calculator probability

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mcarvell

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A shipment of 40 fancy calculators contains 5 defective units. In how many ways can a college bookstore buy 20 of these units and receive:
no defective units
A) one defective unit
B) at least 17 good units
C) What is the probability of the bookstore receiving 2 defective units?
D) Find the probability of receiving at most 2 bad calculators.
E) Find the probability of receiving at least 4 defective units.

I set the first one up as

(35!)/(35-20)!20!
(35!)/(15!)20!

Then I come up with some huge number as I end up doing 35*34*....*21=4.247252019052923e21/130767436800 which comes out to 3247943160. I feel that number is way to large and could use some help setting up the other problems thanks!
 
Your answer is correct; large numbers are endemic to combinatorics. The numbers will get more reasonable when you get into the probabilities.

Let's see what you do for part A.

Now, I'm wondering if you're aware that many calculators can calculate combinations directly, using a button (or menu option) often called nCr. That would make the work a little easier.
 
Thanks Dr. Peterson for the help!

The set up for A was the math I had done out for the first on. So your saying my math was right for A?
 
A should be or the math I did was for no defective units. I guess I just get confused when I start going to one defective unit
 
The wording of the problem as you stated it was confusing; the first question is stated BEFORE "A". That unlabeled question is what you've answered correctly.

Now, what did you try for A (exactly one defective)? Even if you are not sure, give it a try,
 
Here is how I would think about it. You have two piles of calculators. One pile has 35 good calculators while the other pile has 5 defected ones.
Say you want to pick 20 and you want no defective. So you pick 20 from the 35 good calculators and 0 from the 5 defected calculators. You can do this in 35C20 * 5C0 ways

If you wanted to know the number of ways of picking 20 with 3 defected calculators than the answer will be 35C17*5C3 ways.

This method is called hypergeometric. Look it up!
 
You might not have been so shocked by your correct first computation of you 3,247,943,160 had you first computed how many ways you can select 20 from 40, which is 137,846,528,820.

One solution is to leave your answers in factorial format.

So the probability of no defectives is

[MATH]\dfrac{35!}{20! * 15!} \div \dfrac{40!}{20! * 20!} = \dfrac{35!}{40!} * \dfrac{20!}{15!} =[/MATH]
[MATH]\dfrac{20 * 19 * 18 * 17 * 16}{40 * 39 * 38 * 37 * 36} = \dfrac{17 * 16}{2 * 39 * 2 * 37 * 2} =[/MATH]
[MATH] \dfrac{34}{39 * 37} \approx 2.4\%.[/MATH]
 
mcarvell said:
A shipment of 40 fancy calculators contains 5 defective units. In how many ways can a college bookstore buy 20 of these units and receive: no defective units
A) one defective unit
B) at least 17 good units
C) What is the probability of the bookstore receiving 2 defective units?
D) Find the probability of receiving at most 2 bad calculators.
E) Find the probability of receiving at least 4 defective units.
I set the first one up as
(35!)/(35-20)!20!
(35!)/(15!)20!
Then I come up with some huge number as I end up doing 35*34*....*21=4.247252019052923e21/130767436800 which comes out to 3247943160. I feel that number is way to large and could use some help setting up the other problems thanks!
Click to expand...
Do you know the notation (nk)=n!k!(n−k)!\displaystyle \dbinom{n}{k}=\dfrac{n!}{k!(n-k)!} that is n objects choosing k.
Here is a web resource you can use to check your work.
 
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