Three six-sided fair dice are rolled. The six sides of each die are numbered 1,2, ...,6. Let A be the event that the first die shows an even number, let B be the event that the second die shows an even number, and let C be the event that the third die shows an even number.

Question

Mathematics

Jett Moody

18 March, 19:45

Three six-sided fair dice are rolled. The six sides of each die are numbered 1,2, ...,6. Let A be the event that the first die shows an even number, let B be the event that the second die shows an even number, and let C be the event that the third die shows an even number.

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Gentry · Answer 1

The answer is: 27/216 = 0.125

Step-by-step explanation:

Since the dices are rolled and are fair. We can assume randomness of observation.

A is an event of even number from 1st die

B is an event of even number from 2nd die

C is an even of even number from 3rd die.

=> The number of possible outcome or sample space is 6^3 = 216.

=> Since each die has 3 even number each, the possibility of even number showing is 3^3 = 27.

When the dice is rolled, for even outcomes we have the following:

(2,2,2), (2,4,2), (2,6,2), (4,2,2), (4,4,2), (4,6,2), (6,2,2), (6,4,2), (6,6,2), (6,2,2),

(6,4,2), (6,6,2), (2,2,4), (2,4,4), (2,6,4), (4,2,4), (4,4,4), (4,6,4), (6,2,4), (6,4,4), (6,6,4), (2,2,6), (2,4,6), (2,6,6), (4,2,6), (4,4,6), (4,6,6), (6,2,6), (6,4,6), (6,6,6).

If we observe the outcomes, it is 27 outcomes for the even outcomes. And all possible outcomes is 6^3 = 216.

Then, the probability of even outcomes from each die from the roll is:

P (even) = (3^3) / (6^3) = (27) / (216) = 1/8 = 0.125