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I have this question in my textbook I am having trouble answering.

You have a fleet of 19 vehicles, 4 of which are vans and 15 of which are cars. Say you need to choose two vehicles at random. Regardless of which vehicle you choose first, what is the probability that second vehicle you choose is a van?

I've tried several answers but none of them make sense.

The one I thought was correct is to sum the two possible ways that the second vehicle be a van.

$$(4/19)\times(3/18) + (15/19)\times(4/18)$$

But I'm told that's wrong.

Could somebody please explain the way to do this?

Thank you very much

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  • $\begingroup$ This question appears to be off-topic because it is about probability/statistics and has been flagged for migration to Cross Validated. $\endgroup$
    – Thomas
    Commented Mar 3, 2014 at 9:40

2 Answers 2

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Hmm.

P(v2 = van) = sum P(v2 = van, v1 = v) for v in (car, van)
  = sum P(v2 = van | v1 = v) P(v1 = v) for v in (car, van)
  = P(v2 = van | v1 = car) P(v1 = car) + P(v2 = van | v1 = van) P(v1 = van)
  = 4/18 * 15/19 + 3/18 * 4/19

I dunno, looks the same as what you have. By the way, if you spell it out like this, you have a better chance of arguing that your result is correct (if the official answer is something else).

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    $\begingroup$ Because the order of choosing the vehicles does not affect their probabilities, the chance the second vehicle is a van must equal the chance the first is a van and that obviously is $4/19\approx 0.211$. Your solution simplifies to this answer. You can check it with a quick simulation in R, such as mean(replicate(10^4, sample(c(rep(0,4), rep(1,15)), 2)[2]==0)). $\endgroup$
    – whuber ♦
    Commented Mar 25, 2014 at 23:22
  • $\begingroup$ I dunno. It's not clear a priori that order doesn't matter; there are various trick questions floating around which are apparently similar and yet order or something like it does matter. Instead of simply affirming the conclusion, it seems more convincing to work through the details and show it. Anyway, that's my excuse for doing it the hard way. $\endgroup$
    – Robert Dodier
    Commented Mar 26, 2014 at 0:00
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    $\begingroup$ I did not affirm the conclusion: I merely observed that the two values drawn are exchangeable. If this is not perfectly obvious, imagine drawing the two members of the sample without looking at them, and then observing them in random order. A mathematically rigorous account of this is that every two-element sample $(x,y)$ from a set $S$ is in a unique one-to-one correspondence with the sample $(y,x)$. When you assume all samples are equally likely, then clearly the chance that $y$ is the second element drawn equals the chance that $y$ is the first element drawn. $\endgroup$
    – whuber ♦
    Commented Mar 26, 2014 at 14:45
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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).

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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