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+20 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. $\dfrac{7}{2^7}\\$
in Probability by Veteran (52.3k points)
edited by | 4.7k views
@chotu can u describe it an easy way?

2 Answers

+29 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 $=\dfrac{7}{7^{7}}=\dfrac{1}{7^{6}}.$

by Loyal (8.6k points)
edited by

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
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 $\frac{7}{7^7} = \frac{1}{7^6}$.
+24 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 $\frac{1}{7}$)

by Boss (24k points)
edited by

@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

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

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