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.

- anonymous

- katieb

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

What have you tried? Do you want only answers?

- anonymous

@Nurali: which part?

- anonymous

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

- anonymous

Can I have answer of this question as well?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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).

- anonymous

em just asking a guideline to solve these questions
thats it
haven't ask for direct answers

- anonymous

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

- anonymous

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.

- anonymous

.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%

- anonymous

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.

- anonymous

can any one ans

- anonymous

@dinakar
The problem has already been answered. Please, read the thread

- anonymous

answers are very confusing could u please ans them clearly

- anonymous

pls can any one ans clearly

- anonymous

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.

- anonymous

pls some one help me ot

- anonymous

- anonymous

help me out

- anonymous

pllllllllssssssssss some one ans yar

- anonymous

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

- anonymous

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

- anonymous

@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?

- gorv

@dinakar
(1/26)^6

- anonymous

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

- anonymous

Thank u ppl

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