GATE CSE 2024 | Set 2 | Question: 8

Question

​​​​​​When six unbiased dice are rolled simultaneously, the probability of getting all distinct numbers $(i.e., 1, 2, 3, 4, 5, \text{and }   6)$ is

• A. $\frac{1}{324}$
• B. $\frac{5}{324}$
• C. $\frac{7}{324}$
• D. $\frac{11}{324}$

Answer 1

We have 6 unbiased dices that are rolled simultaneously.

As, dices are unbiased. Then,

P(x) = $1/6$ where x $\in$ {1, 2, 3, 4, 5, 6}.

When a dice is rolled then any number out of 6 can show up. So, with 1 dice we have 6 possible outcomes.

And, when we have 6 dices and each dices have 6 choices then total possible outcomes = $6^{6}$.

Total cases : $6^{6}$.

 

Now, we want different numbers on all 6 dices. So, combination like these are our feasible cases :

(1, 2, 3, 4, 5, 6), (1, 3, 5, 2, 4, 6), (2, 4, 6, 5, 3, 1) , (6, 1, 3, 5. 4, 2), - - - - - - - - - - , (6, 5, 4, 3, 2, 1)

How many such feasible case?

It will be 6! = 720

Feasible cases : 6! = 720

 

So, Required Probability : Feasible cases / Total Cases

= 6! / $6^{6}$.

= 5 / 324.

Hence, Option B is correct Answer.

Answer 2

Answer: $6!/6^6 = 5/324.$

Comments

HOW??

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we are rolling 6 dice, total case for 6 dice = 6^6

fav outcome =  6!

prob = 6! / 6^6 = 5/324 (B)