# Understanding probability of independent events

I am trying to learn probability, and having a hard time to be honest. What I try to do is to break a problem in terms of event space. For eg, event space for rolling a die is {1,2,3,4,5,6}. For rolling two dice it would be 36 elements with pairs from 1,1 to 6,6. Then I try to calculate probability of a given event using simple basic formula of occurrences/total events.

On topic of probability of independent events, suppose I roll 2 dice and want to calculate probability of 2 on first die and 5 on second die. Book I am reading uses "And" rule for independent events:

P(2 and 5) = P(2) * P(5)


I am not sure of this approach. When we say event "2 in first roll, 5 in second roll", we basically are talking of an event "2 rolls of die". And so our event space consists of 36 elements, each representing a pair as discussed above. But in above formula, we are multiplying probabilities of events which come from a different event space(containing 6 elements). So basically we are trying to solve problem belonging to some event space using probabilities from different event space.

Am I wrong in my thinking here?

• $P(2\text{ and }5)=P(2)\times P(5)$ must be looked at as an abbreviation of: $$P(\{(i,j)\in\{1,2,3,4,5,6\}^2\mid i=2,j=5\})=$$$$P(\{(i,j)\in\{1,2,3,4,5,6\}^2\mid i=2\})\times P(\{(i,j)\in\{1,2,3,4,5,6\}^2\mid j=5\})$$ Commented Dec 28, 2020 at 12:09

Honestly the simple way is to draw the sample space, that is the following

and immediately you have a clear sight of the elements you need to calculate your probability.

Example 1: sum of the two dice $$\leq 7$$ and first die =1

Result: [$$\frac{6}{36}$$]

Example 2: first die 2 and second die 5, as you requested: result $$\frac{1}{36}$$

Example 3: sum of the two dice $$\leq 7$$ and at least one "1": Result $$\frac{11}{36}$$

and so on...any question you have to answer you have only to "visualize" and count the favourable events in the above table and divide it by $$36$$

PS: the plural of "die" is "dice", I amended you post

Essential is that probability spaces are chosen. This in order to model real-life situations in a natural and convenient way.

If we want to apply probability theory on the throw of a single die then it is quite natural and convenient to go for sample space $$\Omega=\{1,2,3,4,5,6\}$$. In that case the real event that the die shows a $$5$$ corresponds with the theoretical event $$\{5\}\subseteq\Omega$$.

But do not think we are stick to that choice.

If we are interested in a model for the rolling of two dice then it is (again) natural and convenient to go for sample space $$\Omega=\{(i,j)\mid i,j\in\{1,2,3,4,5,6\}\}$$ and this with $$P(\{(i,j)\})=\frac1{36}$$ for every singleton event.

In that case the real event that the first die shows a $$5$$ does corresponds with the theoretical event $$\{(5,j)\mid j\in\{1,2,3,4,5,6\}\}=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\subseteq\Omega$$.

Realize that we could have chosen this space also to manage the first situation.

Only there is no reason to do that, and it is handsome to keep the sample space as simple as possible.

I don't quite understand what you mean but I'll try to explain what's going on here.

In the first roll, the probability(2) = 1/6 (assuming a fair die). In the second roll, the probability of getting any number stays the same 1/6. The two rolls are independent - they don't affect each other's outcome.
So we get this: P(R1 = 2, R2 = Y) = 1/6 Y -> {1,2,3,4,5,6}
Thus, we know that if the second roll is taken in fair circumstances, you are going to have to square the previous probability (1/6)², knowing that the sum of all the possible outcomes in the second roll & P(R1 = 2) is equal to P(R1 = 2)

p.s If someone is willing to format my formulas using latex you're welcome. I hope I answered your question.

In some sense, your understanding is correct -- and your confusion is understandable. Let me explain.

If we let $$X$$ denote the value of the first die, and we let $$Y$$denote the value of the second die, then we may study $$(X,Y)$$, which exactly takes values in $$\{(1,1), (1,2),\ldots,(6,5),(6,6)\}$$.

Independence tells us that $$\mathbb{P}(X = x, Y = y) = \mathbb{P}(X = x)\mathbb{P}(Y=y).$$ Indeed the right hand side involves events from different event spaces, namely the marginal event spaces of $$X$$ and $$Y$$. In this case, both are $$\{1,2,3,4,5,6\}$$.

This shows us that if $$X$$ and $$Y$$ are independent, we only need to know how they act on their individual event space, i.e. we need only to assign a probability to each event $$(X=x)$$ and $$(Y=y)$$.

Of course, we may relate an event such as $$(X=x)$$ to the simultaneous event space. Clearly, $$(X=x) = (X=x, Y \in \{1,2,3,4,5,6\})$$. In particular, \begin{align*} \mathbb{P}(X=x) = \sum_{y=1}^6 \mathbb{P}(X=x, Y = y). \end{align*} (This holds even if $$X$$ and $$Y$$ are not independent.)

I hope this somewhat answers your questions.

I think your reasoning is correct, you just need to push it a little further.

• If you roll a die the sample space is $$\Omega = \{1,2,3,4,5,6\}$$
• If you roll two dice the sample space is $$\Omega = \{(i,j)\ |\ i,j\in \{1,2,3,4,5,6\}\}$$

What follows is to propose that there is a relationship between both sample spaces. The second is an extension of the first. There is a rule for the creation of this new space at the same time that there is a rule for assigning the probability of each new element.

I think it is possible to interpret that what we do when multiplying probabilities of events from a different sample space is to apply this rule. In this case, since these are independent events, we use the multiplication rule.

You can read a better explanation in this post by Terence Tao.

PD: Please correct me if my reasoning is wrong!