Simple probability that I can't get out

  • Thread starter Thread starter rock.freak667
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary

Homework Help Overview

The discussion revolves around probability concepts, specifically focusing on events and their relationships, the birthday problem, and combinatorial choices. Participants are exploring set notation for events, calculating probabilities related to birthdays, and determining combinations of design options.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the interpretation of events in set notation, particularly regarding "at most" and "exactly" conditions. There are attempts to clarify the probability of people sharing birthdays and the correct formulation of combinatorial problems. Some participants express confusion about the relationships between events and how to express probabilities accurately.

Discussion Status

The discussion is active, with participants providing feedback on each other's interpretations and attempts. Some guidance is offered regarding the correct use of set notation and the formulation of probabilities, but there is no explicit consensus on the final answers or methods.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can provide or the depth of their explanations. There is an ongoing exploration of assumptions related to the birthday problem and the definitions of events in set theory.

rock.freak667
Homework Helper
Messages
6,221
Reaction score
31
Well it's been about 4 years since I've done these types of questions, so I need some help in checking if I am overthinking it or if I am just straight wrong.

I did simplify the wording a bit.

Homework Statement


a)Given 2 events, A & B, write in set notation
1)the event that at most A or B occurs
2) The event that exactly A or B occurs

The Attempt at a Solution



n=intersect, u=union

1) P(AuB)=P(A)+P(B)-P(AuB)

2) Not sure


Homework Statement


b)There are n people in a room, what is the probability that at least 2 have the same birthday?

The Attempt at a Solution



P(no 2 having the same birthday)+P(at least 2 having the same birthday)=1

P(no 2 having the same birthday)=(365/365)*(364/365)

P(no 3 having the same birthday)=(365/365)*(364/365)*(363/365)

P(no 4 having the same birthday)=(365/365)*(364/365)*(363/365)*(362/365)
.
.
.
P(no 'n' having the same birthday)=(365/365)*(364/365)*(363/365)*(362/365)*(361/365)*...*(365-n+1)/365)

It looks like the numerator should be 365! and the denominator, 365n, but I am not sure what happens to the 'n' in the numerator.


Homework Statement


c)There's an offer of a choice of 4 designs, 3 different heating systems, a garage or carport, and a patio or screened porch. How many different plans aer available?


The Attempt at a Solution



I think it would be 4*3*2*2 = 48

Homework Statement


d)A fair coin is tossed until a head appears for the first time. The tosses are independent.
Give the sameple space and what is the probability that the first head appears when the toss number is odd.


The Attempt at a Solution



S={H,TH,TTH,TTTH,TTTTH,TTTTTH,...}

Is that how it woud look?

For the odd numbered toss, it would be

P(H),P(TTH),P(TTTTH),...
= 1/2,1/8,1/32,1/128,...

which forms a GP with first term a=1/2 and common ratio r =1/4. So the sum to infinity is a/1-r

=(1/2)(1-0.25)=(1/2)*(4/3)=2/3
 
Physics news on Phys.org
I think you simplified the wording a bit too much on the first one. I'm not sure I understand what e.g. "at most A or B occurs". Do you mean at most one of A or B occurs? That would mean the only events that don't count is where A and B both occur. For the birthday problem, what's the probability that NONE of the n people have the same birthday? c) and d) look ok to me.
 
Dick said:
I think you simplified the wording a bit too much on the first one. I'm not sure I understand what e.g. "at most A or B occurs". Do you mean at most one of A or B occurs? That would mean the only events that don't count is where A and B both occur. For the birthday problem, what's the probability that NONE of the n people have the same birthday? c) and d) look ok to me.

sorry I left out the word one. It should bethe event that at most one of A or B occurs
The event that exactly one of A o B occursAlso if one person has a birthday, then the probability that the rest do not have that birthday is 364/365.
 
rock.freak667 said:
sorry I left out the word one. It should bethe event that at most one of A or B occurs
The event that exactly one of A o B occursAlso if one person has a birthday, then the probability that the rest do not have that birthday is 364/365.

Ok, then your whole event space, call it E, can be divided into four nonoverlapping regions, E-(AuB), A-B, B-A and AnB, right? Which combination is the answer to each part?
 
rock.freak667 said:
Also if one person has a birthday, then the probability that the rest do not have that birthday is 364/365.

Not exactly. If they all have DIFFERENT birthdays, then the first one can choose any day, the second has 364 days to choose from, the third 363, etc etc.
 
Dick said:
Ok, then your whole event space, call it E, can be divided into four nonoverlapping regions, E-(AuB), A-B, B-A and AnB, right? Which combination is the answer to each part?

