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So, as I have understood it, if you have two experiments and you want to know the probability of a set of two outcomes happening concurrently, then you multiply the chance of the first outcome by the chance of the second outcome and voila, you have your probability.

However, I am confused as to when this doesn't work. For example, I just did a problem:

James lives in San Francisco and works in Mountain View. In the morning, he has 3 transportation options (bus, cab, or train) to work, and in the evening he has the same 3 choices for his trip home. What is the probability that he uses the same of mode of transportation twice?

My first inclination was 1/9th but apparently I am wrong. I was told to use a tree to count the favorable outcomes. I did so, and see that the answer is 1/3, but for the life of me I can't see the difference between this question and the first type I mentioned.

I am obviously missing some finer points or nuance in the question which should clue me in. What is it?

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  • $\begingroup$ The problem statement is not complete. what is it asking? $\endgroup$ – Paul Jul 16 '17 at 17:11
  • $\begingroup$ @Paul Sorry, I added it: What is the probability that he uses the same mode of transportation twice? $\endgroup$ – rocksNwaves Jul 16 '17 at 17:13
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    $\begingroup$ It rather depends on whether the morning and evening choices are independent. If he leaves his car at home in the morning, can he use it to go home in the evening? I suspect the real answer is greater than $\frac13$ $\endgroup$ – Henry Jul 16 '17 at 17:15
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    $\begingroup$ @Henry To be fair, it does mention cab, not car (all options are forms of public transportation) $\endgroup$ – Marcus Andrews Jul 16 '17 at 17:19
  • $\begingroup$ @Marcus - I misread it, though the independence point still matters $\endgroup$ – Henry Jul 16 '17 at 17:38
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There are three ways to use the same transportation mode twice: Bus twice, cab twice, or train twice. So the probability of using, say, the bus twice is indeed $\frac{1}{9}$, but there is also the option of using the cab twice, or the train twice, so you have to consider those options as well.

$P(\text{same transportation twice}) = P(BB) + P(CC) + P(TT)$

$P(\text{same transportation twice}) = \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$

$P(\text{same transportation twice}) = 3 \cdot \frac{1}{9}$

$P(\text{same transportation twice}) = \frac{1}{3}$

Remember that "or" implies addition, whereas "and" implies multiplication.

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    $\begingroup$ The "or", "and" explanation is exactly what I needed. I'll keep that in mind from now on. $\endgroup$ – rocksNwaves Jul 16 '17 at 17:31
  • $\begingroup$ Note that using multiplication for "and" requires that the events are independent (valid here, since which morning option he chose doesn't affect how he chooses the evening option), while using addition for "or" requires that the events are mutually exclusive (valid here, since if he takes the bus both times, he can't also take the cab both times or train both times in the same outcome). $\endgroup$ – YawarRaza7349 Jul 17 '17 at 3:23
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1/9 is the probability that a specific mode of transportation is used twice. There are 3 different modes, so the answer is 1/9 + 1/9 +1/9 = 1/3.

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You can just do it by counting. You are correct that he has $3 \cdot 3=9$ choices of morning and evening transportation. There are $3$ cases where he takes the same type both ways, giving a probability of $\frac 39=\frac 13$

Alternately, you can reflect on the fact that whatever choice he makes in the morning, he can still match it. In the evening he has three choices, one of which matches the morning. Again $\frac 13$

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  • $\begingroup$ Could your answer include the same situation in two different ways, i.e. the one where the question merits counting and the other needs the multiplication rule? I'd like to see the two types contrasted so that I can pick out the differences. $\endgroup$ – rocksNwaves Jul 16 '17 at 17:23

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