be the event that the 99 people we asked all have different birthdays It is true that Pr (A)>0.99(that is indeed the probability spoken about in the lecture notes). However, that is the a priori probability, i. e, assuming all the birthdays are uniform and independent, with no other constraints. The argument above makes the erroneous assumption that event a has probability of at least 99% even once we know that event b holds. But once we know that event b holds the birthdays are case, they are actually quite different as will be computed iqual to Pr(A)(in our (b)What is the actual probability that Jane has the same birthday as some other Solution. Let S be the set of birthdays of the 99 people in the room other than Jane By assumption, SI=99. Let b be the date of Jane's birthday. Since b is uniformly distributed over a set of size 365, and b is independent of all the birthdays in s,we nave Pr(A|B)=Pr(b∈S) 27.1% 365365 where B is the event that the 99 people we asked all have different birthdaysRecitation 19 4 Problem 3. Suppose there are 100 people in a room. Assume that their birthdays are in￾dependent and uniformly distributed. As stated in lecture notes, with probability > 99% there will be two that have the same birthday. Now suppose you find out the birthdays of all the people in the room except one—call her “Jane”—and find all 99 dates to be different. (a) What’s wrong with the following argument: With probability greater than 99%, some pair of people in the room have the same birthday. Since the 99 people we asked all had different birth￾days, it follows that with probability greater than 99% Jane has the same birthday as some other person in the room. Solution. Here’s the problem with the argument: Let A be the event that some two people in the room have the same birthday. Let B be the event that the 99 people we asked all have different birthdays. It is true that Pr (A) > 0.99 (that is indeed the probability spoken about in the lecture notes). However, that is the a priori probability, i.e., assuming all the birthdays are uniform and independent, with no other constraints. The argument above makes the erroneous assumption that event A has probability of at least 99% even once we know that event B holds. But once we know that event B holds, the birthdays are no longer independent. Thus Pr (A | B) is not necessarily equal to Pr (A) (in our case, they are actually quite different, as will be computed in part (b)). (b) What is the actual probability that Jane has the same birthday as some other person in the room? Solution. Let S be the set of birthdays of the 99 people in the room other than Jane. By assumption, |S| = 99. Let b be the date of Jane’s birthday. Since b is uniformly distributed over a set of size 365, and b is independent of all the birthdays in S, we have S 99 Pr (A B) = Pr (b ∈ S) = | | | = 365 365 ≈ 27.1%, where B is the event that the 99 people we asked all have different birthdays