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probability problem

Anonymous Fri 15 Oct 23:25:22 2021 No.13755912 View Reply Original Report

Quoted By: >>13756625 >>13756743 >>13756762

/sci/ can solve this easy right?

Anonymous

Anonymous Fri 15 Oct 2021 23:30:35 No.13755925 Report

Quoted By:

q1: 100%
q2: pretty low
q3: paradoxically high

Anonymous

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Anonymous Sat 16 Oct 2021 01:00:27 No.13756142 Report

Quoted By:

Assuming being born on any day is equally likely $p := \frac{1}{4(365) + 1}$

>q1
$P = 22p^1 (1-p)^{21}$

>q2
$P = \sum_{k=1}^22 {22 \choose k} p^k (1-p)^{22-k}$

>q3
$P = {23 \choose 2} p^2 (1-p)^{21}$

How'd I do on your homework, homo?

Anonymous

Anonymous Sat 16 Oct 2021 02:56:05 No.13756395 Report

Quoted By: >>13756872

>Q1
0.068% assuming the other person must also be born on Feb 29th and every day, every 4 years is equally likely
>Q2
1.5%
>Q3
50.02% for an effective 1/365.25 chance for any 2 people to share a birthday and 253 possible pair combinations
No I will not show any more math

Anonymous

Anonymous Sat 16 Oct 2021 04:42:52 No.13756625 Report

Quoted By: >>13756639

>>13755912
can someone explain how to do he last one

Anonymous

Anonymous Sat 16 Oct 2021 04:48:23 No.13756639 Report

Quoted By:

>>13756625
you gotta count the number of distinct pairs, and then for each pair, you have the probability that they they don't share a birthday, then find the complement of that probability

Anonymous

Anonymous Sat 16 Oct 2021 06:15:33 No.13756743 Report

Quoted By: >>13757019

>>13755912
1) 50/50
2) 50/50
3) 50/50
Mathcucks un suicide watch

Anonymous

Anonymous Sat 16 Oct 2021 06:33:04 No.13756762 Report

Quoted By:

>>13755912
Q1: 1/(365*4) = p = 0.068%
Q2: 1 - C(22, 0) * (p)^0 * (1p)^22 = 1.496%
Q4: 1.496% + 365!/(365-22)!/365^22 = 53.896%

Anonymous

Anonymous Sat 16 Oct 2021 07:27:32 No.13756872 Report

Quoted By:

>>13756395
>50.02% for an effective 1/365.25 chance for any 2 people to share a birthday and 253 possible pair combinations
No I will not show any more math
you forgot to factor in that one of the birthdays is already known and is on a rarer day
to solve t was you have to do is
P(bob shares a birthday with someone) + P(2 of the non-bob people share birthdays)

Anonymous

Anonymous Sat 16 Oct 2021 08:32:38 No.13757005 Report

Quoted By:

Bob was born on March 1.

Anonymous

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Anonymous Sat 16 Oct 2021 08:38:45 No.13757019 Report

Quoted By:

>>13756743
cringe

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