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If 7 accidents happen in a week, what is the probability that exactly 2 accidents happen in any day of the week.

I tried to work it this way.

2 accidents out of 7 can be taken in ${7}\choose{2}$$= 21$. Now this can happen in any of the 7 days, so we have $7 \times 21$. Now, the remaining 5 can happen in $6^5$ ways. So the probability is $\frac{7\times 21 \times 6^5}{7^7} = \frac{3 \times 6^5}{7^5}$

But this gives me a value greater than 1. I know I am doing something wrong. Can anyone help me ?

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    $\begingroup$ are the accidents uniformly distributed in the week , or is 7 the average of a Poisson distribution ? $\endgroup$
    – G Cab
    Mar 6 '20 at 22:16
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    $\begingroup$ You say "exactly $2$ accidents happen in a given day" in your question but then say "this can happen in any of the $7$ days" in your suggested answer. I do not think these are consistent. For the former, e.g. the probability of exactly two accidents on the Wednesday of your week of seven accidents, just divide your suggested answer by $7$ for the binomial probability of ${7 \choose 2}\frac{6^5}{7^7}$ . $\endgroup$
    – Henry
    Mar 6 '20 at 22:22
  • $\begingroup$ @Henry, Thanks. But in that case, what will be the probability that exactly 2 accidents happen in any of the seven days ? $\endgroup$
    – Shew
    Mar 6 '20 at 22:35
  • $\begingroup$ @Shrew On at least one day? About $0.826$. On exactly one day? About $0.371$. See my answer $\endgroup$
    – Henry
    Mar 7 '20 at 1:26
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  • The probability of exactly two accidents on say the Wednesday of your week of seven accidents is the binomial ${7 \choose 2}\frac{6^5}{7^7} = \frac{163296}{823543}\approx 0.198$

  • The probability of exactly two accidents on each of say the Wednesday and Thursday of your week of seven accidents is ${7 \choose 2}{5 \choose 2}\frac{5^3}{7^7} = \frac{26250}{823543}\approx 0.032$

  • The probability of exactly two accidents on each of say the Wednesday, Thursday and Friday of your week of seven accidents is ${7 \choose 2}{5 \choose 2}{3 \choose 2}\frac{4^1}{7^7}=\frac{2520}{823543}\approx 0.003$

  • The probability of exactly two accidents on at least one day of your week of seven accidents can be found by inclusion-exclusion and is ${7 \choose 1}{7 \choose 2}\frac{6^5}{7^7} - {7 \choose 2}{7 \choose 2}{5 \choose 2}\frac{5^3}{7^7} + {7 \choose 3}{7 \choose 2}{5 \choose 2}{3 \choose 2}\frac{4^1}{7^7} =\frac{680022}{823543}\approx 0.826$

  • The probability of exactly two accidents on exactly one day of your week of seven accidents can also be found by inclusion-exclusion and is ${7 \choose 1}{7 \choose 2}\frac{6^5}{7^7} - 2{7 \choose 2}{7 \choose 2}{5 \choose 2}\frac{5^3}{7^7} + 3{7 \choose 3}{7 \choose 2}{5 \choose 2}{3 \choose 2}\frac{4^1}{7^7} =\frac{305172}{823543}\approx 0.371$

The inclusion-exclusion calculations are because when you count the individual days having two accidents with ${7 \choose 1}{7 \choose 2}\frac{6^5}{7^7}$, you are double-counting each of the pairs of days having two accidents each, so to take account of this you have to subtract ${7 \choose 2}{7 \choose 2}{5 \choose 2}\frac{5^3}{7^7}$ once or twice depending on which question you are answering; this now means you are wrongly counting the each of the triples of days having two accidents each and need to adjust again.

A simulation in R produces roughly similar results

set.seed(2020)
cases <- 10^5
twos <- numeric(cases)
for (i in 1:cases){ twos[i] <- sum(table(sample(7,7, replace=TRUE)) == 2) }
table(twos)/cases
# twos
#       0       1       2       3 
# 0.17502 0.37036 0.34863 0.10599 
mean(twos >= 1)
# 0.82498
mean(twos == 1)
# 0.37036
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  • $\begingroup$ can you elaborate the last two answers ? $\endgroup$
    – Shew
    Mar 7 '20 at 6:44
  • $\begingroup$ @Shrew - I have added to my answer $\endgroup$
    – Henry
    Mar 7 '20 at 10:52

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