91Ó°ÊÓ

A town contains $4$people who repair televisions.

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

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

$x_i$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 $A⊆S$that contains$n$elements is:

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

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

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

$ii)$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.

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

$iii)$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" $\frac{4!}{2!}$.

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

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

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

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