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Carefully describe the events $A$, $B$, $A\cup B$, and $A\cap B$ in the following sample space:

Six people enter the elevator in the basement of a building with 6 floors. Each states where they will get out.

$A$ = {all six get off on the same floor}, $B$ = {exactly five people get off on floor two}

From my understanding:

$A$ - Everyone says they will get out on a floor

$S$ = Everyone gets off at floor {1,2,3,4,5,6}

$|S| = 6$

$B$ - Exactly five people get off on floor two

$S$ = One person will get off at either floor {1,3,4,5,6}

$|S|$ = 5

But as far as calculating $A\cup B$ or $A\cap B$ of the following, I don't quite understand

Asaf Karagila
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petrov
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3 Answers3

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There may be more than one sample space that is appropriate for an experiment. We may wish to record only how many people get off at each floor. Then the sample space consists of all sextuples $(x_1,x_2,x_3,x_4,x_5,x_6)$ of non-negative integers that add up to $6$, the total number of people.

This would not be my first choice for a sample space. I would call the people U, V, W, X, Y, Z, and record for each person the floor she intends to get off at. The record would be in alphabetical order. So for example $(3,6,3,1,5,4)$ means that U is getting off on floor $3$, V is getting off on $6$, W is getting off on $3$, and so on.

So this sample space is the set of all sextuples $(y_1,y_2,y_3,y_4,y_5,y_6)$, where each $y_i$ is an integer between $1$ and $6$.

We will use this as the sample space from now on.

In terms of this sample space, the event $A$ consists of the ordered sextuples $(1,1,1,1,1,1)$, $(2,2,2,2,2,2)$, and so on up to $(6,6,6,6,6,6)$. You may be expected to write this in set notation.

The event $B$ is, in this sample space, more complicated. It consists of all sextuples where five of the entries are $2$ and the sixth is any of $1,3,4,5,6$. One example of a sextuple in the sample space is $(2,5,2,2,2,2)$. There are $30$ in all.

For $A\cup B$, just list all the elements of $A$ together with all the elements of $B$.

For $A\cap B$, notice that $A$ and $B$ are disjoint (they have no element in common). Thus $A\cap B=\emptyset$, the empty set.

André Nicolas
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You must be clear on how you are defining your sample space. I would say the sample space is the total number of ways that the 6 people can get off the elevator. Let's say the sample space is denoted as $S$. I will let you figure out |S|.

Say $S_1 = A \cup B$. This can be read as the event where either all six get off on the same floor or that exactly five people get off on floor two. This would be just simple addition of the two ideas. Hence, $|S_1|=11$. Remember, $S_1 \subseteq S$.

Say $S_2 = A \cap B$. This can be read as the event where both events A and B happen at the same time. But since you can't have A and B happen at the same time, $|S_2|=\emptyset$. Remember, $S_2 \subseteq S$.

In the cases you have already done ($A$, $B$), make sure you define them as subsets of the sample space.

Rye_yawn
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Each of the six people (numbered from $1$ to $6$) entering the lift has a number from $1$ to $6$ in his mind. Therefore the sample space $S$ is the set of all words $x=(x_1,x_2,\ldots, x_6)$ over the alphabet $\{1,2,3,4,5,6\}$, and $|S|=6^6=46656$.

The event $A$ consists of all constant words. There are six of them, so $P(A)={6\over 46656}={1\over7776}$.

The event $B$ consists of all words $x\in S$ containing five $2$'s and one non-$2$. The non-$2$ can have $5$ different values and can be at six different places. It follows that there are $5\cdot 6$ such words, so that$P(B)={30\over46656}={5\over7776}$.

Obviously $A\cap B=\emptyset$, so $P(A\cap B)=0$. It follows that $A\cup B$ consists of $6+30=36$ words, so $P(A\cup B)={1\over1296}$.

Christian Blatter
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