# Probability in heterogeneous sample space

Consider tossing a fair coin once followed by rolling a fair die, however the die is rolled once only when we get a head in the preceding toss. Need to find out probability of getting a six or a tail. I could derive the answer rather mechanically using a tree diagram. But I'm looking for a more convincing solution perhaps using set notation and formula. Any help would be much appreciated.

Elaboration of my attempt and confusion:

Sample space $\Omega=\{\{H,1\}, \{H,2\}, \{H,3\}, \{H,4\}, \{H,5\}, \{H,6\}, \{T\}\}$ [note that all sample points are not equi-dimensional]

Events are defined for the same sample space as follows:
$A=$the event that we get a H in the toss, i.e. all 6 points starting with H
$B=$the event that we get a six in the roll of a die i.e. a single point ending with 6
$C=$the event that we get a T in the toss i.e. a single point with T
$ℙ$(getting a six or a tail) = $ℙ(B∪C) = ℙ(B) + ℙ(C) - ℙ(B∩C) = ℙ(B) + ℙ(C) – 0$

However, at this point, I’m not sure how to proceed further to calculate $ℙ(B)$?
One, thought could be replacing $B$ with $\{H,6\}$ i.e. $ℙ(B) = ℙ(\{H\}∩\{6\})$. However, here I’ve two confusions –
(1) If I use events like $\{H\}$ and $\{6\}$, then they are scoped within respective sub-experiments of tossing and rolling. Moreover, such decomposition is allowed only-when the final space $\Omega$ can be expressed as Cartesian product space $S_1\times S_2$ where $S_1=$ sample space of tossing and $S_2=$sample space for rolling
(2) How can I define another event say $B^1$ in $\Omega$, such that $B=\{A \cap B^1 \}$

• The probability to get a 6 and a tail... in how many rolls? At a specific roll? You're talking about multiple repeat of the basic experiment (at least 2 are necessary to get the event you're interested in), but you never elaborate on this. Commented Aug 5, 2015 at 16:44
• This needs clarification. I assume you are talking about a sequence of rounds and that your condition on H refers to successive terms in that sequence? So, in particular there is no condition at all on the first round, yes? Are you then asking for the probability that some round in the sequence results in $(6,T)$? As a function of the number of rounds?
– lulu
Commented Aug 5, 2015 at 16:50
• Or did you get the order wrong...the condition on H makes more sense if the problem started "Consider tossing a fair coin followed by rolling a fair die" where the rule then is that you only roll the die if the coin yields H. That's a much simpler problem.
– lulu
Commented Aug 5, 2015 at 16:59
• @Tryss - Edited the question to add more clarification. Anyway, it is a '6 or a tail', I mistakenly wrote '6 and tail'. We toss once and if we get a head, we roll a die once. Hope this clarifies the doubt. Commented Aug 5, 2015 at 17:01
• Half the time you get $(T,*)$. $\frac {1}{12}$ of the time you get $(H,6)$. There is no overlap between these events. So....
– lulu
Commented Aug 5, 2015 at 17:14

You could use conditionnal probability :

$$P(6\ \text{and}\ H) = P(6| H)P(H)$$

And

$P(6| H) = \frac{1}{6}$ as you're in the situation of a fair dice roll

$P(H) = \frac{1}{2}$ as you're in the situation of a fair coin toss

• Sorry if I've confused earlier, but am looking for $P( 6$ or $tail)$ Commented Aug 5, 2015 at 17:36
• @ Tryss – As we can see that $P$(6 or tail) $= P$(6) + $P$(tail) – $P$(6 and tail) $= P$(6) + $P$(tail) $– 0$. We could find the desired answer if we know $P(6)$. Can you provide $P(6)$ so that your above calculation of $P$(6 and $H)$ is not violated? Commented Aug 5, 2015 at 20:59
• @BuckCherry : You have that $P(6)$ is exactly $P(6 \text{ and } H)$ (they are the same event) Commented Aug 5, 2015 at 21:02
• @ Tryss : If$P(6) = P(6$ and $H)$, then $H$ has to be sample space $\Omega$ and thus $P(H)=P(\Omega)=1$ which contradicts your evaluation of $P(H)=1/2$. You see the logic is falling apart- or am I missing something!! Commented Aug 6, 2015 at 9:52
• @BuckCherry : The events A = "a 6" and B = "a Head and a 6" are the same events there is absolutely no difference between them : It's clear that $B\subset A$, but as you can't have a 6 without a head, you have $A \subset B$, hence the equality ... And no, if you have E and F = E, that doesn't means that $F= \Omega$, just that $E \subset F$. Commented Aug 7, 2015 at 18:25

Since there is no overlap between event T and event H followed by 6, it is simply $\frac12 + \frac12 \cdot \frac16$

I think most of your confusion is arising because you are not considering that there are sequential events. The first event is a coin toss. You proceed to the die toss only if the outcome of the 1st event is heads.

• @ true blue anil - The answer of yours is right and in fact I could get the same. The crux of my question is, how to arrive at this result mathematically. For example, your answer takes the form of $P(A)+P(B)P(C)$; and you could perhaps explain the definition of the events $A$, $B$, and $C$. In addition, how did you deduce the final answer to be of this form! Commented Aug 5, 2015 at 19:57
• A = tails, B = heads, C = 6:: Indicated Pr = P(A) + P(B).P(C|B) Commented Aug 5, 2015 at 20:11
• @ true blue anil - Why not, $P$(tail or 6) $= P$(tail $\cup$ 6)$= P(A \cup C)$ - based on your definitions of $A$, and $C$?? Commented Aug 5, 2015 at 20:39
• Make an equi-probable sample space $\Omega=\{\{H,1\}, \{H,2\}, \{H,3\}, \{H,4\}, \{H,5\}, \{H,6\}, \{T\}, \{T\}, \{T\},\{T\},\{T\},\{T\},\}$. Getting a 6 is the same as getting $H\cap6$. Commented Aug 6, 2015 at 5:12
• @ true blue anil - Why would any one make an equi-probable sample when it is not - seems like a desperate effort to match the answer!! Btw, the other anomaly remained unanswered - Why did you equate $P(tail$ or $6) = P(A∪(B \cap C))$ and not to $P(A∪C)$ ?? Commented Aug 6, 2015 at 6:19