Which basic probability theory formula do I use?

Question

I am trying to do the last question shown in red. I would appreciate a walk through. I can understand that the answer is 1/6 by just using my eyes, but I want to use the formula.

I was given this formula but I would never get it to work:

$\Pr[A \cap B] = \Pr[A|B] \cdot \Pr[B]$

I found this but I don't see how it would give me the correct answer:

$\Pr[A \cap B] = \Pr[A] \cdot \Pr[B]$

If someone walks through step by step I would be eternally grateful. It is possible that all of these formulas are the incorrect ones to use...

Answer 1

Our sample space is the six possible outcomes $\{1, 2, 3, 4, 5, 6\}$ that occur when a six-sided die is thrown.

The event $X = \text{even}$ occurs when the throw results in an even number, which is $\{2, 4, 6\}$.

The event that $Y = c$ occurs when the throw results in a number divisible by $3$, which is $\{3, 6\}$.

Since each possible outcome of a fair die is equally likely, $$\Pr(X = \text{even}) = \frac{|\{2, 4, 6\}|}{|\{1, 2, 3, 4, 5, 6\}|} = \frac{3}{6} = \frac{1}{2}$$ and $$\Pr(Y = c) = \frac{|\{3, 6\}|}{|\{1, 2, 3, 4, 5, 6\}|} = \frac{2}{6} = \frac{1}{3}$$ The event that $X = \text{even}$ and $Y = c$ occurs when a $6$ is obtained. Hence, $$\Pr(X = \text{even} \wedge Y = c) = \frac{|\{6\}|}{|\{1, 2, 3, 4, 5, 6\}|} = \frac{1}{6}$$ The conditional probability that $Y = c$ given that $X = \text{even}$ is $$\Pr(Y = c \mid X = \text{even}) = \frac{\Pr(X = \text{even} \wedge Y = c)}{\Pr(X = \text{even})} = \frac{|\{6\}|}{|\{2, 4, 6\}|} = \frac{1}{3}$$ where we take the event that $X = \text{even}$ as our sample space.

Therefore, we have $$\Pr(X = \text{even} \wedge Y = c) = \Pr(Y = c \mid X = \text{even})\Pr(X = \text{even}) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$

The rule that $\Pr(A \cap B) = \Pr(A \mid B)\Pr(B)$ always applies. The rule $\Pr(A \cap B) = \Pr(A)\Pr(B)$ only applies when events $A$ and $B$ are independent, meaning that the occurrence of one event does not affect the probability the other event will occur. In this case, you can verify that the events $\Pr(X = \text{even})$ and $P(Y = c)$ are independent, but that is not always the case.

Answer 2

I think you are just getting confused by the logical operator. The question is asking for the probability that $X(\omega)=even$, and $Y(\omega)=c$ are true. $X(\omega)=even$ is true for $\omega\in \{2,4,6\}$, and $Y(\omega)=c$ is true when $\omega\in \{3,6\}$. Replace the logical statements with their set counterparts and you replace the logical operator with the set operator. $$Pr(\omega \in \{2,4,6\} \cap \omega\in \{3,6\})=Pr(\omega=\{6\})=\frac{1}{6} $$

The probability of the intersection can be done directly, so the first formula you linked isn't necessary, and the second is only true if $A$ and $B$ are independent.