# Two Stat Word Problems

1. Jan 13, 2008

### kuahji

I'm having a bit of difficulty thinking through two problems.

A shipment of 10 television sets includes three that are defective. In how many ways can a hotel purchase four of these sets an receive at least two of the defective sets?

Without making a tree diagram, I'd like to understand the problem, & how to correctly write it. I know the answer is 70, because that is what the book says. I assumed I'd want a permutation, because it appears order will matter. I tried (7!/4!)/(3!/2!) & got 70. But it still doesn't make much sense.

In how many ways can five persons line up to get on a bus? In how many ways can they line up if two of the persons refuse to follow each other?

Here again I'm gonna need to use a permutation. So in part one I did 5! & got 120, which was correct, that makes sense. But for part two the answer is 72. Which I believe will be 5!/?, I tried 5!/2! assuming because of the two people that won't follow each other, but that clearly didn't work. Any ideas how to make sense of these two problems?

2. Jan 13, 2008

### olgranpappy

Well, how many different ways are there to pick 4 televisions out of the ten? There are 210 ways. So, just right down all 210 possibilities and see how many of them contain at least two broken sets.

3. Jan 14, 2008

### kuahji

Thats a bit tedious. There must be a better way to think through the problem then trying to write down 210 possibilities. If I use that method, it'll work, but I'll only be able to work 1 problem in the given amount time when exam comes.

4. Jan 14, 2008

### arunbg

Of course there's an easier way. You need to use combinations here. I'll give you a start.
As mentioned 10C4 = 210 is the total no of ways of picking 4 TVs. Subtract from this the no. of ways of getting all good TVs and the no of ways of getting exactly one bad TV and 3 good TVs and you have your answer. Can you finish it off?

Second question, consider the two not so friendly people as one unit. This should help you find the no of arrangements in which they come together. Subtract this from the total no. of arrangements.

Last edited: Jan 14, 2008
5. Jan 14, 2008

### unplebeian

This is similar to a hypergeometric probability. Try looking that up in your text book.

6. Jan 15, 2008

### silver-rose

It is hypergeometric.

Just find the Probability of the occurrence, then multiply that by the total number of ways you can purchase 4 sets to get the E(x)

=P(two defective) + P(three defective) + P(four defective)
=[ 3C2 * 7C2 ] / [ 10C4] + [ 3C3 * 7C1] / [10C4] + 0

So P(x) * n = E(x), which should amount to 70 (I haven't calculated that yet)

7. Sep 9, 2009

### darkmagicianoc

I know that this is way late in responding, but I figured I'd share the way that I solved the bus problem since I just figured it out myself (and stumbled upon this post for help). since there are a total of 120 ways that all 5 can be arranged in the line, we have to look at the different cases of that where the two people are next to each other. Lets refer to the two people who do not want to be next to each other A and B.

For example, you can have:

A B 3 2 1 = 6 choices (1*1*3*2*1 = 6 permutations with person A first and B second)

Repeat this process for A in slots 1-4 and repeat for B in slots 1-4. Subtracting the total number of ways this can happen from the original 120 should give you the answer you need.

Hope this helps.

Last edited: Sep 9, 2009
8. Sep 9, 2009

### Elucidus

Sometimes people forget to consider the arrangements where B is before A (i.e. BA321).

It is often helpful (as someone mentioned) to think of the two "enemies" as a fused unit (for exclusion purposes) and figure how many ways the 4 objects can be arranged and then multiply this by how many ways the enemies can be arranged within the diad. Subtract this from the original unrestricted orderings and you get your answer.

--Elucidus

9. Sep 10, 2009

### njama

It very easy to solve this problem. Just you need to use combinations.

$$C_7^1*C_3^3 + C_7^2*C_3^2=7+63=70$$