# Am I using the combinatorial methods correctly?

For the following two problems I have worked out the following solutions:

1: A box contains $24$ light bulbs, of which $4$ are defective. If a person selects $4$ bulbs from the box at random, without replacement, what is the probability that all $4$ bulbs will be defective?

solution: there are $4!$ ways to select the $4$ defective bulbs and there are ${24\choose 4}$ ways to select the $4$ defective bulbs from the $24$.

so the probability that all $4$ bulbs will be defective is: $$\frac{4!}{24\choose4}$$

2: Suppose that $n$ people are seated in a random manner in a row of $n$ theater seats. What is the probability that two particular people $A$ and $B$ will be seated next to each other?

There are $n \choose 2$ ways in which $A$ and $B$ can be selected and there are $n^2$ possible orderings so: $$\frac{n\choose 2}{n^2}$$

Where am I going wrong with these?

EDIT:

I have further two questions which I still cannot even attempt to solve but would really like to understand the reasoning behind it:

The United States Senate contains two senators from each of the 50 states.

a)If a committee of 8 senators is selected at random, what is the probability that it will contain at least one of the two senators from a certain specified state?

b)What is the probability that a group of 50 senators selected at random will contain one senator from each state?

I really don't know how to incorporate 'at least one of the two'

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In first question, numerator should be $1$, not $4!$. There is only one way to choose $4$ bads. –  André Nicolas Oct 14 '11 at 19:56
In second question, how many ways to choose an (unordered) pair of seats for A and B? How many of these pairs consist of adjacent seats? (There are other approaches, but if we are seating $n$ people there are $n!$ orders, not $n^2$.) –  André Nicolas Oct 14 '11 at 20:01
(a) For probability of at least one from specified state, easiest to find $q$, the probability of none. Then our answer is $1-q$. Very useful idea. Clearly $q=\binom{98}{8}/\binom{100}{8}$. But could find probability of at least one by adding together the probability $p_1$ of exactly $1$ and $p_2$ of exactly $2$. For $p_1$, the numerator is $\binom{2}{1}\binom{98}{7}$, denominator is $\binom{100}{8}$. For $p_2$ numerator is $\binom{98}{6}$. (b) There are $2^{50}$ ways of selecting $1$ from each state. –  André Nicolas Oct 15 '11 at 14:05

First Question: In many problems where we find probabilities by counting, we produce a sample space with $N$ equally likely outcomes. If for $S$ of these outcomes we have "success," then the probability of success is $S/N$.

In this problem, there are two natural sample spaces of equiprobable outcomes.

(a) Imagine choosing $4$ bulbs and putting them in a bag. There are $\binom{24}{4}$ different possible bags, all equally likely. How many bags have $4$ bad bulbs? It is clear that there is only $1$ such bag. (There is good reason to write instead that there are $\binom{4}{4}$ such bags.) So the probability of "success," if you can call $4$ bad bulbs success, is $$\frac{1}{\binom{24}{4}}.$$

(b) Imagine picking out a bulb, then another, and so on, and putting them down in a row, in the order in which they were selected. Then there are $(24)(23)(22)(21)$ possible such rows, all equally likely. How many possible rows are there with $4$ bad bulbs? Clearly $4!$. So the desired probability is $$\frac{4!}{(24)(23)(22)(21)}.$$

The mistake in the proposed solution is that the denominator is the one we get if we use bags, while the numerator is the one we get if we use rows. The first step in the analysis is to decide on an appropriate sample space. All counting of outcomes refers to that space.

Second Question: Again, there are several ways to get a sample space of equiprobable outcomes, and therefore several ways of getting at the answer.

(a) Imagine that before the performance, someone chooses a pair of seats "at random," and puts Reserved signs on these seats. There are $\binom{n}{2}$ ways of choosing $2$ seats, all equally likely.

How many of these sets of two seats are adjacent? The leftmost one of the two adjacent seats can be any one of seats $1, 2, 3,\dots,n-1$. So there are $n-1$ adjacent pairs, and our probability is $$\frac{n-1}{\binom{n}{2}}.$$ But $\binom{n}{2}=(n)(n-1)/2$, so our probability simplifies to $\frac{2}{n}$.

(b) Forget about theater seats, let's line up $n$ people in a row at random. We ask for the probability that A and B are next to each other. The $n$ people can be lined up in a row in $n!$ different ways, all equally likely. Put A and B in a bag (it's OK, they like to be close). Now we have $n-1$ abstract people, really $n-2$ people and a bag. These $n-1$ objects can be lined up in $19!$ ways. Cut the bag open. Then A can go to the immediate left of B, or to the immediate right, for a total of $2(n-1)!$ arrangements in which A and B are next to each other. Note that all arrangements with A and B next to each other are uniquely obtainable in this way. Thus our probability is $$\frac{2(n-1)!}{n!},$$ which again simplifies to $\frac{2}{n}$.

(c) For a change of pace, we work more directly with probabilities. Imagine that the theater is empty, and A comes in and chooses a seat at random. Then B comes in and chooses a seat at random. We calculate the probability that A and B end up next to each other. This can happen in two ways: (i) A sits at an end and B sits next to A and (ii) A sits at a non-end, and B sits next to A.

The probability that A chooses an end seat is $\frac{2}{n}$. Given that A is in an end seat, the probability that B sits next to A is $\frac{1}{n-1}$, since only $1$ of the remaining $n-1$ seats work.

The probability that A chooses a non-end seat is $\frac{n-2}{n}$. Given that A has chosen a non-end seat, the probability that B sits next to A is $\frac{2}{n-1}$, since there are $2$ seats that work.

Thus the probability that A and B end up next to each other is $$\frac{2}{n}\cdot\frac{1}{n-1}+\frac{n-2}{n}\cdot\frac{2}{n-1}.$$ The above expression simplifies to $\frac{2}{n}$.

Comment: The three solutions to the second problem are not complicated, but in each solution the very simple answer $\frac{2}{n}$ is obtained by algebraically simplifying a more complicated expression. There ought to be an argument that yields $\frac{2}{n}$ directly, but I have not come up with one.

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Very nice writeup. –  Brian M. Scott Oct 14 '11 at 21:08
Thank you for clarifying this, it was really helpful. Please see my edited question, I'm trying to apply this knowledge to another set of question but i'm still stuck. It would be really helpful if you could clarify this further. I'm starting to get a grasp of the problem... –  Dbr Oct 15 '11 at 10:31
is there any way to learn these 'tricks' when calculating such probabilities? any book in particular? also it would nice to have some source with explanations included, because DeGroot doesn't even have solutions for each question... –  Dbr Oct 15 '11 at 20:24
Just do many problems, it will click after a while. Asking on this site can be occasionally useful, answers are sometimes too terse. –  André Nicolas Oct 16 '11 at 3:59