A fair six-sided die is rolled 6 times. What is the probability of get

Question

A fair six-sided die is rolled 6 times. What is the probability of getting 1, 2, 3, 4, 5, and 6 in no particular order ?

A. 1/216
B. 5/216
C. 5/324
D. 1/7,776
E. 1/46,656

gmatophobia · Answer

Option 1

The first roll can be any number.

The second can be any but the first so only has a 5/6 chance of success.

The third roll has only a 4/6 chance and so on until the 6th roll which has only a 1/6 chance.

So, the overall probability is 5/6* 4/6 *3/6 * 2/6*1/6 = 5 / 324

Option 2

The probability of a particular number at any given roll is  \frac{1}{6}

For example, the probability of getting a 1 in the first roll =  \frac{1}{6} , the probability of getting a 2 in the second roll =  \frac{1}{6} , the probability of getting a 3 in the third roll =  \frac{1}{6}  and so on

As there are 6 rolls possible, the net probability = (\frac{1}{6})^6

However, this method fixes the outcome, i.e we are only considering the probability of getting a 1 in the first roll, we are only considering the probability of getting a 2 in the second roll and so on.

As the order is not mentioned, we can have the output in any order. Hence, the actual probability is various arrangements of the above

Net probability =  (\frac{1}{6})^6 * 6! = \frac{5}{324}

IMO - C

ankushanshuman · Answer

Total possible outcome is   6^6  . (1st roll has 6 possibilities,2nd has 6 possibilities and so on.. for 6 rolls)

Number of times 1,2,3,4,5,6 can appear is  6!  ( arranging the 6 different numbers is 6! )

So, probability will be 6!/ 6^6  =  5/324

Regor60 · Answer

The probability of getting any particular set of numbers in a particular order is the same, (1/6)^6.

Given that no particular order matters, any one of the permutations of the set satisfies the question

6!(1/6)^6 = 5!/6^5= 120/7776

= 5/324

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BrushMyQuant · Answer

The probability of getting 1, 2, 3, 4, 5, or 6 in each roll is =   \frac{1}{6}  each

Now, we have to get numbers from 1 to 6 in any order

We can get the first number out of the 6 in 6 ways
We can get the next number out of the remaining 5 in 5 ways
 We can get the next number out of the remaining 4 in 4 ways 
 We can get the next number out of the remaining 3 in 3 ways 
 We can get the next number out of the remaining 2 in 2 ways 
 We can get the next number out of the remaining 1 in 1 ways

=>  We can get the 6 numbers in any order  in 6 * 5 * 4 * 3 * 2 * 1 ways

=>  Probability of getting the numbers from 1 to 6 in any order  = 6 * 5 * 4 * 3 * 2 * 1 * Probability of getting each number
= 6 * 5 * 4 * 3 * 2 * 1 *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  *  \frac{1}{6}  =  \frac{6 * 5 * 4 * 3 * 2 * 1}{6^6}  =  \frac{5}{324}

So,  Answer will be C 
Hope it helps!

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A fair six-sided die is rolled 6 times. What is the probability of get

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