Statistics grocery store question

Statistics Grocery Store Question I am a new online Math Help Forum Friend. I'm back to school after many years out and am taking Statistics. I have a problem that I cannot seem to solve. Can you help?

A grocery produce manager inspects the corn supplied by a nearby farm. In a bushel of 30 ears of corn, the manager inspects 3 ears of corn. If 2 or more of the ears of corn are defective, the entire bushel is rejected. In this bushel, 5 ears of corn are defective. Find the number of ways that 2 or more ears of corn will be found defective by the produce manager.

Thank you.

Re: Statistics grocery store question

The bushel will be rejected in one of two ways:

Case 1: Exactly two ears are defective.

Case 2: Exactly three ears are defective.

Case 1: We want the number of ways to choose three ears of corn such that two of them are defective (given that 5/30 are defective). Here, we are choosing two defective ears out of five, and one normal ear out of the remaining 25. This can occur in $\displaystyle {{5}\choose{2}}{{25}\choose{1}} = 250$ ways.

Case 2: We want to pick all three defective ears out of the five. This can occur in $\displaystyle {{5}\choose{3}} = 10$ ways.

The total number of ways that two or more defective ears is chosen is $\displaystyle 250+10 = 260$.

Re: Statistics grocery store question

Thank you Richard. This was very difficult for me and after several hours of working on it, I went searching for help. I appreciate

your writing out the explanation so that I can augment my learning.

Now I have yet another question. I can see that there is a 5 over 2 in parentheses and a 25 over 1 in parenthesis and a 5 over the 3 in parentheses. How do those numbers get worked in order to equal the sums you found?

Thank you.

Re: Statistics grocery store question

I had one other question. Is there a name for the formula that you used to figure this out? Thank you.

Re: Statistics grocery store question

The numbers in the funny parentheses are "binomial coefficients".

Binomial coefficient - Wikipedia, the free encyclopedia

Re: Statistics grocery store question

So, if I haven't learned those yet is there another way to figure out this problem?

Re: Statistics grocery store question

Quote:

Originally Posted by

**u12480** So, if I haven't learned those yet is there another way to figure out this problem?

If you have not had those yet, then you should not have been asked to do this question.

Re: Statistics grocery store question

Those are *combinations*, and they return *binomial coefficients*.

Basically, the number of ways to choose k objects out of a set of n (order doesn't matter, this is called a "combination") is denoted $\displaystyle {{n}\choose{k}}$ ("n choose k") and is given by

$\displaystyle {{n}\choose{k}} = \frac{n!}{(n-k)!k!}$.

For example, $\displaystyle {{5}\choose{2}} = \frac{5!}{3!2!} = \frac{120}{6(2)} = 10$

Re: Statistics grocery store question

I appreciate your replies. We are just starting to do binomial equations. Thank you for the explanation.