Maths- Need help Q related to Prob (1 Viewer)

[Saturn WY15]

Hi

I need help in solving this question

"A class of twenty people are all born in the same year. Find the probability that atleast two people are born on the same day"

Please provide explanation and steps

Thanks You
W.Y

 

[random-1006]

Hi

I need help in solving this question

"A class of twenty people are all born in the same year. Find the probability that atleast two people are born on the same day"

Please provide explanation and steps

Thanks You
W.Y

this is a classic question!!

probability ( atleast two born on same day) = 1 - probability(none born on same day)

{common technique with "atleast" questions}

prob ( none born same day ) = 365/365 x 364/ 365 x 363/ 365 x ..... 345/365 ( 20 terms )

assuming 365 days to a year.

the first person can be any day, the next person can be any day BUT the day of the first person, the third person can be any day BUT those of the first and second people etc.

since we want the second person AND the third person AND the fourth etc. to all be different to the first we multiply probabilities

REMEMBER WITH PROBABILITY:

"AND" means multiply
"OR" means add

im not going to evaluate it but interestingly it turns out to be more than 50% ( pretty close to it), seems kinda strange aye

 

[ibbi00]

What's the answer?

 

[random-1006]

What's the answer?

use a calculator, im not doing calculator work

 

[shaon0]

I think the above answer is wrong

 

[random-1006]

I think the above answer is wrong

im fairly sure its right, 20 people, first person is any day, the second person should be a different day (ie 364 ways out of 365 ways that are acceptable) , third person a different day again etc.
then 1- (all of that mulitplied) should be the answer

 

[shaon0]

im fairly sure its right, 20 people, first person is any day, the second person should be a different day, third person a different day again etc.
then 1- (all of that mulitplied) should be the answer

Not that. P(X>=2)=1-(P(X=0)+P(X=1))

 

[random-1006]

Not that. P(X>=2)=1-(P(X=0)+P(X=1))

this isnt a binomial probability question, but any way you do it , it will come out with the same final answer

there is no such thing as "one success" ( well probably for one), or "zero success" ( but not for zero, we dont know what date we should be matching until we know what dates there are), remember we want a PAIR, thats it

 

[ibbi00]

Birthday problem - Wikipedia, the free encyclopedia

Random-1006, you're right. Are you sure there's no shorter method of doing this though?
Also, I doubt this is a 2U question.

 

[random-1006]

Birthday problem - Wikipedia, the free encyclopedia

Random-1006, you're right. Are you sure there's no shorter method of doing this though?
Also, I doubt this is a 2U question.

it is a 2 unit question, maybe a slightly harder one, but the concepts are fairly basic

no, unforunately you have to plug in 20 values on calculator, nothing you can do about it

 

[ibbi00]

it is a 2 unit question, maybe a slightly harder one, but the concepts are fairly basic

no, unforunately you have to plug in 20 values on calculator, nothing you can do about it

Just the prospect of inputting all said calculations in a calculator is enough for a rational person to skip the question.

 

[random-1006]

Just the prospect of inputting all said calculations in a calculator is enough for a rational person to skip the question.

its a shame, its not like its an AP or GP , or anything that can be written as a sum of AP and a GP , then you could use those sum formulas

EDIT: you could enter it as 1 - { [ 365! / 344!] / (365)^20 }, i think that should be it

but you have to know what factorials are ( 3 unit) , and must put the brackets EXACTLY where i have

 

[ibbi00]

Oh I see. I knew these '!' weren't so useless after all xD.

 

[Reanna]

this is a classic question!!

probability ( atleast two born on same day) = 1 - probability(none born on same day)

{common technique with "atleast" questions}

prob ( none born same day ) = 365/365 x 364/ 365 x 363/ 365 x ..... 345/365 ( 20 terms )

assuming 365 days to a year.

the first person can be any day, the next person can be any day BUT the day of the first person, the third person can be any day BUT those of the first and second people etc.

since we want the second person AND the third person AND the fourth etc. to all be different to the first we multiply probabilities

REMEMBER WITH PROBABILITY:

"AND" means multiply
"OR" means add

im not going to evaluate it but interestingly it turns out to be more than 50% ( pretty close to it), seems kinda strange aye

^^^^^^^^^^^this answer is right!!! ^^^^^^^^^^

this isnt a 3-unit question just advanced

it is not P(>2) = P(0) + P(1) or whatever was suggested

 

[random-1006]

^^^^^^^^^^^this answer is right!!! ^^^^^^^^^^

this isnt a 3-unit question just advanced

it is not P(>2) = P(0) + P(1) or whatever was suggested

it is a semi understandable mistake ( even though a quick glance at the question would have revealed that this is question is NO WHERE near the format of binomial probability questions, binomial probability questions have a DISTINCTIVE format) , but thats the thing when you do 4 unit maths, if they put a lot of HARD 2 unit questions in the four unit hsc i reckon a fair few would lose marks ( thus why they are not the most suitable people to ask about 2 unit questions), because they dont have to sit a 2 unit paper many of them dont know their 2 unit stuff too well.

Moral of the story : just because you do four unit doesnt mean your shit hot at maths, lots will not be able to do 2 unit questions

 

[Saturn WY15]

this is a classic question!!

probability ( atleast two born on same day) = 1 - probability(none born on same day)

{common technique with "atleast" questions}

prob ( none born same day ) = 365/365 x 364/ 365 x 363/ 365 x ..... 345/365 ( 20 terms )

assuming 365 days to a year.

the first person can be any day, the next person can be any day BUT the day of the first person, the third person can be any day BUT those of the first and second people etc.

since we want the second person AND the third person AND the fourth etc. to all be different to the first we multiply probabilities

REMEMBER WITH PROBABILITY:

"AND" means multiply
"OR" means add

im not going to evaluate it but interestingly it turns out to be more than 50% ( pretty close to it), seems kinda strange aye

Darn, (thump my head) i cant believe it, the classic complementary event so obvious

Yep ur correct, answer is 57.1982416 %( i think i plug in right and yes its seems quite strange)

Well Thanks everyone for the contribution, i post some more question

 

[Affinity]

this is a classic question!!

probability ( atleast two born on same day) = 1 - probability(none born on same day)

{common technique with "atleast" questions}

prob ( none born same day ) = 365/365 x 364/ 365 x 363/ 365 x ..... 345/365 ( 20 terms )

assuming 365 days to a year.

the first person can be any day, the next person can be any day BUT the day of the first person, the third person can be any day BUT those of the first and second people etc.

since we want the second person AND the third person AND the fourth etc. to all be different to the first we multiply probabilities

REMEMBER WITH PROBABILITY:

"AND" means multiply
"OR" means add

im not going to evaluate it but interestingly it turns out to be more than 50% ( pretty close to it), seems kinda strange aye

Problem:
365/365 x 364/ 365 x 363/ 365 x ..... 346/365
I think why you get flamed by jm01 is that you sound cocky