Probability dice roll Question

Lyn05
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Hello! I need help on yet another probability question..

Question:
A red dice has the number 1 on one face, the number 2 on two faces and the number 3 on three faces. Two green dice each has the number 6 on one face and the number 5 on five faces. The three dice are thrown together.

Calculate the probability of obtaining 2 on the red dice, 5 and 6 on the two green dice.


I know that...

P(2 on red dice)= 2/6

P(5 and 6 on the two green dice)
= 2(5/6 X 1/6)
= 5/18

But here's where my confusion sprouts from. The different possible outcomes are:
2 on red, 5 on G-1 and 6 on G-2, or
2 on red, 5 on G-2 and 6 on G-1.

where G-1 and G-2 are the green dices.

I'm not sure if

(A)I should count the "2 on red" once, since the events for the red dice are independent of the other dices,i.e.

P(obtaining 2 on red,5 and 6 on green dice)
= 2/6 X 5/18
= 5/54


Or (B)Add the probabilities of the possible outcomes together, which will get me...

P(obtaining 2 on red,5 and 6 on green dice)
= 2(2/6 X 5/18)
= 5/27

Sorry if this post is confusing..
 
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The probability of a 2 on the red die (not "dice"- that is plural) is 2/6= 1/3 and the probability of 6 on a red dies is 1/6, probability of 5 on the other green die is 5/6 just as you say. So the probability of "2, 5, 6", in that order, is (1/3)(5/6)(1/6)= 5/108. The probability of "2, 6, 5", in that order, is (1/3)(1/6)(5/6)= 5/108 also. The probability of either order for the two green dice is 5/108+ 5/108= 2(5/108)= 5/54.

You had already taken the two different orders into account when you multiplied (1/6)(5/6) by 2. You do NOT have to take different positions of the red die (say, GGR or RGG or GRG) into account because that die is distinguishable from the other two green dice.
 
Oh..I see. Thank you very much! And sorry for the language error..
 
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