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38 votes

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. $\dfrac{7}{2^7}\\$
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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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Probability = No. of favorable cases / Total no. of cases

We want every accident in same day and days are 7. So, 7 ways (Favourable cases)

Now 1 accident can occur in any of the 7 days means 7 ways for each accident and we have 7 accidents. So sample space has to be—

7*7*7*7*7*7*7 = $7^{7}$

Probability = $\frac{7}{7^{7}}$

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Each accident can occur in any of the 7 days in a week, therefore, the probability of occurrence of an accident in a week is, P(a)=1/7

There are a total of 7 accidents, therefore the sample space is (1/7)^7

Now, all these accidents are occurring on the same day, so now, we must choose 1 day out of 7 days, ie, 7C1

The required probability is (7C1*((1/7)^7)), ie (1/7)^6
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4 Answers

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46 votes
Best answer

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

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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 $\frac{7}{7^7} = \frac{1}{7^6}$.
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Can anyone explain how the sample space is $7^7$
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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.
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40 votes
40 votes

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 $\frac{1}{7}$)

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4 Comments

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.
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Please read question carefully they have mentioned SEVEN car accidents occurred in a WEEK. Each day in a week has seven possibilities.
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Because 7 accident are unique...

Chocolate problem apply for similar things
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7 votes
7 votes

 

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

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4 Comments

Yes understood Ty :)
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you are so great mota bhai

 

how did you develop such a great intuiton?
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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 $e^{-1} \frac{\lambda^7}{7!} = \frac{1}{e} \frac{1}{7!}$
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6 votes
6 votes

I followed this approach

Answer:

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