If three dice are rolled under the condition that no two dice show the same face, then what is the probability that one of the faces is having the number 6?

Concept:

Probability of an event happening = (Number of ways it can happen) / (Total number of outcomes)

If three dice trhown, number of sample space = 63 = 216

Calculation:

Three dice are rolled, number of sample space = 216

Two dice show the same face =

{(1, 1, 1), (1, 1, 2), (1,1, 3), (1, 1, 4), (1, 1, 5), (1, 1, 6), (1, 2, 1),(1, 3, 1), (1, 4, 1), (1, 5, 1), (1, 6, 1), (2, 1, 1), (3, 1, 1), (4, 1, 1), (5, 1, 1), (6, 1, 1).......

(2, 2, 2).....,

(3, 3, 3)......,

(4, 4, 4)......,

(5, 5, 5)......,

(6, 6, 6).......}

n = 16 × 6 = 96

Number of no two dice show the same face, n(S) = 216 - 96 = 120

Faces having no number 6,

{(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 2), (1, 3, 4) , (1, 3, 5), (1, 4, 2), (1, 4, 3), (1, 4, 5), (1, 5, 2), (1, 5, 3), (1, 5, 4)

(2, 1, 3), .......

..........

.........

..........(5, 4, 3)}

n = 12 × 5 = 60

So the number of one of the faces is having the number 6 = 120 - 60 = 60

∴ The probability that one of the faces is having the number 6 = 60/120 = 1/2

Hence, option (3) is correct.

A die is thrown: Sample space S = {1, 2, 3, 4, 5, 6}

Now, possible outcomes of three dice the condition that no two dice show the same face = -  -  - = 6 × 5 × 4 = 120

Now, the number of one of the faces is having the number 6 = 1 × 5 × 4 × 3! = 60

∴ The probability that one of the faces is having the number 6 = 60/120 = 1/2