Question

Question…
In a group of 100 people, 20 are albino.
If I choose 5 people at random, what is the chance:
0 are albino
1 is albino
2 are albino
3 are albino
4 are albino
5 are albino

Answer 1

Am I correct to say, that all the above cases should add up to a probability of 1?

What I have:

for 0 albino:
80/100*79/99*78/98*77/97*76/96 = 0.3193…

for 1 albino:
20/100*80/99*79/98*78/97*77/96 = 0.0840…

for 2 albino:
20/100*19/99*80/98*79/97*78/96 = 0.0207…

you get the picture.

When I add them up, I don't get anything near 1 (or even 0.5)

Answer 2

1 in 5 are albino

5 are randomly chosen

0 are albino, 4/5
1 albino 1/5
2,3,4,5 albino = no chance

Answer 3

i think

Answer 4

\[P(0)=\frac {\left(\begin{matrix}20 \\ 0\end{matrix}\right)\left(\begin{matrix}80 \\ 5\end{matrix}\right)}{\left(\begin{matrix}100 \\ 5\end{matrix}\right)}\]

Answer 5

\[P(1)=\frac{\left(\begin{matrix}20 \\ 1\end{matrix}\right)\left(\begin{matrix}80 \\ 4\end{matrix}\right)}{\left(\begin{matrix}100 \\ 5\end{matrix}\right)}\]

Answer 6

Do you know what was wrong with my method? I can't see it

Answer 7

This problem is sampling without replacement. The hypergeometric disgtribution applies.
\[P(2)=\frac{\left(\begin{matrix}20 \\ 2\end{matrix}\right)\left(\begin{matrix}80 \\ 3\end{matrix}\right)}{\left(\begin{matrix}100 \\ 5\end{matrix}\right)}\]

Answer 8

sorry, i don't understand

Answer 9

*distribution

Answer 10

kropot took my idea, the hypergeometric distribution

Answer 11

can you please explain the hypergeometric distribution

Answer 12

@psujono Do you understand the binomial coefficient\[\left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}\]

Answer 13

ya

Answer 14

Well you can calculate the values of P(0), P(1) and P(2) from the equations I have already given.

Answer 15

i love the word "hyper-geometric"

Answer 16

Oh yeah, sure thanks...
I did a little experiment, it turns out you get the same results if you had a group of 50, and 10 albino. (obviously)
Could we generalise this into a formula such:
You have that x/y (<-- fraction in simplest form) are albino
And we want to pick a group of n, where p are albino...

Answer 17

Sorry to say I think it is a horrible word. Just about as bad as the word statistics :(

Answer 18

@psujono there is nothing wrong with your method, so long as you are careful

Answer 19

^ actually, not so obviously it turns out I was wrong. The percentages change slightly.

@satellite73, how come in my first post I got such different answers?

Answer 20

take for example your calculation of the probability you get two albino
you write
20/100*19/99*80/98*79/97*78/96 which makes sense, but this is the probability you pick the two albino first

Answer 21

to adjust, multiply by the numbers of ways to arrange the two albino in the 5 picks, namely \(\binom{5}{2}=10\)

Answer 22

For 2 albino, I got 0.0207...
with @kropot72's method, I got 0.2073...

Answer 23

How come this is necessary?

Answer 24

right, as i said, you computed "albino, albino, not, not, not" in that order

Answer 25

Oh.. I get it :D

Answer 26

P(0) = 0.3193
P(1) = 0.42
P(2) = 0.207
P(3) = 0.0478
P(4) = 0.00515
P(5) = 0.0002

Answer 27

but there are ten different ways to arrange those two albinos
all with the same probability
so your method is intuitively correct, you just need to multiply by the possible number of ways to arrange each outcome

Answer 28

Sure thanks!

Answer 29

compute for example
\[\frac{\dbinom{20}{3}\times \dbinom{80}{3}}{\dbinom{100}{5}}\] and
\[\dbinom{5}{2}\times \frac{20}{100}\times \frac{19}{99}\times \frac{80}{98}\times \frac{79}{97}\times \frac{78}{96}\] without a calculator. i.e. by factoring and canceling
you will see not only that you get the same thing, but maybe you will get some insight in to why you get the same thing (maybe)

Answer 30

ok that way a typo, first line should have been
\[\frac{\dbinom{20}{2}\times \dbinom{80}{3}}{\dbinom{100}{5}}\]

Answer 31

Great, I see why it works now. Thanks

Answer 32

yw

Answer 33

so when 1 in 4 cereal boxes have a token, and there are 10000 cereal boxes,
If you buy 4 boxes, there is a 68.37% chance you will get at least 1 token.
How Insightful!

Answer 34

really?