M41-72

Question

What is the probability of rolling four fair six-sided dice and getting exactly two 6s and two 1s?

A.  \frac{1}{6^3} 
B.  \frac{1}{6^2*3^2} 
C.  \frac{1}{2*6^3} 
D.  \frac{1}{3*6^3} 
E.  \frac{1}{6^4}

Bunuel · Answer

Official Solution:

What is the probability of rolling four fair six-sided dice and getting exactly two 6s and two 1s?

A.  \frac{1}{6^3} 
B.  \frac{1}{6^2*3^2} 
C.  \frac{1}{2*6^3} 
D.  \frac{1}{3*6^3} 
E.  \frac{1}{6^4}

The probability of obtaining two 6s and two 1s is  \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{4!}{2! * 2!}=\frac{1}{6^3} . We multiply by   \frac{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   \frac{4!}{2! * 2!}=6  cases has a probability of  \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} . Since there are 6 such arrangements, the total probability is:
 
 \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{1}{6} * \frac{4!}{2! * 2!}=\frac{1}{6^3}

Answer: A

Slimeto21 · Answer

I did not quite understand the solution. Wondering how you use 4C2 to account for the all possible arrangements?

Bunuel · Answer

4C2 counts the ways to place two 6s among four dice, with the remaining spots automatically assigned to 1s. Here you can find more solutions: https://gmatclub.com/forum/what-is-the- ... 14317.html

JuDC · Answer

I like the solution - it’s helpful.

SM15 · Answer

Why do we need to arrange the numbers, we are only asked to calculate the probability right. I marked 1/6^4

Bunuel · Answer

The four dice are  different  dice, Die 1, Die 2, Die 3, and Die 4.

So 1/6^4 gives only one specific outcome, for example:

Die 1 = 6, Die 2 = 6, Die 3 = 1, Die 4 = 1

But the question allows other distribution with exactly two 6s and two 1s. That is why we must count all possible arrangements. The solution above lists all 6 arrangements, so the probability is 6 * 1/6^4 = 1/6^3.

M41-72

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