Sample Spaces, Events and the Axioms, Worked Solutions

Ques­tion 1: Writ­ing sam­ple spaces#

The prob­lem#

Write the sam­ple space for each exper­i­ment:

(a) a coin tossed three times;
(b) a die rolled twice (how many points?);
(c) a fam­i­ly's two chil­dren, by sex, older child first.

Under­stand­ing the prob­lem#

The sam­ple space $S$ is the list of all pos­si­ble out­comes of the exper­i­ment, with noth­ing repeated and noth­ing miss­ing. In each part the order mat­ters: the first toss is dif­fer­ent from the sec­ond, and the older child is writ­ten first.

The idea#

Build each out­come stage by stage. If the first stage has $m$ pos­si­ble results and the sec­ond has $n$, there are $m \times n$ com­bined out­comes. List them in a sys­tem (for exam­ple alpha­bet­i­cal) so none is missed.

Step-by-step solu­tion#

Part (a)

Step 1. Each toss gives H (head) or T (tail): $2$ choices per toss.

Step 2. Three tosses give $2 \times 2 \times 2 = 8$ out­comes. Write them in order, the first let­ter for the first toss:

$$S = \{HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT\}$$

Part (b)

Step 1. Record an out­come as an ordered pair $(i, j)$, where $i$ is the first roll and $j$ the sec­ond. Each of $i$ and $j$ can be $1, 2, 3, 4, 5, 6$.

$$S = \{(i, j) : i, j \in \{1, 2, 3, 4, 5, 6\}\}$$

Step 2. Count: $6$ choices for $i$ and $6$ for $j$.

$$n(S) = 6 \times 6 = 36$$

Note that $(1, 2)$ and $(2, 1)$ are dif­fer­ent out­comes, because the rolls are in order.

Part (c)

Step 1. Write B for boy and G for girl; the first let­ter is the older child.

Step 2. Each child is B or G, giv­ing $2 \times 2 = 4$ out­comes:

$$S = \{BB,\ BG,\ GB,\ GG\}$$

Here $BG$ (older boy, younger girl) and $GB$ (older girl, younger boy) are dif­fer­ent out­comes.

Check­ing the answer#

The counts match the mul­ti­pli­ca­tion rule: $2^3 = 8$, $6^2 = 36$, $2^2 = 4$. Each list has no repeats.

Answer#

(a) $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$, $8$ points; (b) the $36$ ordered pairs $(i, j)$ with $i, j \in \{1, \ldots, 6\}$; (c) $\{BB, BG, GB, GG\}$.

Ques­tion 2: A two-stage exper­i­ment where the sec­ond stage depends on the first#

The prob­lem#

A coin is tossed and, if it shows a head, a die is rolled. Describe the sam­ple space $S$.

Under­stand­ing the prob­lem#

The die is rolled only when the coin shows a head. If the coin shows a tail, the exper­i­ment stops there. So the out­comes are not all of the same shape.

The idea#

Fol­low each branch of the exper­i­ment sep­a­rately (a tree dia­gram in your head): the head branch splits into six, the tail branch ends at once.

Step-by-step solu­tion#

Step 1. Head branch: the die then shows $1$ to $6$. Write these out­comes as $H1, H2, H3, H4, H5, H6$.

Step 2. Tail branch: noth­ing more hap­pens, so there is a sin­gle out­come, $T$.

Step 3. Put both branches together.

$$S = \{H1,\ H2,\ H3,\ H4,\ H5,\ H6,\ T\}$$

Step 4. Count: $6 + 1 = 7$ sam­ple points.

Check­ing the answer#

A com­mon wrong answer is $12$ points ($H1, \ldots, H6, T1, \ldots, T6$). That would mean the die is rolled after a tail too, which the ques­tion rules out.

Answer#

$S = \{H1, H2, H3, H4, H5, H6, T\}$, with $7$ sam­ple points.

Ques­tion 3: Events when two dice are rolled#

The prob­lem#

Two dice are rolled. List the events $A$ = the sum is $8$; $B$ = both dice show the same num­ber; $C$ = the first die shows $5$. Find $A \cap B$, $A \cap C$, and $B \cup C$ in terms of counts.

Under­stand­ing the prob­lem#

The sam­ple space is the $36$ ordered pairs $(i, j)$ from Ques­tion 1(b). Each event is a sub­set of these. $A \cap B$ means "both $A$ and $B$ hap­pen", and $B \cup C$ means "$B$ or $C$ (or both) hap­pens". For $B \cup C$ you are asked for the num­ber of points.

The idea#

List each event in full, then read off com­mon points for the inter­sec­tions. For the union, use the count­ing rule

$$n(B \cup C) = n(B) + n(C) - n(B \cap C),$$

sub­tract­ing the over­lap so it is not counted twice.

