## TayyabaRawjani Group Title There are 5 people in a room. Make the following simplifying assumptions about their birthdays: Ignore leap years; assume 365 days in the year. For each person, assume that the chance he/she is born on a specified day is 1/365. No twins! Assume that the chance that someone is born on a specified day is not affected by other people’s birthdays. one year ago one year ago

1. TayyabaRawjani Group Title

1.Find the chance that the 5 people have 5 different birthdays.

2. TayyabaRawjani Group Title

2.Find the chance of a "match": that is, at least two people in the room have the same birthday.

3. TayyabaRawjani Group Title

3.Repeat 4B for a room in which there are 400 people.

4. drawar Group Title

What have you tried? Do you want only answers?

5. drawar Group Title

@Nurali: which part?

6. drawar Group Title

Can you copy-paste the message I've sent to you?

7. karan81 Group Title

Can I have answer of this question as well?

8. whatname Group Title

they are cheating at Stat2.2x (edx.org)

9. drawar Group Title

@whatname: don't worry, doing this will get them nothing, it's just going against the purpose of education :)

10. hlpwntd Group Title

5 different bdays = (364/365) * (.../...) * (361/365) at least 2 same bday = 1 - (5 different bdays)

11. hlpwntd Group Title

400 in the room: 400>365 the probability is 1 or 100%

12. creator26 Group Title

hello folks. please don't provide answers to questions directly, but you can at least give an explanation to questions asked. because some of the questions asked are from an online programme (Stat 2.2x edx).

13. TayyabaRawjani Group Title

14. whatu Group Title

again 400 is greater than 365 so the it is either 1/365 or 100/365?

15. hlpwntd Group Title

ok let me explain in details: 1. you want 5 people to have 5 different bdays. 1st person has 365 chances out of 365 so you can drop the first bday, second person get 364 chances out of 365 because 1 bday is already taken, 3rd person 363, and so on 5 different bdays = (364/365) * (364/365) * ..... 2. is the opposite of the previous question 3. if x > y, then x/y > 1, then chance is > 1. if chance is > 1 then check where you are located at in the probability spectrum. hint: probability spectrum is 0 to 1. the answer is in the question.

16. CarlosGP Group Title

.Just imagine a specific group of different dates. For instance: 1st Jan, 2nd Jan, 3rd Jan, 4th Jan and 5th Jan and a group of five people {A,B,C,D,E}. What is the probability of having in the room 5 people that were born in each of these different dates? For the case of A born in date 1, B in date 2, C in date 3, D in date 4 and E in date 5, it is $P_1=1/365^5$However we can assign each person to each one of the different dates which means that we have "5!" possible assignments. We can state that the probability of having 5 people born in different dates (1, 2, 3, 4 and 5 of Jan) is $P_2=5!·P_1=\frac{ 5! }{ 365^5 }$This probability is valid for the five dates we have chosen. Now you have to see, how many groups of five different dates you can make with 365 dates. Order does not matter, then it is $N=C^{365}_5=\frac{ V^{365}_5 }{ 5! }$As you have to consider all these possibilities, the solution of the problem is: $P_3=N·P_2=\frac{V^{365}_5 }{ 5!}·\frac{ 5! }{ 365^5 }=\frac{ 365·364·363·362·361 }{ 365^5 }=0.973$that is a 97.3% and the chance of a match 100%-97.3%=2.7%

17. dinakar Group Title

There are 5 people in a room. Make the following simplifying assumptions about their birthdays: Ignore leap years; assume 365 days in the year. For each person, assume that the chance he/she is born on a specified day is 1/365. No twins! Assume that the chance that someone is born on a specified day is not affected by other people’s birthdays. PROBLEM 4A This Problem is worth 1 Point Find the chance that the 5 people have 5 different birthdays.

18. dinakar Group Title

can any one ans

19. CarlosGP Group Title

20. dinakar Group Title

21. dinakar Group Title

pls can any one ans clearly

22. dinakar Group Title

There are 26 letters in the English alphabet. Letters are picked one by one at random, so that each letter has the same chance of appearing as any other letter, regardless of which letters have appeared or not appeared. PROBLEM 5A This Problem is worth 1 Point Six letters are picked. Find the chance that the sequence that appears is RANDOM, in that order.

23. dinakar Group Title

pls some one help me ot

24. dinakar Group Title

There are 26 letters in the English alphabet. Letters are picked one by one at random, so that each letter has the same chance of appearing as any other letter, regardless of which letters have appeared or not appeared. PROBLEM 5A This Problem is worth 1 Point Six letters are picked. Find the chance that the sequence that appears is RANDOM, in that order.

25. dinakar Group Title

help me out

26. dinakar Group Title

pllllllllssssssssss some one ans yar

27. jaysmittt Group Title

5a it goes like this 1/26*1/25.....

28. jaysmittt Group Title

u should be able to guess the other 4 by now..

29. CarlosGP Group Title

@dinakar When you say the answers are confusing, no doubt you mean you do not understand the answers. What is the part you do not understand?

30. gorv Group Title

@dinakar (1/26)^6

31. jaysmittt Group Title

4b= 1-(365-1/365) 4c= 1/1

32. TayyabaRawjani Group Title

Thank u ppl