4

A town contains 4 people who repair televisions. If 4 sets break down, what is the probability that exactly 2 of the repairers are called?

I know that the solution is $$\frac{6 * (2^4-2)}{4^4}$$

But I don't understand why we do $-2$ the the numerator.

My understanding:

6 is the amount of ways you can combine 4 workers along 2 choices. (4 choose 2).

$2^4$ is the amount of times a house can choose 2 repairers.

So, in my mind, the solution is: $$\frac{6 * (2^4)}{4^4}$$

a6i09per5f
  • 347
  • 1
    For what its worth, rather than calling it $2^4-2=14$ we could have used Stirling Numbers of the Second Kind and done the calculation as ${4 \brace 2}\cdot 2! = 7\cdot 2 = 14$. This is perhaps the preferred approach when trying to generalize to $n$ tv's breaking down, $r$ repairers available, the probability exactly $k$ of the repairers having been called being $\dfrac{\binom{r}{k}{n \brace k} k!}{r^n}$ which would have otherwise been difficult to write without them. – JMoravitz Sep 13 '22 at 02:55

4 Answers4

6

But I don't understand why we do $-2$ the the numerator.

If Bill and Ben are the repairers for the four broken sets, you do not want Bill hogging all of the jobs; nor Ben. Those two ways must be excluded.

$2^4-2$ counts the ways to assign two repairers to four sets, excluding ways just one repairer is assigned to all four sets.

Graham Kemp
  • 129,094
3

There are $4\choose2$ ways that two repairers can be chosen, and then there are $4\brace2$ ways that the two chosen repairers can be assigned to repair four televisions. Here $4\brace2$ is the Stirling number of the second kind, which may be read as "four subset two".

The Stirling number gives us the number of ways to break four labelled items into two non-empty unlabelled subsets. Let's say we call the customers a, b, c, and d, then we can break them up as

  1. {{a},{b,c,d}}
  2. {{a,c,d},{b}}
  3. {{a,b,d},{c}}
  4. {{a,b,c},{d}}
  5. {{a,b},{c,d}}
  6. {{a,c},{b,d}}
  7. {{a,d},{b,c}}

So in this case, ${4\brace2}=7$ and this is multiplied by 2 since each pair of subsets of customers can go to either of the two repairers.

This is the origin of the $2^4-2=14$ term. Another way to see this is that you can break up the sets of televisions in sixteen different ways, but two of those ways are of the form {{a,b,c,d},$\emptyset$}, in the above notation, and result in one of the repairmen not being assigned, and the other being assigned all four, which contradicts that two people were chosen.

So the overall probability is $$ p={{4\choose2}{4\brace2}2\over4^4}={84\over256} $$

In case you don't believe this or think it is all mumbo-jumbo, here is a computer program which explicitly counts all the ways:

package main

import "fmt"


func main() {
    verbose := false
    nPairs := 0
    for a := 0; a < 4; a++ {
        for b := 0; b < 4; b++ {
            for c := 0; c < 4; c++ {
                for d := 0; d < 4; d++ {
                    val := []int{a, b, c, d}
                    vals := make([]int, 4)
                    for i := 0; i < 4; i++ {
                        vals[val[i]]++
                    }
                    pairs := 0
                    triples := 0
                    for i := 0; i < 4; i++ {
                        if vals[i] == 2 {
                            pairs++
                        }
                        if vals[i] == 3 {
                            triples++
                        }
                    }
                    if pairs == 2 {
                        if verbose {
                            fmt.Printf("pairs %v\n", val)
                        }
                        nPairs++
                    }
                    if triples == 1 {
                        if verbose {
                            fmt.Printf("triples %v\n", val)
                        }
                        nPairs++
                    }
                }
            }
        }
    }
    fmt.Printf("%d\n", nPairs)
}

Run it online here.

There is a variable called verbose which, if set to true, will print all the pairs for you.

Suzu Hirose
  • 11,660
1

Reverse engineering the answer, once you use the $~\displaystyle \binom{4}{2}~$ factor to narrow the selection to $(2)$ of the $(4)$ repair people, you have to deduct the $(2)$ cases where exactly $(1)$ of the $(2)$ candidate repair people is called.

user2661923
  • 35,619
  • 3
  • 17
  • 39
  • So let's say we have the repairers ${x1, x2}$. You are saying that we have to take out the choices ${x1,x1}$ and ${x2,x2}$, correct? – a6i09per5f Sep 13 '22 at 02:00
  • @swisstackle Close. Include in the $2^4$ enumeration are the choices $(x_1,x_1,x_1,x_1)$ and $(x_2,x_2,x_2,x_2)$. Of the $(16)$ choices, these are the two that have to be eliminated. – user2661923 Sep 13 '22 at 04:45
1

My solution strategy would be to eliminate this kind of error by being more specific in the coverage of the $4$ houses by $2$ TV repairmen. That is, there are only $2$ possible assignment sets to cover the $4$ houses. $1$ and $3$, and $2$ and $2$. There are $8$ ways for the former and $6$ ways the latter making a total of $14$.

Phil H
  • 5,579