Question

you need 5 men that are monochromatic colorblind to participate in a study. what is the probability you have to sample 20 men to find these 5 that are color blind?

Answer 1

\[\left(\begin{matrix}20 \\ 5\end{matrix}\right) .01^5 (1-.01)^15\]

Answer 2

i got .000001
but my professor took off two points

Answer 3

on the 20 choose 5 and i dont understand why

Answer 4

what is the chance that someone is colorblind?

Answer 5

The question you answered was "If I choose 5 men from a group of 20, what's the probability that they are all color blind."

Answer 6

1% @Miracrown

Answer 7

ok
so how could i have written it to say that 5 are blind from the 20?

Answer 8

reverse?

Answer 9

5 choose 20?

Answer 10

i don't think it's that straightforward

Answer 11

i'm thinking kinda like a geometric distribution

Answer 12

if its the geometric then it would be \[\theta(1-\theta) ^{x-1}\]

Answer 13

so it would be \[.01(1-.01)^{15}\]

Answer 14

Well, that's the probability of finding 15 normal sighted before your first color-blind.

Answer 15

so i have to change the 15?

Answer 16

Oh, right. Ok. So, I don't remember the reason for this, but your binomial coefficient should be:
\[\left(\begin{matrix}N-1 \\ k-1\end{matrix}\right)\]

Answer 17

The reason has something to do with having k-1 successes before the Nth trial...or something.

Answer 18

yes i recognize that equation

Answer 19

Ah, right. The probability of getting 4 color blind in 19 tries, then the probability of the last one being color blind.

Answer 20

so 19 choose 4?

Answer 21

\[\left(\begin{matrix}N-1 \\ k-1\end{matrix}\right) p^k (1-p)^{N-k}\]

Answer 22

ok let me solve it

Answer 23

33.3359?

Answer 24

I get something similar, but MUCH smaller.

Answer 25

3.33359?

Answer 26

I ended up with 3.33359 x 10^(-7)

Answer 27

yes I got the same

Answer 28

thank you so much! :)

Answer 29

K. It was just that 10^-7 I was talking about. Same result, but smaller :P

Good work!