math qns( probability)?

Question

three unbiased dice are thrown find the probability that they all show different numbers. do show the working,thank you

Answer 1

The first die can be any of 6 numbers, the second any of 5, the third any of 4, so the total number of throws with all dice different is 6x5x4=120. Total number of possible throws is 6x6x6=216. So your answer is 120/216 = 5/9.

Answer 2

The total number of outcomes is: 6 x 6 x 6 =216
To get different combinations is: 6 x 5 x 4 =120

So the answer is 120/216 = .5555.

Answer 3

let the event be A
n(s)=6^3
to get the different nos in the dice
first dice gets one of the 6 nos i.e, in 6 ways
second in 5 ways i.e, except the no used in first one
third one in 4 ways
these three can be rearranged in themselves in 3!
totally n(A)=6*5*4*3!=6!
hence n(E)=6!/6^3

Answer 4

P(any no)*P( any no. but the first)*P(any no. but first 2)
= 6/6*5/6*4/6
= 120/216

Answer 5

we can do this in more better way that prob.
p=1-(show same no.)
prob.of same nos.=6/216=.028
so p=1-.028=.972
ans=.972