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I have two exercises where I have got troubles with define probability/measurable space $(\Omega,S)$.

($1$) We have two dices. One dice is fair, it means that each number from $1,2,3,4,5,6$ falls with probability $\frac{1}{6}$. The second dice is not fair. The probability of falling 6 is $\frac{1}{2} $ and the probability of falling number from $ 1,2,3,4,5 $ is $\frac{1}{10}$. Randomly we choose one dice and we roll it $4$-times. The number $6$ falls twice. What is the probability that chosen dice is not fair.

I can compute probablity but I have been trying to build $(\Omega,S)$. I have been trying somethink like that $$\Omega=\{(D_F; 1,1,1,1),(D_F; 2,1,1,1),(D_F; 1,2,1,1),\dots,(D_F; 6,6,6,5),(D_{F}; 6,6,6,6),(D_{notF}; 1,1,1,1),(D_{notF}; 2,1,1,1),(D_{notF}; 1,2,1,1),\dots,(D_{notF}; 6,6,6,5),(D_{notF}; 6,6,6,6)\},$$ where $D_F$-fair dice, $D_{notF}$-not fair dice.

I would like to now if I am correct.

($2$) Two shooters, independently of each other, shoot at a common target, one shot each. The probability that the fisrts shooters shoot the target is $\frac{8}{10}$. The probability that the second shooters shoot the target is $\frac{4}{10}$. We know that only one hit the target. What is the probability, that the target was hit by first shooter.

I think that $\Omega$ can look likes $$\Omega=\{(A_H,B_H),(A_H,B_{notH}),(A_{notH},B_H),(A_{notH},B_{notH})\},$$ where $A_{H}$ - the first shooter hits the target, $A_{notH}$ - the first shooter does not hit the target, $B_{H}$ - the second shooter hits the target, $B_{notH}$ - the second shooter does not hit the target.

Probability that the first ($A$) shooter hits target is $\frac{8}{10}$, it means that $$P\left(\{(A_H,B_H),(A_H,B_{notH})\}\right)=P\left(X\right)=\frac{8}{10},$$ and $$P\left(\{(A_{notH},B_{H}),(A_{notH},B_{notH})\}\right)=P\left(X^c\right)=\frac{2}{10},$$ Probability that the second ($B$) shooter hits target is $\frac{4}{10}$, it means that $$P\left(\{(A_H,B_H),(A_{notH},B_H)\}\right)=P\left(Y\right)=\frac{4}{10},$$ and $$P\left(\{(A_{notH},B_{notH}),(A_{H},B_{notH})\}\right)=P\left(Y^c\right)=\frac{6}{10},$$ The probability, that the target was hit by first shooter if we know that only one shooter hit formally is $$P(\{(A_H,B_H),(A_H,B_{notH})\}|\{(A_H,B_{notH}),(A_{notH},B_H)\})=P\left(X|Z\right).$$

I would like to compute this probability by using Bayes theorem, but I am little bit confused. I have been trying somthink like that $$P(X|Z)=\frac{P(Z|X)P(X)}{P(Z|X)P(X)+P(Z|X^c)P(X^c)},$$ but I am stuck.

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  • $\begingroup$ This question is too broad as stated , there are three different questions. I think you will benefit from having the second and third as separate questions to the first, which seems to be your big doubt here. Talking merely about the first, your sample space is correct, and I think the calculations work out as well. The point is , that the result of your experiment is what you see in two draws, and each draw can be white or blue. So your sample space is correctly designed and the calculations are in fact correct, I checked twice to make sure. $\endgroup$ Oct 15, 2021 at 9:13
  • $\begingroup$ Since the first one is correct and I can assure you of this, you can probably edit it out and focus upon the other two questions. $\endgroup$ Oct 15, 2021 at 9:14
  • $\begingroup$ Thank you very much. In each exercise I can compute the probability, but I have troubles with how the probability space looks like. I am going to edit my question. Probabily I am going to delet the first part which is correct.I am apologize for this. I only want to show that in some cases I can make the probability space but sometimes I am little bit confused. $\endgroup$
    – Waney
    Oct 15, 2021 at 9:22
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    $\begingroup$ Please feel free to edit. Since I've been of help, I'll probably place the feedback for the third question into the answer box itself once you're done editing and we'll take the discussion from there. $\endgroup$ Oct 15, 2021 at 10:03
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    $\begingroup$ It's time I write an answer. I'll do that shortly. $\endgroup$ Oct 15, 2021 at 14:51

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So your first question (the one on the fair and unfair dice) is correctly attempted. To make some comments : the question clearly says that a dice is chosen of the two, and then four rolls are performed. Therefore, your sample space can also be written more neatly as : $$ \Omega = \{(D , r_1,r_2,r_3,r_4) : D \in \{D_{\text{F} , }D_{\text{not F}}\} , r_1,r_2,r_3,r_4 \in \{1,2,3,4,5,6\}\} $$

Note that $S$ is just $2^{\Omega}$, the set of subsets of $\Omega$. This is always true in the finite discrete setting, because you can consistently assign probabilities to every possible event by summing over the individual probabilities of the finitely many outcomes it contains.

