Question

A town contains $4$people who repair televisions. If$4$sets break down, what is the probability that exactly$i$of the repairers is called? Solve the problem for$i=1,2,3,4.$What assumptions are you making?

Short answer

$$\begin{gathered} P(1repairer)=0.015625 \\ P(2repairers)=0.328125 \\ P(3repairers)=0.5625 \\ P\left(4repairers\right)=0.09375 \end{gathered}$$

Step-by-step solution

Given Information.

A town contains $4$people who repair televisions.

Explanation.

Translated into mathematical notation, sample space $S$is:

$S=\left\{\left(x_1,x_2,x_3,x_4\right):x_i\in \{1;2;3;4.\}\right\}$

$x_i$is a repairer a numerated representation of a repairer that repairs the $i$-th problem?

Presumption: Each element $S$is equally likely. Each house will equally likely call any of the repairers. A number of elements in $S$is $4^4$Probability of an event $A\subseteq S$that contains$n$elements is:

$P\left(A\right)=\frac{n}{4^4}$

Explanation.

$i)$What is the probability that exactly $1$the repairer is called:

This event contains only $4$possibilities from the sample space -$(1;1;1;1),(2;2;2;2),(3;3;3;3),(4;4;4;4)$

So by the formula above$P\left(1\text{repairer}\right)=\frac{4}{4^4}=0.015625$.

Explanation.

$ii)$What is the probability that exactly $2$what repairers are called:

There are $\left(\begin{array}{l}4\\ 2\end{array}\right)=6$choices for the repairers that will repair something.

And then each house can choose any of the two repairers$-2^4$. But if we want only those where both workers fix a house, so subtract$2$from that number. Because every household can choose the same repairer.

So the number of sample events in this event is$6·\left(2^4-2\right)$repairers.

$P\left(\text{2 repairers}\right)=\frac{6·\left(2^4-2\right)}{4^4}=0.328125$

Explanation.

$iii)$What is the probability that exactlylocalid="1648995415309" $3$what repairers are called:

There are $\left(\begin{array}{l}4\\ 3\end{array}\right)=4$subsets of the repairers that may be in play here.

If two of them are the same, choose that repairer in$3$ways the number of permutations with repetitionlocalid="1648996017325" $\frac{4!}{2!}$.

$P\left(3\text{repairers}\right)=\frac{4·3·4!}{4^4·2!}=0.5625$

Explanation.

$iiii)$What is the probability that exactly $4$what repairers are called:

Each repairer repairs one problem, the number of permutations is$4!$.

$P\left(4\text{repairers}\right)=\frac{4!}{4^4}=0.09375$