Q. 2.4 - A First Course in Probability

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?

Step-by-Step Solution

Answer

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

Step 1 Given Information.

A town contains $4$people who repair televisions.

Step 2 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:

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

Step 3 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$P\left(1\text{ repairer }\right)=\frac{4}{4^4}=0.015625$.

Step 4 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.

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

Step 5 Explanation.

$iii)$What is the probability that exactly$3$ 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 repetition$\frac{4!}{2!}$.

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

Step 6 Explanation.

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

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

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