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Dilip Sarwate
Dilip Sarwate
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Here is another way of doing the problem (which I hope will persuade other people that what whuber says in a comment on Robert Dodier's answer is correct).

First, I hope that it is OK to assert thatMake a list of all possible outcomes of the probability thatassignment of vehicles to the firstfirst vehicle picked is a vanand second choice. A typical outcome is of the form $\frac{4}{19}$$(X,Y)$ where $X \in \Omega = \{C_1, C_2, \ldots, C_{15}, V_1, V_2, V_3, V_4\}$ and this probability is not affected by whether$Y \in \Omega - \{X\}$. Now we pick asort the list lexicographically by first entry and then by second vehicle from those that remain or notentry. Now considerThe result is shown in the table below where pickingeach of the second vehicle and make$19$ rows has $18$ entries on it. $$\begin{array}{cccccccccc}(C_1, C_2)& (C_1, C_3) & \ldots &(C_1, C_{15}) &(C_1, V_1)&(C_1, V_2) &(C_1, V_3) &(C_1,V_4)\\ (C_2, C_1)& (C_2, C_3) & \ldots &(C_2, C_{15}) &(C_2, V_1)&(C_2, V_2) &(C_2, V_3) &(C_2,V_4)\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\vdots\\ (C_{15}, C_1)& (C_{15}, C_2) & \ldots &(C_{15}, C_{14}) &(C_{15}, V_1)&(C_{15}, V_2) &(C_{15}, V_3) &(C_{15},V_4)\\ (V_1, C_{1})& (V_1, C_2) & \ldots &(V_1, C_{14}) &(V_1, C_{15})&(V_1, V_2) &(V_1, V_3) &(V_1,V_4)\\ (V_2, C_{1})& (V_2, C_2) & \ldots &(V_2, C_{14}) &(V_2, C_{15})&(V_2, V_1) &(V_2, V_3) &(V_2,V_4)\\ (V_3, C_{1})& (V_3, C_2) & \ldots &(V_3, C_{14}) &(V_3, C_{15})&(V_3, V_1) &(V_3, V_2) &(V_3,V_4)\\ (V_4, C_{1})& (V_4, C_2) & \ldots &(V_4, C_{14}) &(V_4, C_{15})&(V_4, V_1) &(V_4, V_2) &(V_4,V_3) \end{array}$$

Since the last four rows have a list of all possibilities that we getvan in the first column, we confirm suchwhat we already "know" viz. the probability of van as the first pick is $(C_3, C_5), (C_4, V_1), (V_2, C_8), (V_1, C_3)$, etc$\frac{4~\text{rows}}{19~\text{rows}} = \frac{4\times 18~\text{pairs}}{19\times 18~\text{pairs}} = \frac{4}{19}$.

But, if we sort our list by second entry first and then first entry, There arethe table above gets $19\times 18$ such pairs andre-arranged with the $4\times 18$ of them havelast four rows becoming

$$\begin{array}{cccccccccc}(C_1, V_1)& (C_2, V_1) & \ldots &(C_{14}, V_1) &(C_{15},V_1)&(V_2, V_1) &(V_3, V_1) &(V_4, V_1)\\ (C_1, V_2)& (C_2, V_2) & \ldots &(C_{14}, V_2) &(C_{15},V_2)&(V_1, V_2) &(V_3, V_2) &(V_4, V_2)\\ (C_1, V_3)& (C_2, V_3) & \ldots &(C_{14}, V_3) &(C_{15},V_3)&(V_1, V_3) &(V_2, V_3) &(V_4, V_3)\\ (C_1, V_4)& (C_2, V_4) & \ldots &(C_{14}, V_4) &(C_{15},V_4)&(V_1, V_4) &(V_2, V_4) &(V_3, V_4) \end{array}$$

These last four rows are the only rows with a $V$$V_i$ in the first position. These are of the formsecond $(V_1, \star), (V_2, \star), (V_2, \star), (V_4, \star)$ wherecolumn $\star$ can be(be sure you understand why) any of the remaining $18$ vehicles. OK, so far,and so good: we seeget again that the probability probability of seeing a $V$van as the first choice issecond $\frac{4\times 18}{19\times 18} = \frac{4}{19}$ exactly as stated earlierpick is $\frac{4}{19}$.

Now take the $19\times 18$ pairs of vehicle choices, which are of the form $(A,B)$, and swap the entries, transforming $(A,B)$ into $(B.A)$. We still have exactly the same collection of $19\times 18$ pairs, just in a different order, right? We didn't lose any pair that was in the original set and we didn't generate any new pair that was not in the original set. So the number of pairs having a $V$ in the first position is exactly the same as before and the probability of having a $V$ in the first position (used to be second position before we did the swap) is still the same $\frac{4}{19}$. Thus, the probability of getting a van as the second choice is $\frac{4}{19}$ also; same as the probability of getting a van as the first choice.

Here is another way of doing the problem (which I hope will persuade other people that what whuber says in a comment on Robert Dodier's answer is correct).

First, I hope that it is OK to assert that the probability that the first vehicle picked is a van is $\frac{4}{19}$ and this probability is not affected by whether we pick a second vehicle from those that remain or not. Now consider picking the second vehicle and make a list of all possibilities that we get, such as $(C_3, C_5), (C_4, V_1), (V_2, C_8), (V_1, C_3)$, etc. There are $19\times 18$ such pairs and $4\times 18$ of them have a $V$ in the first position. These are of the form $(V_1, \star), (V_2, \star), (V_2, \star), (V_4, \star)$ where $\star$ can be any of the remaining $18$ vehicles. OK, so far, so good: we see that the probability of seeing a $V$ as the first choice is $\frac{4\times 18}{19\times 18} = \frac{4}{19}$ exactly as stated earlier.

