What is the probability of obtaining two 6s and two 1s when four fair

Question

What is the probability of obtaining two 6s and two 1s when four fair six-sided dice are rolled?

A. 1/6^3

B. 1/(6^2*3^2)

C. 1/(2*6^3)

D. 1/(3*6^3)

E. 1/6^4

Bunuel · Accepted Answer

The dice are NOT identical.

The probability of obtaining two 6s and two 1s is (1/6) * (1/6) * (1/6) * (1/6) * (4!/(2! * 2!)). We multiply by 4!/(2! * 2!) to account for all possible arrangements of (6, 6, 1, 1) across the four dice:

Die 1  
 Die 2  
 Die 3  
 Die 4  
 
  6 
 6 
 1 
 1 
 
  6 
 1 
 6 
 1 
 
  6 
 1 
 1 
 6 
 
  1 
 6 
 6 
 1 
 
  1 
 6 
 1 
 6 
 
  1 
 1 
 6 
 6

Each of these  4!/(2! * 2!)  = 6 cases has a probability of (1/6) * (1/6) * (1/6) * (1/6). Since there are 6 such arrangements, the total probability is:

6 * (1/6)^4 = 6/1296 = 1/216.

gmatophobia · Answer

The probability of getting 6 on the first dice =  \frac{1}{6}  
 The probability of getting 6 on the second dice =  \frac{1}{6}  
 The probability of getting 1 on the third dice =  \frac{1}{6}  
 The probability of getting 1 on the fourth dice =  \frac{1}{6}

We have fixed the position of the numbers in the above case, however, the numbers can appear in various arrangements of 6 6 1 1

Probability =  \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6 }* \frac{4!}{2!*2!}

=  \frac{1}{6} * \frac{1}{6} * \frac{1}{6}  =  (\frac{1}{6})^3

Option A

Archit3110 · Answer

gmatophobia 
check highlighted part
(1/6)^4 * 4!/2! 2!
ways to arrange 6611

(1/6)^3
option A

gmatophobia · Answer

Added the missing 2! in the denominator.

I couldn't understand the logic of the highlighted portion (not sure if you meant that's the answer). IMO the answer should be A.

Adarsh_24 · Answer

Bunuel

Sorry One more final probability doubt.
Not urgent.

Should we remove repetition unless it’s mentioned they are identical. Because in many questions where dies added up to a certain sum, I saw dies were considered as different colours. So let’s say to get sum of 6, (2,4) and (4,2) were valid.

Here if dies are not identical,
Wouldn’t d1:d2:d3:d4 = 6:6:1:1 be different from 6:1:1:6 ?

In this question we considered the dice as identical. But, below one we considered dice as possibly different.
I got this answer wrong as I followed the assumption that dice may have been different.

https://gmatclub.com/forum/a-6-sided-cu ... s%205%2F36.

RicharddelPino · Answer

Possible arrangements of (6, 6, 1, 1), (6, 1, 6, 1), (6, 1, 1, 6), (1, 1, 6, 6), (1, 6, 1, 6), (1, 6, 6, 1).
Each number has 1/6 of probability, 1/6^4 is the probability of each arragment. 
As if there are 6 possible arragementes: 1/6^4 * 6 = 1/6^3

What is the probability of obtaining two 6s and two 1s when four fair

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What is the probability of obtaining two 6s and two 1s when four fair

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