## 4 Answers

**Answer - B **

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

### 5 Comments

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 $\frac{1}{7}$)

### 8 Comments

@Rajesh Pradhan @ankitrokdeonsns @Arjun

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

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.

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**

### 7 Comments

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) . . .`