Three fair dice are rolled, one red, one green and one blue. What is the probability that the upturned faces of the three dice are all of different numbers?

Answer:   [tex]\dfrac{5}{9}[/tex]

Step-by-step explanation:

When we throw a die , Total outcomes =6

When we throw 3 dice , Total outcomes = 6 x 6 x 6 = 216 [by fundamental counting principle]

Given : Three fair dice are rolled, one red, one green and one blue.

Favorable outcomes : When the upturned faces of the three dice are all of different numbers i.e. no repetition of numbers allowed

By Permutations , the number of favorable outcomes = [tex]^6P_3=\dfrac{6!}{(6-3)!}=\dfrac{6!}{3!}=6\times5\times4=120[/tex]

The probability that the upturned faces of the three dice are all of different numbers = [tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]

[tex]=\dfrac{120}{216}=\dfrac{5}{9}[/tex]

The probability that the upturned faces of the three dice are all of different numbers  is [tex]\dfrac{5}{9}[/tex] .

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Fiona and her friends are playing a game by guessing where a coin will land when it is randomly dropped inside the square shown below. fiona guesses that the coin is likely to land in the blue area. which explains whether or not fiona is correct and why?

Please solve and show work