Question

1. Dicey Odds [2 Points]You roll two six-sided dice. Die 1 is fair. Die 2 is unfair such that the probability of rolling anodd number is23and the probability of rolling an even number is13, though the probabilityrolling of each odd number is the same, and the probability of rolling each even number isthe same. What is the probability of:a) rolling a number less than 4 on Die 1 and rolling a 5 on Die 2.b) the sum of both dice adding up to 4.

Answer 1

a) P(rolling a number less than 4 on Die 1 and rolling a 5 on Die 2) =
`(1)/(9)`

b) P( the sum of both dice adding up to 4) =
`(5)/(54)`

Explanation:

Given - You roll two six-sided dice. Die 1 is fair. Die 2 is unfair such that the probability of rolling an odd number is
`(2)/(3)` and the probability of rolling an even number is
`(1)/(3)` , though the probability rolling of each odd number is the same, and the probability of rolling each even number is the same.

To find - What is the probability of:

a) rolling a number less than 4 on Die 1 and rolling a 5 on Die 2.

b) the sum of both dice adding up to 4.

Proof -

The sample space , S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

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

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

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

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

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

Now,

Given that , Dice 1 is fair

⇒P₁(1) = P₁(2) = P₁(3) = P₁(4) = P₁(5) = P₁(6) =
`(1)/(6)`

Also, Given Dice 2 is unfair

⇒P₂(1) = P₂(3) = P₂(5) =
`(2)/(3).(1)/(3) = (2)/(9)`

P₂(2) = P₂(4) = P₂(6) =
`(1)/(3).(1)/(3) = (1)/(9)`

Now,

a)

P(rolling a number less than 4 on Die 1 and rolling a 5 on Die 2) = P₁(1)P₂(5) + P₁(2)P₂(5) + P₁(3)P₂(5)

= P₂(5) [ P₁(1) + P₁(2) + P₁(3) ]

=
`(2)/(9)` [
`(1)/(6)` +
`(1)/(6)` +
`(1)/(6)` ] =
`(2)/(9)[ (3)/(6)]\\` =
`(1)/(9)`

⇒P(rolling a number less than 4 on Die 1 and rolling a 5 on Die 2) =
`(1)/(9)`

b)

P( the sum of both dice adding up to 4) = P₁(1)P₂(3) + P₁(2)P₂(2) + P₁(3)P₂(1)

=
`(1)/(6)` ×
`(2)/(9)` +
`(1)/(6)` ×
`(1)/(9)` +
`(1)/(6)` ×
`(2)/(9)` =
`(5)/(54)`

⇒P( the sum of both dice adding up to 4) =
`(5)/(54)`