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

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

total number of ways in which accidents can be assigned to days = 77

probability of accidents happening on a particular day = 1/77

we can choose a day in 7 ways

hence probability = 7/77= 1/76

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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

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