MHB What is the Probability of a Tetrahedron Landing on a Specific Colored Side?

[mathmari]

Hey!

Let one of the four sides of a tetrahedron be red, one blue, one green and the fourth side painted with all three colours. We consider the following events :
A: = The tetrahedron falls on a side with red colour.
B: = The tetrahedron falls on a side with a blue colour.
C: = The tetrahedron falls on a side with a green colour

Give a suitable probability space and give the events A, B and C as subsets of the result set.
(a) Show that two of the events A, B and C are independent.
(b) Are A, B, C independent? Explain. Does it hold that $P(A)=P(B)=P(C)=\frac{2}{4}$ because there is one side for each colour the the 4th side contains also each colour?
The probability space is the set of colours, or not?

:unsure:

 

[I like Serena]

Does it hold that $P(A)=P(B)=P(C)=\frac{2}{4}$ because there is one side for each colour the the 4th side contains also each colour?
The probability space is the set of colours, or not?

Hey mathmari!

I believe the probabilities are indeed $\frac 24$ each. (Nod)

A probability space is supposed to be a triplet of a sample space, an event space, and a probability function.
You only mentioned the sample space. (Worried)

Either way, I think that the outcome space must contain each of the sides. We only have 3 colors but 4 sides... (Worried)
We could pick for instance $\Omega=\{red,green,blue,all\}$.

 

[mathmari]

I believe the probabilities are indeed $\frac 24$ each. (Nod)

A probability space is supposed to be a triplet of a sample space, an event space, and a probability function.
You only mentioned the sample space. (Worried)

Either way, I think that the outcome space must contain each of the sides. We only have 3 colors but 4 sides... (Worried)
We could pick for instance $\Omega=\{red,green,blue,all\}$.

So the probability space consists of the outcome space $\Omega=\{red,green,blue,all\}$ and each outcome has the same probability $\frac{2}{4}=\frac{1}{2}$, right? :unsure:

We have that \begin{align*}&A = \{red, \ all\}\subseteq \Omega \\ &B = \{blue, \ all\}\subseteq \Omega\\ &C = \{green, \ all\}\subseteq \Omega\end{align*}

(a) We have that $A\cap B = A\cap C=B\cap C=\{all\}$ so we get that \begin{align*}&P(A)=\frac{1}{2} \\ &P(B)=\frac{1}{2} \\ &P(C)=\frac{1}{2} \\ &P(A\cap B) = \frac{1}{4} \\ &P(A\cap C)= \frac{1}{4} \\ &P(B\cap C)= \frac{1}{4} \\ & \Rightarrow P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}=P(A\cap B) \\ & \Rightarrow P(A)\cdot P(C)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}=P(A\cap C) \\ & \Rightarrow P(B)\cdot P(C)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}=P(B\cap C)\end{align*} That means that each two of the events $A$, $B$ and $C$ are independent. (b) We have that $A\cap B \cap C=\{all\}$ so we get that \begin{align*}&P(A)=\frac{1}{2} \\ &P(B)=\frac{1}{2} \\ &P(C)=\frac{1}{2} \\ &P(A\cap B\cap C) = \frac{1}{4} \\ & \Rightarrow P(A)\cdot P(B)\cdot P(C)=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}\neq \frac{1}{4}=P(A\cap B\cap C) \end{align*} That means that the events $A$, $B$ and $C$ are not independent. Is everything correct and complete? :unsure:

 

[I like Serena]

All correct and complete. (Sun)