Double Dice Probability: Event A and B in Sample Space | Solved

[Mark53]

Homework Statement

In a probability experiment, a fair die is rolled twice.
• If the first roll is odd, the outcomes are recorded as they appear.
• If the first roll is even, the recorded outcome for the second die is doubled. For example, if the first die was 2 and the second 4, the recorded outcome would be (2,8).

Let A be the event that the first recorded outcome is even and B be the event that the second is even.

(a) Write down the sample space S for this experiment.
(b) Express A, B and A∩B as subsets of S.
(c) Find P(A) and P(B).
(d) Given that the second recorded outcome is even, what is the probability that the first roll was also even?

The Attempt at a Solution

[/B]
I am not sure if the the numbers that I have used below are correct any help is much appreciated

a)
s={(2,4),(2,8),(2,12),(4,4),(4,8),(4,12),(6,4),(6,8),(6,12)}

b)

A={2,4,6}
B={4,8,12}
A∩B = {4}

c)

P(A)=1/2
P(B)=1/2

d)

P=1/4

 

[Orodruin]

s={(2,4),(2,8),(2,12),(4,4),(4,8),(4,12),(6,4),(6,8),(6,12)}

This set is far from complete. What happened to all of the cases where either the first or second die gave 1, 3, or 5?

A={2,4,6}
B={4,8,12}
A∩B = {4}

These are not subsets of your S.

 

[Mark53]

This set is far from complete. What happened to all of the cases where either the first or second die gave 1, 3, or 5?

These are not subsets of your S.

S={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12),(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5)(3,6),(5,1),(5,2),(5,3)(5,4),(5,5),(5,6)}

A={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12)}

B={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12),(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)}

A∩B = {(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12)}

P(A)=1/2
P(B)=3/4

d)
P=1/2

Would this be correct now

 

[PeroK]

S={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12),(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5)(3,6),(5,1),(5,2),(5,3)(5,4),(5,5),(5,6)}

A={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12)}

B={(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12),(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)}

A∩B = {(2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(4,2)(4,4),(4,6),(4,8),(4,10),(4,12),(6,2),(6,4),(6,6),(6,8),(6,10),(6,12)}

P(A)=1/2
P(B)=3/4

d)
P=1/2

Would this be correct now

a) - c) are correct. How did you calculate d)?

 

[Mark53]

a) - c) are correct. How did you calculate d)?

I used A∩B how should I be calculating it?

 

[PeroK]

I used A∩B how should I be calculating it?

$P(A \cap B)$ is the probability that both are even. The probability that the first is even given that you know the second is even is something else. Can you see that?

 

[Mark53]

$P(A \cap B)$ is the probability that both are even. The probability that the first is even given that you know the second is even is something else. Can you see that?

Does that mean that it is 1/4 I am still unsure

 

[PeroK]

Does that mean that it is 1/4 I am still unsure

You're not going to learn just by guessing a number. Have you heard the term "conditional probability" or seen the term $P(A|B)$?

https://en.wikipedia.org/wiki/Conditional_probability

 

[Mark53]

You're not going to learn just by guessing a number. Have you heard the term "conditional probability" or seen the term $P(A|B)$?

https://en.wikipedia.org/wiki/Conditional_probability

I forgot about that so that means that

$P(A|B)$=(1/2)/(3/4) = 2/3

 

[PeroK]

I forgot about that so that means that

$P(A|B)$=(1/2)/(3/4) = 2/3

Yes.