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Unfortunately, I couldn't solve this question. Actually, I am very suspicious about whether the question is correct or not.

An experiment has a sample space that consists of 8 equally likely outcomes. $ S = \{x_1, x_2, ..., x_8\} $. Three events are defined as $ A = \{x_2, x_4, x_6, x_8\}, B = \{x_1, x_3, x_6, x_7\}, C = \{x_1, x_2, x_3, x_5\} $ . Find the probabilities of the following events.

a)$ A \cap B$

b) $ \overline{A \setminus B}$

c)$ A \cap (B\cup\overline{A})$

d $\overline{A \cap B \cap C} $

I thought for question in a $ A \cap B = \{x_6\} $ so $ P(A \cap B) = 1/8 $ but i am not sure

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    $\begingroup$ If the elements are equally likely then $P(A\cap B)=P(\{x_6\})=\frac{1}{8}.$ $\endgroup$ – zoli Mar 18 '15 at 17:44
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    $\begingroup$ Why do you think that $P(\{x_6\})=\frac12$? $\endgroup$ – drhab Mar 18 '15 at 17:45
  • $\begingroup$ Hint: each singleton event (an event that contains exactly one outcome, e.g. $\{x_6\}$) has probability $\frac 18$ since you are told that the outcomes are equally likely and that there are $8$ outcomes. $\endgroup$ – Dilip Sarwate Mar 18 '15 at 17:45
  • $\begingroup$ Hint: singleton event is a synonym for outcome. Don't know which word you saw in class. $\endgroup$ – Pedro Mar 18 '15 at 17:51
  • $\begingroup$ i am so sory, i accidentally wrote @drhab $\endgroup$ – verdery Mar 18 '15 at 18:04
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Question seems correct. If you intersect events $A$ and $B$ you get the event $\{x_6\}$. The probability of event $\{x_6\}$ happening is just $\frac{1}{8}$, because there are 8 outcomes en all are equally likely (thus every outcome has probability $\frac{1}{8}$). The event $\{x_6\}$ consists of only 1 outcome, namely the outcome $x_6$ so the probability is just $\frac{1}{8}$.

Hint for following exercises: probability of an event consisting of 2 outcomes is $\frac{2}{8}$, probability of an event consisting of 3 outcomes is $\frac{3}{8}$, the probability of an event consisting of 4 outcomes is $\frac{4}{8}$, ... . You thus need to do the set operations first and see how much outcomes are still left. Count these outcomes and this will give you the probability.

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  • $\begingroup$ @verdery $P\left(A\right)=P\left(\left\{ x_{2}\right\} \cup\left\{ x_{4}\right\} \cup\left\{ x_{6}\right\} \cup\left\{ x_{8}\right\} \right)=P\left(\left\{ x_{2}\right\} \right)+P\left(\left\{ x_{4}\right\} \right)+P\left(\left\{ x_{6}\right\} \right)+P\left(\left\{ x_{8}\right\} \right)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$ $\endgroup$ – drhab Mar 18 '15 at 18:23
  • $\begingroup$ @verdery Here you can see information about the probility axioms. In short: probabilities are always assigned to subsets. In words it says that: (1) every subset is assigned a positive number (=> because we don't want something to have a negative probability, that would be silly), (2) the probability of all outcomes together is 1, (3) the probability of a union of things in the sum of those things, thus the probability of A or B happening, is just probability of A + probability of B. :) $\endgroup$ – Pedro Mar 18 '15 at 20:44
  • $\begingroup$ @verdery your question was absolutely not a bad question, these sort of questions are very nice because it explains very good that we always take probabilities of subsets, of the subset of all possible outcomes. Btw: if you work with random variables, you do actually the same thing, since technically, a random variable is a function from the set of all outcomes to the set of values the variable can take (often these are the same sets). When you write then something as $P(X = 3)$ you actually mean $P(X(\omega) = 3)$ and you view P as a function. $\endgroup$ – Pedro Mar 18 '15 at 20:48
  • $\begingroup$ @verdery and since P is a function, you first need to evaluate it's input (don't know whether you already knew this fact, but you always need to evaluate first the input of a function before you can put it into the function itself), which is actually solving the equation $X(\omega) = 3)$ which gives as a result back a set which contains just one outcome. The result of the equation is in fact the $\omega$ which solves the given expression, which is in fact again a subset. So even with random variables, you are actually getting a subset between your $P( )$ ;) $\endgroup$ – Pedro Mar 18 '15 at 20:54
  • $\begingroup$ @verdery if you don't already get the random variable part, don't worry, it will come. But actually probability theory has always to do with sets. You always, always need to have a set between your $P( )$. And from that set you can say what the probability is. And you can use the 3 axioms I typed above, especially the third that the union is just adding, and intuitively this is also what we want. Because the probability of A or B happening, should just be P(A) + P(B). (one last remark A and B need to be non-overlapping sets!!! I forgot to add that to the information of the 3rd axiom) $\endgroup$ – Pedro Mar 18 '15 at 20:59

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