Respuesta :
Answer:
a) P(rolling a number less than 4 on Die 1 and rolling a 5 on Die 2) = [tex]\frac{1}{9}[/tex]
b) P( the sum of both dice adding up to 4) = [tex]\frac{5}{54}[/tex]
Step-by-step 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 [tex]\frac{2}{3}[/tex] and the probability of rolling an even number is [tex]\frac{1}{3}[/tex] , 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) = [tex]\frac{1}{6}[/tex]
Also, Given Dice 2 is unfair
⇒P₂(1) = P₂(3) = P₂(5) = [tex]\frac{2}{3}.\frac{1}{3} = \frac{2}{9}[/tex]
P₂(2) = P₂(4) = P₂(6) = [tex]\frac{1}{3}.\frac{1}{3} = \frac{1}{9}[/tex]
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) ]
= [tex]\frac{2}{9}[/tex] [ [tex]\frac{1}{6}[/tex] + [tex]\frac{1}{6}[/tex] + [tex]\frac{1}{6}[/tex] ] = [tex]\frac{2}{9}[ \frac{3}{6}]\\[/tex] = [tex]\frac{1}{9}[/tex]
⇒P(rolling a number less than 4 on Die 1 and rolling a 5 on Die 2) = [tex]\frac{1}{9}[/tex]
b)
P( the sum of both dice adding up to 4) = P₁(1)P₂(3) + P₁(2)P₂(2) + P₁(3)P₂(1)
= [tex]\frac{1}{6}[/tex] × [tex]\frac{2}{9}[/tex] + [tex]\frac{1}{6}[/tex] × [tex]\frac{1}{9}[/tex] + [tex]\frac{1}{6}[/tex] ×[tex]\frac{2}{9}[/tex] = [tex]\frac{5}{54}[/tex]
⇒P( the sum of both dice adding up to 4) = [tex]\frac{5}{54}[/tex]
