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Q1) Consider random class 0/ 23 persons born In 2000 (hds 366 days; Leap year). Delerinine the probabillty that two persons or more have (he same blithday...

Question

Q1) Consider random class 0/ 23 persons born In 2000 (hds 366 days; Leap year). Delerinine the probabillty that two persons or more have (he same blithday

Q1) Consider random class 0/ 23 persons born In 2000 (hds 366 days; Leap year). Delerinine the probabillty that two persons or more have (he same blithday



Answers

If 25 people are randomly selected, find the probability that no 2 of them have the same birthday. Ignore leap years.

We have a sample of 45 people and we want the probability that no two of them have. It's the same birthday. What is the number of days? Let's say that I picked the first person. Now. How many days are possible for him to have a birthday? We have 365 days. He can have a birthday on any one of these days. So this is 365 divided by 365. Now comes the second person. Now this guy has already taken one day, so his choices are 364. He can have a birthday on 364 days out of 365 days. This will go all the way up to 365 minus 25. That is 340 340 divided by 365. This will be the 25th guy. How many options are available for him? 340 days. And what is the total number of days in the year 365? And this value actually turns out to be 0.43 if you do all of this calculation? This will turn out to be 0.4313 This is how we do this question.

In question 16. It says suppose there are 30 people at a party. Do you think any of them have the same birthday? So we're gonna use a random number table to simulate the birthdays of the 30 people at the party. We're gonna ignore leap year so that there's this Onley 365 possible days. We're gonna let one represent January 1 to represent January 2 and so on so that the last day of the year, December 31st, is 365. So these days represented birthdays that the people of the people at the party or any of the two birthdays the same. Compare your results with those obtained by other students in the class. Would you expect the results to be the same or different? So there's a few different ways. If you're doing the random number table, you can look across at, um, 33 digit values. So we have the values 001 through 3 65 representing all of the days of the year because 3 65 is a three digit value. We must have everything as a three digit value as we look from left to right across the random number table. We're gonna look at each three digit number and just determine if we can use that or not. We can use it if it's in the strange. If we get the values 000 or 3. 66 through 9 99 we can't use these values. We must skip them. So, um, as you look through and you pick out 30 different values that represent these birthdays, um, I did this in a few different trials. So trial one, I found zero matching birthdays. But in a second trial, uh, found zero matching birthdays in a third trial. I did found one matching birthday. So if you were looking at this or comparing this with your class, it's not unusual to find. Imagine birthday in a random sample of 30 people. Eso The question ultimately is, Would you expect the results to be the same or different as you compare them with anybody in your class? So you would expect that each student would get different and you would expect to have 30 different values for each of the students. But it wouldn't be uncommon to find that multiple students did find a match in terms of birthdays within the groups, so I'll say, if anything, other people to have results that have repeated birthdays.

Now, this is again a class activity. What I want to do is I want a random number generator. Okay, So what I'm doing is I'm using this, uh, random table. Right? And I am generating random numbers. Okay. A random number table is used. Toe. Make this simulation where my one stands for 1st January. And this goes all the way till 3. 65. Where 3. 65 is 30. 1st off. December, Right. 31st off. December. This is 1st January. Okay, so after you perform this activity, everyone will get different answers. Okay. So is it possible to get any two off the birthdays? Same, That is. Is it possible to get to same numbers? It is quite possible. Okay, we cannot completely rule it out. For example, if I have 30 simulations and the first I get one, then I get to or lets it. Then I get 21. Okay, Then I get 38 Then I get 1 96. So is it possible that I will again repeat these values or any one of these values? For example, I get 38 again out of these 30. Is it possible to have 2 38. It is possible. Although the probability is very low, it is still possible. Okay. And after doing all of these simulations, when we compare our results with somebody else, would we expect the results to be the same? Well, we would not expect the results to be the same, because in order to be the same results, in order to get the same set of values, the probability will be very low. Okay, somebody gets 30 values, somebody gets 30 values and I get the same 30 values. All right, so this probability is very, very low. So we do not really expect the results to be same or we can say that like, we expect the results to be different.

So we want to find the smallest number of people so that the probability that at least two of them are born on April 1st is creator that 1/2. So we're given the information about 366 days are and one year old birthdays are equally likely. And all birthday months are you really likely? And the number of people that we have is an and we want to find the smallest and so that the probably that told them are born on April 1st. At least two of them are born in April 1st is greater than 1/2. That is the same. A subtracting from one. The probability that only one person is born on April 1st and the probability that no one is born on April 1st. Well, let me be the event that two people how are born on April 1st that e prime do the events that one person is born on April 1st and let e prime prime beauty events that there are people are born on April 1st. To find the probability of e, we have to subtract from one the probability of the prime and the probability of the double prime. Well, if we just one for one person to be born on this day, let's look at the probability Bubby Prime. Let's look at it on the next page so we can have more space to work. The probability of e crime, that is the probability that one person is born on April 1st would be equal to the one combination of people because we need to pick one person to be 1 April first times all possible outcomes for the for the birthday off well and minus one people. And this is 365 because we need to. We need only one person to be born on April 1st, and everybody else can only be born on the remaining days of the year. And we divide this by all possible outcomes, which is 366 to the end. And that is equal to 365 to the N minus one times and over 366 to the end. Now, for the probability of a prime prime, that is the probably that no one is born on April 1st we need to Everybody is born on one of the 365 remaining days of the year, so we would have 365 days to the end power and we will divide by all possible outcomes, which is 366 to the end. And the problem Therefore, the probability that to people, at least people are born on April 1st is one minus. The probability of e prime finals of probability, you prime prime. And that is equal to one minus 3 65 to the N minus one over 3 66 times and linus 3 65 over 3 66 to the end power and for probability of e to be greater than or equal to 1/2. If we use a calculator, we can find out that when you end to be greater than or equal to 613.


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