If 6 fair 6-sided dice are rolled, what is the probability that at lea

Question

If 6 fair 6-sided dice are rolled, what is the probability that at least two of them show the same face?

A.  \frac{1}{6^6}

B.  \frac{57}{6^6}

C.  \frac{5!}{6^5}

D.  1 - \frac{5!}{6^5}

E.  1 - \frac{1}{6^5}

Danush649 · Answer

ans= 1-(all are different)
 = 1-(6!/6^6)
 =1-(5!/6^5)
IMO option D

Bunuel · Answer

FYI, this question was written for the competition but then I thought it was too easy for a question for which one has 24 hours...

ThatDudeKnows · Answer

We can either find the probability of getting at least two of the same, or we can find the probability of getting no repeats and subtract that from one. The latter is easier, so let's do that. As soon as I decide to do a problem in a way that requires one last little step at the end that I might otherwise forget to do (we've all made that mistake, so don't pretend you've never done it!!

), I flip my computer mouse over and then get to work solving the problem. That way, when I go to click my answer, I have a little reminder to make sure I've done that last step. What's the probability of getting SOMETHING on the first die? 1 What's the probability of getting something DIFFERENT on the second die? 5/6 What's the probability of getting something DIFFERENT on the third die? 4/6 ...3/6 ...2/6 ...1/6 So, the probability of getting all different is 1*\frac{5}{6}*\frac{4}{6}*\frac{3}{6}*\frac{2}{6}*\frac{1}{6} \frac{5!}{6^5} 1-\frac{5!}{6^5} Answer choice D.

BrushMyQuant · Answer

If 6 fair 6-sided dice are rolled, what is the probability that at least two of them show the same face?

P(Something Happening) = 1 - P(Thing not happening)

=>  P(At least two will show same face) = 1 - P(Every die show a different face)

=>  P(Every die show a different face)  = P(Get any number on the  1^{st}  die) * P(Get any number apart from previous number in the  2^{nd}  die) * P(Get any number out of the 4 remaining number in the  3^{rd}  die) * P(Get any number out of the 3 remaining number in the  4^{th}  die) * P(Get any number out of the 2 remaining number in the  5^{th}  die) * P(Get any number out of the 1 remaining number in the  6^{th}  die)
=  \frac{6}{6}   *  \frac{5}{6}   *  \frac{4}{6}   *  \frac{3}{6}   *  \frac{2}{6}   *  \frac{1}{6}  
=  \frac{5*4*3*2*1}{6^5}   =  \frac{5!}{6^5}

=>  P(At least two will show same face)  = 1 -  \frac{5!}{6^5}

So,  Answer will be D 
Hope it helps!

Playlist on Solved Problems on Probability  here

Watch the following video to MASTER Dice Rolling Probability Problems

https://www.youtube.com/watch?v=5PVZYQXet18

sujoykrdatta · Answer

The easiest way of doing this is: 
1 - p(all faces are different)
= 1 - (6!)/6^6 ------> Since there are 6 faces, total cases for 6 rolls is 6^6
= 1 - 5!/6^5
 Ans D

If 6 fair 6-sided dice are rolled, what is the probability that at lea

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If 6 fair 6-sided dice are rolled, what is the probability that at lea

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