# What is the probability of exactly two accidents on at least one day of a week, in case of 7 indistinguishable accidents per week [duplicate]

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 ?

• I cant follow your reasoning. Can you explain why this formula should be correct? Mar 19 '20 at 11:52
• @Nurator included the information.
– Shew
Mar 19 '20 at 12:10
• 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. Mar 19 '20 at 12:16
• @JMoravitz, yes in practice uniform at random assumption might not hold. But in my case, I assume it. Thanks
– Shew
Mar 19 '20 at 12:19
• I think the word "probability" should have some unusual meaning, so that your solution was correct.
– user
Mar 19 '20 at 12:33

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: $$P={{7\choose 1}{10\choose 5} -{7\choose 2} {7\choose 4}+{7\choose 3}{4\choose 1}\over {12\choose 6}}$$

• @Aqua linked a duplicate above. Note the usage of $7^7$ as the sample space size, rather than $\binom{13}{6}$ Mar 19 '20 at 12:18