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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. $\dfrac{7}{2^7}\\$

edited | 4k views
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@chotu can u describe it an easy way?

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 (8.7k points)
edited
+9

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.

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i missed that i'll edit the answer

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 (23.5k points)
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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

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same doubt here.please someone explain this.
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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.
+1

Simplest Approach!

+2
Shouldn't it be (6/7)^6 for binomial distribution in the 3rd product term?
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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

+1 vote
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