What is the Probability of 3 Defective Plastic Containers in a Sample of 10,000?

[rsala004]

Homework Statement

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).

Homework Equations

binomial?

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 doesn't take into account that 10000 choose 3 is humungo

help?
the answer is .224, but i can't match it

 

[HallsofIvy]

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 10^{11- 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.

 

[rsala004]

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 a lot though