Thread: Hyper geometric Distribution

There is a box of 12 flares, four are selected at random. If the box contains four flares that don't work:

a- what is the probability that all four will not work?

b- What is the probability that, at most two will not work?

MY ANSWERS:

a = p(x=4)= C(4,4) x C(8,0)
/divided C(12,4)

24x1/19958400
0.0000012025 = 0.00012%

(that seems like a very low number for me)

b= x=equal or less 2
= C(4,2) x C(8,2) C(12,4) + C(4,1) x C(8,3) + (4,0) x (8,4)
/divided C(12,4)

= 403392/19958400 = 0.02021164 = 2%

That seems off to me as well...where am I going wrong????

Originally Posted by staceylynnxx

There is a box of 12 flares, four are selected at random. If the box contains four flares that don't work:
a- what is the probability that all four will not work?
b- What is the probability that, at most two will not work?

MY ANSWERS:
a = p(x=4)= C(4,4) x C(8,0)
/divided C(12,4)
24x1/19958400
0.0000012025 = 0.00012%
(that seems like a very low number for me)

You method in a) is correct, but your arithmetic is lousy.
.

For part b)

Thank you again.

I don't know how to do the math signs on the forum.

Originally Posted by staceylynnxx

I don't know how to do the math signs on the forum.

Use LaTeX tags.
[TEX]\frac{1}{C(12,4)}=0.002020202020202[/TEX] gives

[TEX]\sum\limits_{k = 0}^2 {\frac{C(4,k)\cdot C(8,4-k)}{C(12,4)}} [/TEX] gives