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+3 votes

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. $\frac{7}{2^7}$
asked in Probability by Veteran (64.8k points)   | 1.3k views

2 Answers

+12 votes
Best answer

answer - B [EDIT]

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


answered by Boss (9.1k points)  
selected 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
+7 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 $\frac{1}{7}$)

answered by Veteran (18.7k 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

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