Question

A peculiar die has the following properties: on any roll the probability of rolling either a 2, a 3, or a 5 is 1/2, just as it is with an ordinary die. Moreover, the probability of rolling either a 5, a 6, or a 4 is again 1/2. However, the probability of rolling a 5 is 3/16, not 1/6 as one would expect of an ordinary fair die.

From what you know about this peculiar die, compute
A) the probability of rolling either a 2, a 3, or a 1;
B) the probability of rolling anything but a 1.

Answer 1

A. P( 2∪3∪1 )=1/2

B. P( 2∪3∪4∪5∪6 )= 13/16

Explanation:

A. At first we need to get all the information that we can from the question, so we focus on the probabilities given P(2∪3∪5)=1/2 and P(4∪5∪6)=1/2 and we need to use P(5)=3/16.

Notice that when you have a dice, it´s impossible to get two values at the same try. Because of this and the property of probability for events:

P(2∪3∪5)=P(2) + P(3) + P(5) - P(2∩3)(=0) - P(3∩5)(=0) - P(2∩5)(=0) + P(2∩3∩5)(=0)

P(2∪3∪5)=P(2) + P(3) + P(5)

1/2=P(2) + P(3) + 3/16

P(2)+P(3)=5/16

If we do the same with P(4∪5∪6), we get:

P(4)+P(6)=5/16

Now, with this new information we can find P(1):

P(1)= 1 - P(2∪3∪4∪5∪6)

P(1)= 1 - (P(2) + P(3) + P(4) + P(5) + P(6)) (The probability of two of them happening at the same time is zero)

P(1)= 1 - (5/16 + 5/16 + 3/16)

P(1)= 1 - 13/16

P(1)= 3/16

And with it, we can find P(2∪3∪1):

P(2∪3∪1)=P(2)+P(3)+P(1)

P(2∪3∪1)=5/16+3/16

P(2∪3∪1)=1/2

B. Notice if you want to roll anything but a 1, then the probability you are looking for is "P(2∪3∪4∪5∪6)" but we found that in part A

P(2∪3∪4∪5∪6)=13/16