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

Best answer
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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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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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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

Answer:

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