Question

I have a probability question:
4 dice are rolled, what is the probability exactly 3 of the dice end up being different numbers?

Answer 1

6x5x4/6^3

Answer 2

there are 4 dice not 3

Answer 3

\[(6\times5\times4\times6)\div(6\times6\times6\times6) \]
Thus there is a 5/9 probability.

Answer 4

exactly 3 dice r diffenrt so the 4th one has only 3 choices thus
p= (6x5x4x3)/6^4

Answer 5

the same problem was asked by @ffm and we all concluded answer was 6x5x4/6^3

that is probability that it has exactly 3 different numbers (6x5x4x(6-3))/4

and probability that is has exactly 4 different numbers 6x5x4x3/6^4

add those two you will get 6x5x4/6^3

Answer 6

I disagree experiment, you arent counting the 3 possibilities the last die could be, since it has to be one of the other three, i agree with advait, which is what i got originally

Answer 7

care to explain your side experiment?

Answer 8

sorry ... I misread the question. I thought it was at least three dices.

Answer 9

thats what i thought you were thinking

Answer 10

ill give you a chance to redeem yourself with another probability question in a sec, i'll close this now