Statistics probability

Question

A peculiar die has the following properties: on any roll the probability of rolling either a 1, a 5, or a 4 is 1/2, just as it is with an ordinary die. Moreover, the probability of rolling either a 4, a 2, or a 6 is again 1/2. However, the probability of rolling a 4 is 1/16, not 1/6

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

Answer 1

1. Using Complement Rule, P (3,2, or 6) = 1 - P(1,4, or 5) = 1-1/2=1/2
2. Let A = event of rolling either a 1, a 5, or a 4

B = rolling either a 4, a 2, or a 6

Note that A ∪ B = rolling either a 1, a 5, 4, a 2, or a 6

P(A ∪ B) = P(A) + P(B) - P ( A ∩ B)

= 1/2 + 1/2 - 1/16

= 15/16

P (rolling a 3) = 1 - P(A ∪ B) = 1- 15/16 = 1/16

P(rolling neither a 2 nor a 6) = P( rolling 1,3,4 or 5) = P(A) + P(rolling a 3) = 1/2 +1/16 = 9/16