I think at most one of A or B would be E(AuB), and exactly one of A or B would by E(B) or E(A) ?

Sorry I am a bit confused with it.


Dick said:
Not exactly. If they all have DIFFERENT birthdays, then the first one can choose any day, the second has 364 days to choose from, the third 363, etc etc.

Yeah, that is what I was trying with my attempt, I just didn't subtract it from 1.

1-(365!/365n)
 
rock.freak667 said:
I think at most one of A or B would be E(AuB), and exactly one of A or B would by E(B) or E(A) ?

Sorry I am a bit confused with it.

The are asking for sets, not probabilities. I am thinking of E as the set of ALL events, so A and B are subsets of E. I don't know E(A) would mean. If you mean E-(AuB) for the 'at most' part, then sounds ok. Wouldn't 'exactly one' be (AuB)-(AnB)?
 
rock.freak667 said:
Yeah, that is what I was trying with my attempt, I just didn't subtract it from 1.

1-(365!/365n)

If that's what you mean, then 1-365!/365^n is a pretty sloppy way to write it. It doesn't mean what you are describing.
 
Dick said:
The are asking for sets, not probabilities. I am thinking of E as the set of ALL events, so A and B are subsets of E. I don't know E(A) would mean. If you mean E-(AuB) for the 'at most' part, then sounds ok. Wouldn't 'exactly one' be (AuB)-(AnB)?

Then would I not have to subtract the complement of of AuB? Unless the entire event is just A and B, which I don't think I can assume.


Dick said:
If that's what you mean, then 1-365!/365^n is a pretty sloppy way to write it. It doesn't mean what you are describing.


so if (365n-365!)365n is incorrect, how do I go about doing it correctly?
 
  • #10
rock.freak667 said:
Then would I not have to subtract the complement of of AuB? Unless the entire event is just A and B, which I don't think I can assume.





so if (365n-365!)365n is incorrect, how do I go about doing it correctly?

Sure, you can subtract complement of AuB. What makes you think that would change anything? On the second one, take n=1. That's 1-365!/365. 365! is MUCH larger then 365. So that expression is negative. 1-365!/365^n is not a correct expression for your answer.
 
  • #11
Dick said:
Sure, you can subtract complement of AuB. What makes you think that would change anything? On the second one, take n=1. That's 1-365!/365. 365! is MUCH larger then 365. So that expression is negative. 1-365!/365^n is not a correct expression for your answer.

Well I am seeing that I can write the equations in my first post as factorials.


P(no 2 having the same birthday)=(365/365)*(364/365)=365!/363!3652

P(no 3 having the same birthday)=(365/365)*(364/365)*(363/365)=365!/362!3653

which can be extended to 'n' such that

P(no 'n' having the same birthday)=365!/(365-n)!365n= 365Pn/365n

So P(at least two having the same birthday)=1- 365Pn/365n

I think this should always give a number less than 1.
 
  • #12
rock.freak667 said:
Well I am seeing that I can write the equations in my first post as factorials.


P(no 2 having the same birthday)=(365/365)*(364/365)=365!/363!3652

P(no 3 having the same birthday)=(365/365)*(364/365)*(363/365)=365!/362!3653

which can be extended to 'n' such that

P(no 'n' having the same birthday)=365!/(365-n)!365n= 365Pn/365n

So P(at least two having the same birthday)=1- 365Pn/365n

I think this should always give a number less than 1.

That looks better.
 
  • #13
Dick said:
That looks better.

so it's correct then? thanks!
 

Similar threads

Birthday problem (probability)
  • · Replies 9 ·
Replies
9
Views
4K
Sharing a birthday - what's wrong in my approach?
  • · Replies 3 ·
Replies
3
Views
2K
Simple Interest help -- investment in his savings account matured to $4,106.32
  • · Replies 2 ·
Replies
2
Views
2K
Birthday problem with repetitions
  • · Replies 2 ·
Replies
2
Views
3K
Counting problem relatively hard I think.
  • · Replies 7 ·
Replies
7
Views
3K
Simple Interest Calculator for a 5% Rate - Help with Homework Statement
  • · Replies 7 ·
Replies
7
Views
2K
Please help with probability and statistics problem
  • · Replies 3 ·
Replies
3
Views
6K
Multi-event probability puzzle - is my answer correct?
  • · Replies 7 ·
Replies
7
Views
2K
Probability of selecting two white balls from two bags
  • · Replies 2 ·
Replies
2
Views
3K
Probability Word Problem: Aptitude Question
  • · Replies 8 ·
Replies
8
Views
3K