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6 stops on the line, and a car with 4 passengers. Assume they are equally likely to get off at any stop. What is the probability that A $2$ passengers get off stop 2, and $2$ get off stop 4? B $2$ get off stop 1, and $2$ get of at another stop (but they get off at the same stop as each other?)

A I get $1/81$ which is wrong?

B I get $5/6^4$ which is also wrong?

Why are both of these wrong?

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    I don't think anyone can say **why** these answers are wrong unless you explain why you thought they are right. If you just want the correct answers, that's a different question. – 2017-01-12
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    If they are equally likely to get off at any stop, then shouldn't the answer for **A** and **B** be the same? – 2017-01-12
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    @ Shraddheya Shendre -- By addition of mutiually exclusive events, the answer for (B) should be 5 times the answer for (A). – 2017-01-12
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    Yes, I see what you are saying. My bad. – 2017-01-12

1 Answers 1

2

For both the parts, the total number of possible ways by which the passengers can exit is $6^4$.

For part A,
the ways by which $2$ passengers can be grouped is $\binom{4}{2}$ (and the other $2$ are automatically grouped so there's nothing else to do). So the number of ways in which the exiting of passengers can be done is $\binom{4}{2}$. So the required probability in part A is $$\frac{\binom{4}{2}}{6^4}$$

For part B,
the number of ways of grouping remains same: $\binom{4}{2}$; but now the exit stop can be assigned in $5$ ways (one group exits at stop $1$ and other can exit on remaining $5$ stops). So the required probability is $$\frac{5 \times \binom{4}{2}}{6^4}$$

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    Since, as you say, the other $2$ passengers are automatically grouped, this means that you're double counting. – 2017-01-12
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    So you are saying that I shouldn't count for the switching of the groups (and hence not multiply by $2$)? – 2017-01-12
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    Right -- don't multiply by 2. The ordered 4-tuple is completely determined once you select the two passengers who get off at stop 2. – 2017-01-12
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    Yes, I corrected the answer. Thanks for pointing out the mistake. – 2017-01-12
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    No problem -- it looks fine now. – 2017-01-12