MHB Calculating Probability for Random Variables with Two Dice

[mathmari]

Hey!

Two usual six-sided dice are thrown. We are given the probability space $\Omega:=\{(1,1),(1,2),..., (1,6), (2,1),..., (6,1)\}$ and the probability function $ P:Pot (\Omega)\rightarrow [0,1], \ P (\{(a,b)\})=\frac{1}{36}$.
Calculate for the following random variable $X_i:\Omega->R$ the probability $P(X_i=7)$.

a)$X_1 ((a,b))=a+b$

b)$X_2 ((a,b))=2a+1$

c) $X_3 ((a,b))=6+a-b$

Could you give me a hint what I am supposed to do? (Wondering)

 

[I like Serena]

Hey!

Two usual six-sided dice are thrown. We are given the probability space $\Omega:=\{(1,1),(1,2),..., (1,6), (2,1),..., (6,1)\}$ and the probability function $ P:Pot (\Omega)\rightarrow [0,1], \ P (\{(a,b)\})=\frac{1}{36}$.
Calculate for the following random variable $X_i:\Omega->R$ the probability $P(X_i=7)$.

a)$X_1 ((a,b))=a+b$

b)$X_2 ((a,b))=2a+1$

c) $X_3 ((a,b))=6+a-b$

Could you give me a hint what I am supposed to do? (Wondering)

Hey mathmari! (Smile)

For the first sub question we're supposed to calculate (since the probability for each individual outcome is the same):
$$P(X_1=7) = P(a+b=7) = \frac{\text{#outcomes such that a+b=7}}{\text{#outcomes in total}}$$
where $a$ is the value of the first die, and $b$ is the value of the second die. (Thinking)

 

[mathmari]

Hey mathmari! (Smile)

For the first sub question we're supposed to calculate (since the probability for each individual outcome is the same):
$$P(X_1=7) = P(a+b=7) = \frac{\text{#outcomes such that a+b=7}}{\text{#outcomes in total}}$$
where $a$ is the value of the first die, and $b$ is the value of the second die. (Thinking)

So, we get that $$P(X_1=7) = P(a+b=7) = \frac{(1,6), (6,1)}{\#\Omega}=\frac{2}{36}=\frac{1}{18}$$ or not? (Wondering)

 

[I like Serena]

Don't we have more elements in the numerator than just $(1,6), (6,1)$? (Wondering)

Oh, and shouldn't the last element in $\Omega$ in the problem statement be (6,6) instead of (6,1)?

 

[mathmari]

Don't we have more elements in the numerator than just $(1,6), (6,1)$? (Wondering)

So, we have at the numerator the elements $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$.

Therefore we get $$P(X_1=7) = P(a+b=7) = \frac{\#\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}}{\#\Omega}=\frac{6}{36}=\frac{1}{6}$$ right? (Wondering)

Oh, and shouldn't the last element in $\Omega$ in the problem statement be (6,6) instead of (6,1)?

Oh yes (Blush)

 

[I like Serena]

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Thank you! (Sun)