/sci/ - Science & Math

[Anonymous No.11462320]

there is a lottery, where are 1000 tickets
only 3 tickets win
my friend bought 4 tickets
what is the probability he has 2 at least winning tickets?

is it
[math]3! \times \frac{4*3*2}{1000*999*998} + 3 \times \frac{4*3*996}{1000*999*998}[/math]
??

[Anonymous No.11462591]

>>11462320
it's 1/3

[Anonymous No.11462700]

>>11462591
No

[Anonymous No.11463228]

>>11462320
dump

[Anonymous No.11463237]

>>11462320
0.012012%

[Anonymous No.11463254]

Learn to speak like a normal person, you spastic.

[Anonymous No.11463330]

>>11463237
you aren't considering all the cases and the position of the extracted tickets
I wrote a program and the probability turns to be around 0,0047%

[Anonymous No.11463381]

got it
[math]
3! \times \frac{4*3*2}{1000*999*998} + 4 \times \frac{4*3*996}{1000*999*998}
[/math]

[Anonymous No.11463455]

>>11463381
Wrong

[Anonymous No.11464115]

0.024024

[Anonymous No.11464206]

>>11464115
>0.048048

[Anonymous No.11464220]

the closest value I got

-3 \times \frac{4*3*2}{1000*999*998} + 4 \times \frac{4*3*996}{1000*999*998}

[Anonymous No.11464227]

the closest value I got

[math]-3 \times \frac{4*3*2}{1000*999*998} + 4 \times \frac{4*3*996}{1000*999*998}[/math]

[Anonymous No.11465348]

>>11464227

[Anonymous No.11465351]

>>11462320
>>11463381
>>11464227
Stfu you already wrote the same shit 3 times, you genetic dead end.

[Anonymous No.11465464]

>>11465351
help instead of complaining

[Anonymous No.11465527]

(4/1000)/2 = 0.002%

[Anonymous No.11465801]

>>11462320
I get 0.00359879%.

[Anonymous No.11465869]

>>11465801
how did you get it?

[Anonymous No.11465971]

>>11465869
By lurking more in Academia. Memes apart, find in Wikipedia “Hypergeometric distribution”.
If I remember well, the book “Statistical Inference” by Casella & Berg should also have a nice overview about some of the most common and used statistical distributions, both discrete and continuous ones.

[Anonymous No.11466139]

>>11465801
0.003596382%
[math]\frac{6*996*6}{1000*999*998}[/math]

[Anonymous No.11466278]

I got the same by doing:

1 - (997 choose 4)/(1000 choose 4) - ((3 choose 1)*(997 choose 3)/(1000 choose 4))

But I only learned this a few weeks ago by someone who taught it to me in 2 hours so be sure to tell me why it's wrong (probably is)

[Anonymous No.11466329]

>>11466139
That's the probability of winning twice by owning 3 tickets. Not getting equal or more than 2 from 4 tickets.

Your result is really close to his because the odds of winning 3 times is low/negligible so if you find the odds of winning 2 times and 3 times, the answer is almost the exact same as 2 times.

[Anonymous No.11466347]

>>11466139
>>11466329
Also I think that # losing tickets is # total - # winning which is 1000-3 =/= 996. Its 997, no?

[Anonymous No.11466361]

>>11466278
From what I see it should be correct. At most you should write in the first subtraction also (3 choose 0) to improve clarity, but since it is equal to 1 your formula is not wrong. What you are doing is basically [math]F(X\ge 2) = 1 - F(X<2) = 1 - [P(X=0) + P(X=1)][/math], so be more confident.

[Anonymous No.11466400]

>>11466329
It's the probability of winning twice, with owning 4 tickets and with 3 tickets extracted

[Anonymous No.11466415]

>>11466329
yeah

this is the correct answer >>11465801

[Anonymous No.11466503]

>>11466278
Latex version assuming I correctly wrote it:
[eqn]\frac{C^{997}_{2}C^{3}_{2} + C^{997}_{1}C^{3}_{3}}{C^{1000}_{4}} = 0.00003598789[/eqn]
Or
[eqn]1 - \frac{C^{997}_{4}C^{3}_{0} + C^{997}_{3}C^{3}_{1}}{C^{1000}_{4}} = 0.00003598789[/eqn]