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From: Nmacgre on 4 Aug 2010 09:22
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
From: Nmacgre on 4 Aug 2010 09:26
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)^2

oops, typo

C(20,2)*(4/20)^2*(16/20)^18
From: Henry on 4 Aug 2010 09:40
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. 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: Nmacgre on 4 Aug 2010 09:44
On Aug 4, 6:40 am, Henry <s...(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.  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

Ah, of course, thank you.
From: Ray Vickson on 4 Aug 2010 12:03
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?

Because the probabilities change: P{first defective} = 4/20. P{2nd def|
1st def} = 3/19. Thus, P{first two def} = (4/20)*(3/19) = 3/95.

R.G. Vickson


>
> C(20,2)*(4/20)^2*(16/20)^2

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