How to solve the probability that customer C1 is the last to withdraw from the bank?

Question

Consider a bank with two clerks. Three people, $C_1$, $C_2$, and $C_3$, enter simultaneously. $C_1$ and $C_3$ go directly to the clerks, and $C_3$ waits until either $C_1$ or $C_2$ leaves before he begins service. What is the probability that $C_1$ is still in the post office after the other two have left when the service times are $i$ with probability $\frac{1}{10}$ , $i = 1, 2,...,10$?
I know that when $i$ is equal to $\{1,2,3\}$ with probability $\frac{1}{3}$, the probability becomes $$P(T_1>T_2+T_3)=P(T_1=3,T_2=1,T_3=1)=P(T_1=3)P(T_2 =1)P(T_3=1)=\frac{1}{3}*\frac{1}{3}*\frac{1}{3}=\frac{1}{27},$$ but when you have $\frac{1}{10}$ probability, what is the difference? I don't know how the idea of calculating the probability could change.

Answer 1

$(2/10)$ of the time, $C_1$ will be processed in $T \leq (2)$ which implies that it is impossible for him to finish last. That is, it is assumed that if $C_1$ finished with time $T=2$ and $C_2, C_3$ each finish with time $T=1$, that this tie does not constitute $C_1$ being last.

For each of $i \in \{3,4,\cdots,10\}$ there is a $(1/10)$ chance of $C_1$ finishing at time $i$. Let $p(i)$ denote the probability that $C_1$ is last, assuming that $C_1$ finishes at time $i$.

Then, the desired computation will be

$$\sum_{i=3}^{10} \left[\frac{1}{10} \times p(i)\right] = \frac{1}{10} \times \left[\sum_{i=3}^{10} p(i)\right].$$

Therefore, the problem has been reduced to computing $p(i)$ for each $i$ in $\{3,4,\cdots,10\}.$

For each $i$ in $\{3,4,\cdots,10\}$ what is required is to consider each of the equally likely $(100)$ ordered pairs $(x,y)$ where $x,y$ are each in $\{1,2,\cdots,10\}$.

For a given $i$, if the count of satisfactory ordered pairs is $n$, then you will have that $~\displaystyle p(i) = \frac{n}{100}.$

When $i$ is odd, $= (2k+1)$, $C_2$ can go from $1$ through $2k-1$, inclusive. For each such value of $C_2$, you will have $2k-C_2$ possible values for $C_3$. So, the number of satisfactory combinations for $i$ odd, $= (2k+1)$ will be

$$\sum_{C_2=1}^{2k-1} (2k - C_2) = 2k(2k-1) - \frac{(2k-1)(2k)}{2} = k(2k-1).$$

When $i$ is even $= (2k)$, $C_2$ can go from $1$ through $(2k-2)$, inclusive. For each such value of $C_2$, you will have $(2k-1)-C_2$ possible values for $C_3$. So, the number of satisfactory combinations for $i$ even, $= (2k)$ will be

$$\sum_{C_2=1}^{2k-2} (2k-1 - C_2) = (2k-2)(2k-1) - \frac{(2k-1)(2k-2)}{2} = (k-1)(2k-1).$$

These computations result in the following chart of values for $i$, and corresponding counts of satisfactory combinations:

$i = 3 ~: ~$ Count = $1.$
$i = 4 ~: ~$ Count = $3.$
$i = 5 ~: ~$ Count = $6.$
$i = 6 ~: ~$ Count = $10.$

$i = 7 ~: ~$ Count = $15.$
$i = 8 ~: ~$ Count = $21.$
$i = 9 ~: ~$ Count = $28.$
$i = 10 ~: ~$ Count = $36.$

So, the probability that $C_1$ finishes last will be

$$\frac{1}{10} \times \left\{ ~\frac{1}{100} \times [1 + 3 + 6 + 10 + 15 + 21 + 28 + 36] ~\right\}$$

$$= \frac{1}{10} \times \frac{120}{100} = \frac{3}{25}.$$