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If 7 accidents (all are indistinguishable i.e we care only about whether accident happen or not, not the details of it) happen in a week. What is the probability of exactly two accidents on at least one day of a week.

My solution is given below

$$ \frac{1}{13 \choose 6} \Bigg[{7 \choose 1}{10 \choose 5} - {7 \choose 2} {7 \choose 4} + {7\choose 3} 4 \Bigg] = 0.681 $$

I used the equation that number of ways $r$ balls can be arranged in $n $ cells in ${n+r-1 \choose n-1}$ distinguishable ways.

Is it correct ?

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  • $\begingroup$ I cant follow your reasoning. Can you explain why this formula should be correct? $\endgroup$
    – Nurator
    Mar 19 '20 at 11:52
  • $\begingroup$ @Nurator included the information. $\endgroup$
    – Shew
    Mar 19 '20 at 12:10
  • $\begingroup$ In order to have an unambiguous way of answering, we need to know how the accidents themselves are distributed. In a similar "balls-in-bins" problem, we might say "take a ball, throw it at the bins so that it randomly lands in one of the bins uniformly at random. Then take another ball and throw it in as well, so that where that ball lands is uniformly distributed and independent to the previous." As alluded to below, it is very uncommon for the $\binom{13}{6}$ outcomes to be equally likely in practice, making this an uncommon assumption. $\endgroup$
    – JMoravitz
    Mar 19 '20 at 12:16
  • $\begingroup$ @JMoravitz, yes in practice uniform at random assumption might not hold. But in my case, I assume it. Thanks $\endgroup$
    – Shew
    Mar 19 '20 at 12:19
  • $\begingroup$ I think the word "probability" should have some unusual meaning, so that your solution was correct. $\endgroup$
    – user
    Mar 19 '20 at 12:33
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The number of all possible distributions of accidents is (stars and bars method) $${13\choose 6}$$

The number of all possible distributions of accidents if 2 happend on monday is $${10\choose 5}$$

The number of all possible distributions of accidents if 2 happend on monday and 2 on thuesday is $${7\choose 4}$$

The number of all possible distributions of accidents if 2 happend on monday and 2 on tuesday and 2 on wednesday is $${4\choose 1}$$ So the naswer is, by PIE: $$P={{7\choose 1}{10\choose 5} -{7\choose 2} {7\choose 4}+{7\choose 3}{4\choose 1}\over {12\choose 6}}$$

So, your answer maches with my answer.

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  • $\begingroup$ why it is downvoted ? $\endgroup$
    – Shew
    Mar 19 '20 at 12:10
  • $\begingroup$ That makes the wild assumption that each of the possible outcomes where you treat days as distinct and accidents as indistinct are equally likely to have occurred. Do you really think it is equally likely that all seven accidents occur on monday as it is that one accident happens each of the days? $\endgroup$
    – JMoravitz
    Mar 19 '20 at 12:10
  • $\begingroup$ Despite the fact that we say the accidents are indistinguishable, under most sensible interpretations of this problem and problems like it, it makes the most sense to treat this behind the scenes as each of the accidents (or balls for a balls-in-bins problem) as being distinct so that we can work within an equiprobable sample space. The probability of occurrence isnt simply "total number of good outcomes divided by total number of outcomes." That only works if the outcomes are equally likely to have occurred, which the problem statement does not imply that the outcomes you use are. $\endgroup$
    – JMoravitz
    Mar 19 '20 at 12:13
  • $\begingroup$ @JMoravitz Will you provide a correct solution? $\endgroup$
    – Aqua
    Mar 19 '20 at 12:15
  • $\begingroup$ @Aqua linked a duplicate above. Note the usage of $7^7$ as the sample space size, rather than $\binom{13}{6}$ $\endgroup$
    – JMoravitz
    Mar 19 '20 at 12:18

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