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Suppose event $A$ is flipping a coin, $A = \{H, T\}$

Suppose Event $B$ is rolling a die, $B = \{1, 2, 3, 4, 5, 6\}$.

Next, suppose we flip the coin and roll the die. If we want to find the probability of getting, say, a head $H$ (denoted by $A_{H}$) and a $5$ (denoted by $B_{5}$), we write,

$Pr(A_{H} \cap B_{5})$.

What's troubling me is:

What does $A_{H} \cap B_{5}$ mean? To me at least, the intersection of the two events is the empty set $\{\}$.

I know the it used for a substitute for 'AND' in plain English, which makes sense, it just doesn't make sense from its traditional sense where the intersection means 'the set of shared elements'.

Retty
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  • You haven't said what the trial is. Presumably you flip the coin and roll the die, yes? If so then they are asking for the probability of the outcome $(H,5)$. – lulu Mar 11 '17 at 22:51
  • Yes, sorry. I made an edit to the question. – Retty Mar 11 '17 at 22:53
  • Right, but that answers your question, yes? The coin outcome is independent of the die so you just multiply the probabilities. Alternatively, there are $12$ possible outcomes and they are all equally likely so... – lulu Mar 11 '17 at 22:56
  • But I'm talking strictly of their intersection. Which, to me, appears to be the empty set. What am I missing here? – Retty Mar 11 '17 at 23:04

3 Answers3

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Let's pause time before we flip the coin and roll the die, and take the set $\Omega$ of 'possible continuations' (or 'possible worlds').

Then $A_H\subseteq\Omega$ is the set of those 'worlds' where the coin is flipped to the head.

Berci
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  • Do you mean the set of possible worlds is ${(1,H), (2,H), (3,H), (4,H), ... , (5, T) , (6, T)}$? – Retty Mar 11 '17 at 23:10
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You have written down the sample space for the coin flip $A= \{H,T\}$ and the die roll $B=\{1,2,3,4,5,6\}.$ When you consider a situation where both a coin is flipped and a die is tossed, the natural sample space to consider is the cartesian product $\Omega = A\times B$ where the outcomes consist of a pair $(a,b)\in \Omega$ of outcomes, one from the coin flip and one from the die roll. So if you flip heads and roll a $2$ then that is the outcome $\omega = (H,2).$

The event that you get a head (without regard to how the dice roll comes out) is just the the set $$A_H = \{(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)\} \subset \Omega$$ the event that you get a $5$ for the die roll can be written $$B_5 = \{(H,5),(T,5)\}.$$

The interection, which consists of all the outcomes where you both flip a head and roll a $5$ is then $$ A_H\cap B_5 = \{(H,5)\}.$$

Say you also spin a spinner that can come out red blue or green $C=\{R,B,G\}$ and you want to add it to the mix. Then you'd have $ \Omega = A\times B\times C$ and writing down everything would be more tedious, but you would end up with $$ A_H\cap B_5 = \{(H,5,R),(H,5,B),(H,5,G)\}$$ as the set of outcomes where you flip a head and roll a 5.

spaceisdarkgreen
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The key here is to recognize that a "plain-English" definition of the event can only be made precise once you specify a sample space. So, with respect to the sample space $A$, the "coin flip is heads" event, $A_H$, is $\{H\}$. But wait! This sample space is clearly insufficient for your purposes: $B_5$ cannot be expressed in terms of the elements of that sample space. Similarly, $A_H$ cannot reify w.r.t. $B$, but $B_5=\{5\}$ in the sample space $B$.

To ask the question you want to ask, you need to use the sample space $A\times B$. With respect to $A\times B$, the event $A_H=\{H\}\times B$, because those are the outcomes in $A\times B$ for which the coin flip was heads. Similarly, $B_5$ reifies to $A\times\{5\}$, and hence $$A_H\cap B_5=(\{H\}\times B)\cap(A\times\{5\})=\{(H,5)\}$$ as you expect.

Jacob Manaker
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