Prev: easy probability question
Next: inverting elliptic integral second kind
From: A N Niel on 4 Aug 2010 15:00
In article
<1fdc1c38-f976-405f-b4ff-0ca6d36a706f(a)m1g2000yqo.googlegroups.com>,
Henry <se16(a)btinternet.com> wrote:

> On 4 Aug, 14:26, Nmacgre <natemacgre...(a)gmail.com> wrote:
> > On Aug 4, 6:21�am, Nmacgre <natemacgre...(a)gmail.com> wrote:
> >
> > > Hello:
> >
> > > There is a bucket of 20 golfballs, 4 of which are defecive. If you
> > > select two at random, what is the probability that they are both
> > > defective?
> >
> > > I get 3/95. This right? Also why does the binomial distribution not
> > > work here?
> >
> > C(20,2)*(4/20)^2*(16/20)^18
>
> 3/95 is correct
> Sampling without replacement:
> Prob(2 defective of 2) = C(2,2)*(4*3)/(20*19) = 3/95
> Prob(1 defective of 2) = C(2,1)*(4*16)/(20*19) = 32/95
> Prob(0 defective of 2) = C(2,0)*(16*15)/(20*19) = 60/95
>
> The binomial distribution does not work because you are sampling
> without replacement.

He didn't say "without replacement" ... maybe he is doing it WITH
replacement?

> In any case your corrected calculation would be
> wrong.
> Sampling with replacement
> Prob(2 defective of 2) = C(2,2)*(4/20)^2 *(16/20)^0 = 1/25
> Prob(1 defective of 2) = C(2,1)*(4/20)^1 *(16/20)^1 = 8/25
> Prob(0 defective of 2) = C(2,0)*(4/20)^0 *(16/20)^2 = 16/25
>
>
>
From: bacle on 4 Aug 2010 15:19
> Hello:
>
> There is a bucket of 20 golfballs, 4 of which are
> defecive. If you
> select two at random, what is the probability that
> they are both
> defective?
>
> I get 3/95. This right? Also why does the binomial
> distribution not
> work here?
>
> C(20,2)*(4/20)^2*(16/20)^2

This is what I get:

Informally, if you want to form all possible 2-ples of
defectives, you can choose the first ball in 4 ways,
then the second one (assumming no replacement), can be
chosen in 3 ways, out of the three defectives left.
We divide 12/2=6, since choosing (d1,d2) is the same
as choosing (d2,d1), etc.

So you can choose a pair of defective balls in 6
ways. And there is a total of C(20,2)=190 ways of
choosing any pair of balls.

i)There are C(20,2) ways of choosing 2 balls out of
the twenty.

ii)There are C(4,1)=4 ways of choosing the first ball
to be defective , and (again, assumming no replacement)
there are C(3,1)=3 ways of choosing the second ball
to be defective

So my answer agrees with yours: 6/190=3/95.
First  |  Prev  | 
Pages: 1 2
Prev: easy probability question
Next: inverting elliptic integral second kind