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Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?

1. $\dfrac{1}{7^7}\\$
2. $\dfrac{1}{7^6}\\$
3. $\dfrac{1}{2^7}\\$
4. $\dfrac{7}{2^7}\\$

@chotu can u describe it an easy way?
The hard part of this question is deciding the sample space. Once you have done it, the rest of the work is easy.
Consider an array A[i] of length 7, where 1<=i<=7 and 1<=A[i]<=7, i denotes a distinct accident, and A[i] the day on which the accident took place. Therefore the array can take 7^7 values, out of which in 7 cases A[i]=A[j] for all 1<=i,j<=6.

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for every car accident we can pick a day in $7$ ways

total number of ways in which accidents can be assigned to days $= 7^{7}$

probability of accidents happening on a particular day $=\dfrac{1}{7^{7}}$

we can choose a day in $7$ ways.

hence probability $=\dfrac{7}{7^{7}}=\dfrac{1}{7^{6}}.$

the answer is B. The number of ways you can choose the "same day" is 7. The probability of all the accidents happening on same day is 1/77 . So 7*(1/77) is 1/76.

i missed that i'll edit the answer
edited by
So this is like 7 distinct balls going into 7 distinct bins, for which there are $7^7$ total ways (meaning the sample space has $7^7$ outcomes), and out of which there are 7 ways for all balls to get slotted into one bin, one for each bin (seven outcomes for which all accidents happened on the same day, one per day). Thus $\frac{7}{7^7} = \frac{1}{7^6}$.
Can anyone explain how the sample space is $7^7$
edited by
Why can’t we use Poisson distribution here? Here $\lambda$ for 1 day is 1. Now probability of 7 accidents happening on the same day is $e^{-1} \frac{\lambda^7}{7!} = \frac{1}{e} \frac{1}{7!}$

Edit: We are just told that 7 accidents occurred in a week. This is not the normal rate of accidents per week. Meaning: nowhere is it said that the rate of accidents per week is 7. So no question of using Poisson distribution here.

P(accident on a single day out of 7 days)=$\frac{1}{7}$

P(all accident occurred on the same day Out of 7 Day)= $\binom{7}{1}$ $(\frac{1}{7})^{7}$ $(\frac{6}{7})^{0}$   = $\frac{7}{7^{7}}$ = $\frac{1}{7^{6}}$  //Binomial Distribution So Option B is Correct Ans

(Means selecting single day out of 7 days and all the 7 accident should happen on same day with probability $\frac{1}{7}$)

edited

Just wanted to know whether we can calculate Sample space by using 'Combinations with repetition' ( sum of non negative integral solutions)concept which says

If x1 + x2+......xn= r( xi >=0) then no. of combination possible is C(n-1+r, r).

If so, then in above problem

x1+ x2 +x3+x4+x5+x6+x7= 7

which gives us sample space= C(7-1+7, 7) Therefore required probability is 7C1/ 13C7

If this logic cant be applied please explain why?

Thank you

edited
same doubt here.please someone explain this.
I understood why my above mentioned approach is wrong.

First of understand the solution.

7 accidents occurred in a week say a1, a2,....a7.

No of different ways accident can occur is all accidents (a1......a7) have 7choices. Therefore required probability = $\dfrac{7}{7^7}\\$

The approach mentioned in my above comment is wrong because,each accident is unique(a1....a7) which can occur in any day of week.

So if apply x1+x2 ..+x7 = 7 here, 1 combination will be x1= 5, x2= 2 amd rest all equal to 0. But x1 = 5 have 7C5 combination as all accidents are unique and any 5 out of 7 accidents can go in x1. So using this approach is wrong for this problem.

Simplest Approach!

I m not getting this question because we have 7 accidents and 7 days of week and now ACC to me permutations can be7!

First day 7 accidents can occur and if one accident occurs on first day the remaining 6 can occur on 2nd day and so on why this approach does not happen,can anyone explain
why D can't be the answer,for each day we can choose either accident will occur on this day or will not occur and hence 2 possibilities for each day and in this way D can be the answer.Anybody plz clear my doubt.
Please read question carefully they have mentioned SEVEN car accidents occurred in a WEEK. Each day in a week has seven possibilities.
Because 7 accident are unique...

Chocolate problem apply for similar things

The question has already been answered however one very simple way to approach the problem is as follows:

What is happening in the question is that there are reports of accidents occurring per day, which means that for one accident to occur, chances are $1\over7$.

Now if an accident has occurred then we want the remaining $6$ accidents to occur on the same day. Since each accident has a probability of $1\over7$, therefore for $6$ consecutive accidents it will be $\left(1\over7^6\right)$ which is option number B

by

Pls tell me how 6 consecutive accidents?It’ll be 7 consecutive accidents according to your explanation?so answer will be 1/7^7?

I think it’s already mentioned in the answer, “Now if an accident has occurred then we want the remaining $6$ accidents to occur on the same day.”

If you think carefully then $1\over7^ 7$ is the probability of 7 accidents occurring together.

However, the questioner wants to know about the probability of those accidents occurring on the same day. The only way we can ensure that they occur on the same day is by letting one accident to occur and then make the remaining accidents to take place on that day. Therefore, for the first accident, the probability will be $1\over7$, and the probability for the remaining accidents occurring on that day (this is what the questioner is essentially asking) will be $1\over7^6$.

Please note that my answer was only based on intuition. I would suggest that you first learn its quantitative approach and then think about my answer that would make more sense.

As a last note: you can easily relate this to conditional probability.

P(accidents occurring on the same day | accident has occurred) . . .

thank you @stoneheart :)
Yes understood Ty :)
you are so great mota bhai

how did you develop such a great intuiton?
Why can’t we use Poisson distribution here? Here $\lambda$ for 1 day is 1. Now probability of 7 accidents happening on the same day is $e^{-1} \frac{\lambda^7}{7!} = \frac{1}{e} \frac{1}{7!}$

I followed this approach