Now you need the following events to even define your question : $E_1 = \{D = D_{\text{not F}}\}$ and $E_2 = \{|\{i \in \{1,2,3,4\} : r_i = 6\}| = 2\}$. Then you have to find $P(E_1 | E_2)$. Now, the point is that you can't quite calculate $P(E_2)$ since it depends upon whether or not the first dice is chosen or the second dice is chosen. i.e. upon $E_1$ and $E_1^c = \{D = D_{\text{F}}\}$. But then, Bayes' rule applies and we have : $$ P(E_1 | E_2) = \frac{P(E_2|E_1)P(E_1)}{P(E_2 | E_1)P(E_1) + P(E_2 | E_1^c)P(E_1^c)} $$

These conditional probabilities are easier to calculate and you can finish the problem.


So we come to the second question on target shooting. Once again, your sample space is correct (if anything, going over your answers it seems that creation of sample space is something that is not going to trouble you in an exam!) because it contains all possible outcomes of everything we want to see to solve the question. Of course, $S$ in this case will also be $2^{\Omega}$.

Now let us quickly see the events of interest to define the question. These are $E_1 = \{A \text{ hit the target}\}$ and $E_2 = \{\text{Only one person hit the target}\}$. We need to find $P(E_1 | E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)}$, so the question is, can we describe $E_1 \cap E_2$ and $E_2$ and get their probabilities directly?


The point is, Bayes' rule is used when either of these quantities is really difficult to find, usually $P(E_2)$ because to find $P(E_2)$ itself would require you to go through various sub-cases, involving $E_1$ itself (I mean, the denominator of Bayes' rule is in fact equal to $P(E_2)$ so the formula just contains an easier expression for the actual definition).

In our case, neither $E_1 \cap E_2$ nor $E_2$ is particularly difficult to describe in isolation.

For example, $E_1 \cap E_2$ means that only one person hits, but also that $A$ hits. Hence, $$ E_1 \cap E_2 = \{(A_{H}, B_{\text{not }H})\} \implies P(E_1 \cap E_2) = \frac{8}{10} \left(1-\frac {4}{10}\right) = 0.48 $$

and $E_2$ also contains the outcome that $B$ hits and $A$ misses in addition to $E_1$. Hence, $$ E_2 = \{(A_{H}, B_{\text{not }H}) , (A_{\text{not }H}, B_{H})\} \implies P(E_2) = 0.48 + \left(1-\frac{8}{10}\right)\left(\frac 4{10}\right) = 0.48 +0.08 = 0.56 $$

Therefore the answer is $\frac{0.48}{0.56} = \frac 67$.


A couple more tips that will simplify life for you.

  • I've seen Bayes' rule used heavily when the experiment happens in (strictly) chronological order during your recording, and you are asked to predict an older event based on a newer event. For example, in question 1 you saw the dice rolls , which comes strictly after you chose the dice, so predicting the dice based on the rolls is Bayes' territory. The second question can also be interpreted as a Bayes' rule application because you can imagine that $A$ shoots before $B$ and interpret the problem like this, but you only need the formula if any quantity is hard to find, and in this case it wasn't.

  • In any scenario, instead of finding the probability of events, find the probability of each of the outcomes. So for example, in question $2$, before you even see the events of interest, make sure that you have the probabilities of each outcome down like $P(\{(A_H,B_H)\})$ and so on. Then, to find the probability of any event, you just need to add the probabilities of the outcomes inside that event and you are done. This is faster and rigorous, but will also give you a good grasp of everything related to that experiment.

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  • $\begingroup$ You are very nice. Thank you very much. It was/is very useful answer, which help me a lot. $\endgroup$
    – Waney
    Oct 15, 2021 at 19:50
  • $\begingroup$ @Waney Thanks, hope to be of help next time as well! $\endgroup$ Oct 15, 2021 at 20:12

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