A peculiar die?

Question

A peculiar die has the following properties: on any roll the probability of rolling either a 4, a 5, or a 3 is 12, just as it is with an ordinary die. Moreover, the probability of rolling either a 3, a 2, or a 1

 is again 12. However, the probability of rolling a 3is 14, not 16 as one would expect of an ordinary fair die.

From what you know about this peculiar die, compute the following.

1. The probability of rolling either a 6, a 2, or a 1
2. The probability of rolling neither a 2 nor a 1

Answer 1

Let X = result of rolling the die. We are told that

P(X ∈ {4, 5, 3}) = 1/2

P(X ∈ {3, 2, 1}) = 1/2

P(X = 3) = 1/4

Therefore P(X ∈ {2, 1}) = P(X ∈ {3, 2, 1}) - P(X = 3) = 1/4

Also,

P(X ∈ {1, 2, 3, 4, 5}) = P(X ∈ {4, 5, 3}) + P(X ∈ {3, 2, 1}) - P(X ∈ {4, 5, 3} intersect X ∈ {3, 2, 1})

= P(X ∈ {4, 5, 3}) + P(X ∈ {3, 2, 1}) - P{X = 3)

= 1/2 + 1/2 - 1/4 = 3/4

So P(X=6) = 1 - P(X ∈ {1, 2, 3, 4, 5}) = 1/4

Answers:

(a) P(X ∈ {1, 2, 6}) = P(X ∈ {1, 2}) + P(X = 6) = 1/4 + 1/4 = 1/2

(b) We already showed that P(X ∈ {1, 2}) = 1/4