Now take the $19\times 18$ pairs of vehicle choices, which are of the form $(A,B)$, and swap the entries, transforming $(A,B)$ into $(B.A)$. We still have exactly the same collection of $19\times 18$ pairs, just in a different order, right? We didn't lose any pair that was in the original set and we didn't generate any new pair that was not in the original set. So the number of pairs having a $V$ in the first position is exactly the same as before and the probability of having a $V$ in the first position (used to be second position before we did the swap) is still the same $\frac{4}{19}$. Thus, the probability of getting a van as the second choice is $\frac{4}{19}$ also; same as the probability of getting a van as the first choice.

Here is another way of doing the problem (which I hope will persuade other people that what whuber says in a comment on Robert Dodier's answer is correct).

Make a list of all possible outcomes of the assignment of vehicles to the first and second choice. A typical outcome is of the form $(X,Y)$ where $X \in \Omega = \{C_1, C_2, \ldots, C_{15}, V_1, V_2, V_3, V_4\}$ and $Y \in \Omega - \{X\}$. Now we sort the list lexicographically by first entry and then by second entry. The result is shown in the table below where each of the$19$ rows has $18$ entries on it. $$\begin{array}{cccccccccc}(C_1, C_2)& (C_1, C_3) & \ldots &(C_1, C_{15}) &(C_1, V_1)&(C_1, V_2) &(C_1, V_3) &(C_1,V_4)\\ (C_2, C_1)& (C_2, C_3) & \ldots &(C_2, C_{15}) &(C_2, V_1)&(C_2, V_2) &(C_2, V_3) &(C_2,V_4)\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\vdots\\ (C_{15}, C_1)& (C_{15}, C_2) & \ldots &(C_{15}, C_{14}) &(C_{15}, V_1)&(C_{15}, V_2) &(C_{15}, V_3) &(C_{15},V_4)\\ (V_1, C_{1})& (V_1, C_2) & \ldots &(V_1, C_{14}) &(V_1, C_{15})&(V_1, V_2) &(V_1, V_3) &(V_1,V_4)\\ (V_2, C_{1})& (V_2, C_2) & \ldots &(V_2, C_{14}) &(V_2, C_{15})&(V_2, V_1) &(V_2, V_3) &(V_2,V_4)\\ (V_3, C_{1})& (V_3, C_2) & \ldots &(V_3, C_{14}) &(V_3, C_{15})&(V_3, V_1) &(V_3, V_2) &(V_3,V_4)\\ (V_4, C_{1})& (V_4, C_2) & \ldots &(V_4, C_{14}) &(V_4, C_{15})&(V_4, V_1) &(V_4, V_2) &(V_4,V_3) \end{array}$$

Since the last four rows have a van in the first column, we confirm what we already "know" viz. the probability of van as the first pick is $\frac{4~\text{rows}}{19~\text{rows}} = \frac{4\times 18~\text{pairs}}{19\times 18~\text{pairs}} = \frac{4}{19}$.

But, if we sort our list by second entry first and then first entry, the table above gets re-arranged with the last four rows becoming

$$\begin{array}{cccccccccc}(C_1, V_1)& (C_2, V_1) & \ldots &(C_{14}, V_1) &(C_{15},V_1)&(V_2, V_1) &(V_3, V_1) &(V_4, V_1)\\ (C_1, V_2)& (C_2, V_2) & \ldots &(C_{14}, V_2) &(C_{15},V_2)&(V_1, V_2) &(V_3, V_2) &(V_4, V_2)\\ (C_1, V_3)& (C_2, V_3) & \ldots &(C_{14}, V_3) &(C_{15},V_3)&(V_1, V_3) &(V_2, V_3) &(V_4, V_3)\\ (C_1, V_4)& (C_2, V_4) & \ldots &(C_{14}, V_4) &(C_{15},V_4)&(V_1, V_4) &(V_2, V_4) &(V_3, V_4) \end{array}$$

These last four rows are the only rows with a $V_i$ in the second column (be sure you understand why) and so we get again that the probability of van as the second pick is $\frac{4}{19}$.

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Dilip Sarwate
Dilip Sarwate
  • 47.8k
  • 4
  • 124
  • 235

Here is another way of doing the problem (which I hope will persuade other people that what whuber says in a comment on Robert Dodier's answer is correct).

First, I hope that it is OK to assert that the probability that the first vehicle picked is a van is $\frac{4}{19}$ and this probability is not affected by whether we pick a second vehicle from those that remain or not. Now consider picking the second vehicle and make a list of all possibilities that we get, such as $(C_3, C_5), (C_4, V_1), (V_2, C_8), (V_1, C_3)$, etc. There are $19\times 18$ such pairs and $4\times 18$ of them have a $V$ in the first position. These are of the form $(V_1, \star), (V_2, \star), (V_2, \star), (V_4, \star)$ where $\star$ can be any of the remaining $18$ vehicles. OK, so far, so good: we see that the probability of seeing a $V$ as the first choice is $\frac{4\times 18}{19\times 18} = \frac{4}{19}$ exactly as stated earlier.

Now take the $19\times 18$ pairs of vehicle choices, which are of the form $(A,B)$, and swap the entries, transforming $(A,B)$ into $(B.A)$. We still have exactly the same collection of $19\times 18$ pairs, just in a different order, right? We didn't lose any pair that was in the original set and we didn't generate any new pair that was not in the original set. So the number of pairs having a $V$ in the first position is exactly the same as before and the probability of having a $V$ in the first position (used to be second position before we did the swap) is still the same $\frac{4}{19}$. Thus, the probability of getting a van as the second choice is $\frac{4}{19}$ also; same as the probability of getting a van as the first choice.