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There were four accidents in a town during a seven-day period. Would you be surprised if all four occurred on the same day? If each of the four occurred on a different day?

I couldn't figure out how to solve this problem. It looks like that it is somewhat related to Poisson distribution.

sv_jan5
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    There are $7^4$ ways of distributing the accidents between the days. Are they equally probable? How many have all the accidents on the same day? On different days? – Henry Sep 14 '16 at 10:17
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    Probability of happening on same day $ = \frac{7}{7^4}$. And the probability of happening on different days $ = \frac{765*4}{7^4}$. Please check if it is correct – sv_jan5 Sep 14 '16 at 10:50
  • I wouldn't be surprised. For one thing, they could all be related, for example a particular weather condition on that day. Apart from that, it isn't that unlikely, even if the accidents were totally independent. All on the same day is about one in 2500 - things have to happen in the world that are less likely - ok so you might check if there WAS a connection between them I suppose - it's not 1 in 2500, it's the one you got 1/7^3 – Cato Sep 14 '16 at 11:01
  • I agree with your calculations. I would say case 1 - cursory check that there was not a common cause. Case 2 : check it wasn't a 1 a day lunatic causing the problem – Cato Sep 14 '16 at 11:03
  • @Henry I wanted to ask if the op has calculated the correct probability of accidents occurring on different days. Wouldnt it be ${7}\choose{4}$ in the numerator? – Daman Aug 22 '18 at 19:53
  • @Damn1o1 - no, sv_jan5 was correct in the comment with $7 \times 6 \times 5 \times 4$, making the probability about $0.349854$ and so not an extreme result – Henry Aug 22 '18 at 22:05

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The probabilities are:

  • all the same day different days $\frac{7}{7^4}= \approx 0.00292$ as you say in a comment - this might be described as surprising if they are unconnected (perhaps people only drive through the village on a particular day when the by-pass was closed)
  • all different days $\frac{7\times 6 \times 5\times 4}{7^4} \approx 0.34985$ as you say in a comment - this is not particularly surprising
  • on two days $\frac{7 \times 6\times 4 + 7\times 6 \times 3}{7^4}\approx 0.12245$ since you can either have three on one day and one another or two on each of two days
  • on three days $\frac{7 \times 6\times 5 \times 6}{7^4}\approx 0.52478$

and, as a check, these probabilities add up to $1$

Henry
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  • I am not getting numerator of all different days. Can you explain it to me ? Also tell me 7 choose 4 will also choose days with gap in it . So are we talking about consecutive days? – Daman Aug 23 '18 at 06:43
  • @Damn1o1 - For different days: (a) the first accident on the list can be any of 7 days, the second any of the other 6, the third any of the the other 5, and the fourth any of the other 4, then multiply; or (b) the accidents can be spread across the days ${7 \choose 4}$ ways but can be ordered $4!$ ways within the selected days, then multiply. Bot approaches give the same result – Henry Aug 23 '18 at 07:24
  • I got that part. And in your third case(on two days), I am having $\frac{7 \times 6\times 3}{7^4}$ because there are three possibilities $(1,3),(3,1),(2,2)$. The first possibility implies 1 accident on one day and 3 accidents on another day. How did you get $\frac{7 \times 6 \times 4}{7^4}$ ? – Daman Aug 23 '18 at 08:15
  • @Damn1o1 - $7 \times 6 \times 3$ is just the $(2,2)$ cases. $7 \times 6 \times 4$ is for the $(3,1)$ cases with $7$ possibilities for the three, $6$ other possibilities for the one and $4$ possible ways of choosing which three out of four – Henry Aug 23 '18 at 15:03
  • You went this far so help me a little more please tell me how did we get $3$ in $(2,2)$ case. The way I am thinking is that $2$ accidents can happen in $7$ days making it$ 7$ ways and other$ 2$ can happen in any other 6 days. That makes $7\times 6$ . – Daman Aug 23 '18 at 15:19
  • @Damn1o1 - in the $(2,2)$ case a particular accident can happen on any of $7$ days and the other day with accidents can be any of the other $6$ days, and the accident which happens on the same day as the first particular accident can be any of the other $3$ accident – Henry Aug 23 '18 at 16:14