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Probability: If 4 fair dice are thrown simultaneously

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Probability: If 4 fair dice are thrown simultaneously

by ChessWriter » Sun Mar 11, 2012 11:37 am
If 4 fair dice are thrown simultaneously, what is the probability of getting atleast one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18

This is a question from the Veritas Prep Combinatorics and Probability book. I got this one wrong when I solved it. But, The book contains only the answer but not the detailed explanation for it.

I will post the Official Answer after some time
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by ChessWriter » Sun Mar 11, 2012 11:49 am
Immediately after I hit the "post" button after typing the above question, the method to solve this came to me. So, I am sharing my method and the official answer below.

One way to solve it is by considering the opposite - That on throwing the dice 4 times, you never get a pair.

In this case, the probability of what happens on the first throw does not matter. But the probability of getting a different number on the second, third and fourth throw are 5/6, 4/6 and 3/6 respectively.

So on multiplying we get 6/6*5/6*4/6*3/6 = 5/18

On substracting it from 1, we get 1-(5/18) = [spoiler]13/18[/spoiler]

Therefore [spoiler]E. 13/18[/spoiler] is the correct answer

But, I wonder, Is there any other way to solve this question?
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by pemdas » Sun Mar 11, 2012 2:46 pm
ChessWriter wrote:If 4 fair dice are thrown simultaneously, what is the probability of getting atleast one pair?

A. 1/6
B. 5/18
C. 1/2
D. 2/3
E. 13/18

This is a question from the Veritas Prep Combinatorics and Probability book. I got this one wrong when I solved it. But, The book contains only the answer but not the detailed explanation for it.

I will post the Official Answer after some time
total possible outcomes for dice 6*6*6*6 or 6^4
favorable outcomes or possible pairs =(4C2)*6=6^2=36

Thus, P(at least one pair)=(36+35+34+...1)/6^4, find mean of consecutive no-s 1...36, and calculate the sum -> P(at least one pair)= 666/6^4 = 6*111/6^4=111/216
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