# Find P(X = 3) porbability

1. Jul 7, 2009

### rsala004

1. The problem statement, all variables and given/known data

Suppose that 0.03% of plastic containers manufactured by a certain process have small
holes that render them unfit for use. Let X represent the number of containers in a
random sample of 10,000 that have this defect. Find P(X = 3).

2. Relevant equations
binomial?

3. The attempt at a solution

.97^9997
*
.03^3
*
10000 choose 3

i bet this actually does equal the answer maybe, but my calculator of course spits out 0 since .97^9997 is really really tiny...and doesnt take into account that 10000 choose 3 is humungo

help?
the answer is .224, but i cant match it

2. Jul 7, 2009

### HallsofIvy

Staff Emeritus
Re: probability

It's not that hard to work it out "by hand".

$$\left(\begin{array}{c}10000 \\ 3\end{array}\right)= \frac{10000!}{3!9997!}$$
$$= \frac{10000(9999)(9998)}{6}=\frac{10000}{2}\frac{9999}{3}(9998)$$
$$= 166616670000$$

Now, .033=0.000027 and (this is the hard one) .979997 is about .6 x 10-133. Since 166616670000 is about 1.7 x 1011 that product will be on the order of 1011- 5- 133= 10-127.

However, is the proportion of of "unfit" cups .03% or 3%? You say ".03%" but you use 3%. If the correct value is .03%, then you want
$$\left(\begin{array}{c}10000 \\ 3\end{array}\right)(.0003)^3)(.9997)^{9997}$$
Now that would be about .224.

Last edited: Jul 7, 2009
3. Jul 8, 2009

### rsala004

Re: probability

i was sitting in class today and wondered if i can use poisson distribution on this?

o and yes that is my mistake that problem is copy and pasted, so yes it is actually 0.03% PERCENT...i didnt notice

edit: ah yes i figured it out, $$\lambda$$ = .0003*10000 = 3

P(X=3)

$$\frac{e^{-3}*3^{3}}{3!}$$ = .224

thanks alot though

Last edited: Jul 8, 2009