Step-by-step solu­tion#

Step 1. Event $A$: pairs whose sum is $8$. Go through the first die from $1$ to $6$; the sec­ond must be $8 - i$, which must lie between $1$ and $6$.

$$A = \{(2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2)\}, \qquad n(A) = 5$$

Step 2. Event $B$: dou­bles.

$$B = \{(1,1),\ (2,2),\ (3,3),\ (4,4),\ (5,5),\ (6,6)\}, \qquad n(B) = 6$$

Step 3. Event $C$: first die shows $5$.

$$C = \{(5,1),\ (5,2),\ (5,3),\ (5,4),\ (5,5),\ (5,6)\}, \qquad n(C) = 6$$

Step 4. $A \cap B$: look for pairs in both lists. Only $(4,4)$ has sum $8$ and is a dou­ble.

$$A \cap B = \{(4,4)\}, \qquad n(A \cap B) = 1$$

Step 5. $A \cap C$: pairs with first die $5$ and sum $8$. The sec­ond die must be $3$.

$$A \cap C = \{(5,3)\}, \qquad n(A \cap C) = 1$$

Step 6. For $B \cup C$, first find the over­lap: a dou­ble whose first die is $5$ is $(5,5)$, so $n(B \cap C) = 1$. Then

$$n(B \cup C) = 6 + 6 - 1 = 11$$

Check­ing the answer#

List $B \cup C$ directly: the $6$ dou­bles plus the $5$ pairs $(5,1), (5,2), (5,3), (5,4), (5,6)$ that are not already dou­bles. That is $11$ points.

Answer#

$A \cap B = \{(4,4)\}$ ($1$ point); $A \cap C = \{(5,3)\}$ ($1$ point); $n(B \cup C) = 11$.

Ques­tion 4: Mutu­ally exclu­sive pairs#

The prob­lem#

Which pairs of events from Ques­tion 3 are mutu­ally exclu­sive?

Under­stand­ing the prob­lem#

Two events are mutu­ally exclu­sive when they can­not hap­pen together, that is, when their inter­sec­tion is the empty set. There are three pairs to test: $(A, B)$, $(A, C)$ and $(B, C)$.

The idea#

Use the inter­sec­tions already found in Ques­tion 3. A pair is mutu­ally exclu­sive only if its inter­sec­tion is $\varnothing$.

Step-by-step solu­tion#

Step 1. $A \cap B = \{(4,4)\} \ne \varnothing$. Rolling $(4,4)$ makes both hap­pen, so $A$ and $B$ are not mutu­ally exclu­sive.

Step 2. $A \cap C = \{(5,3)\} \ne \varnothing$. Not mutu­ally exclu­sive.

Step 3. $B \cap C = \{(5,5)\} \ne \varnothing$. Not mutu­ally exclu­sive.

Check­ing the answer#

For each pair you can name one actual roll that makes both events hap­pen: $(4,4)$, $(5,3)$ and $(5,5)$. That set­tles it.

Answer#

None of the three pairs is mutu­ally exclu­sive.

Ques­tion 5: Events from a pack of cards#

The prob­lem#

A card is drawn from a well-shuf­fled pack of $52$. Let $A$ = the card is a heart, $B$ = the card is a king, $C$ = the card is red. Describe $A \cap B$, $A \cup C$ and $C - A$.

Under­stand­ing the prob­lem#

A pack has four suits of $13$ cards: hearts and dia­monds are red, spades and clubs are black. There are four kings, one in each suit. $A \cap B$ means "heart and king", $A \cup C$ means "heart or red", and $C - A$ means "red but not heart".

The idea#

Trans­late each sym­bol into words, then think about which cards fit. Notice that every heart is red, so $A$ sits inside $C$.

Step-by-step solu­tion#

Step 1. $A \cap B$: a heart that is also a king. Only one card fits.

$$A \cap B = \{\text{king of hearts}\}$$

Step 2. $A \cup C$: a card that is a heart or red. Since every heart is already red, $A \subset C$, and adding the hearts to the red cards adds noth­ing new.

$$A \cup C = C = \text{all } 26 \text{ red cards}$$

Step 3. $C - A = C \cap A'$: red cards that are not hearts. The red cards are hearts and dia­monds, so remov­ing the hearts leaves the dia­monds.

$$C - A = \text{the } 13 \text{ diamonds}$$

Check­ing the answer#

Counts: $n(A \cap B) = 1$; $n(A \cup C) = 13 + 26 - 13 = 26$; $n(C - A) = 26 - 13 = 13$. All con­sis­tent.

Answer#

$A \cap B$ is the king of hearts; $A \cup C$ is the set of all $26$ red cards; $C - A$ is the set of $13$ dia­monds.

Ques­tion 6: Test­ing a prob­a­bil­ity assign­ment#

The prob­lem#

Check whether $\displaystyle P(a) = \tfrac{1}{3}$, $\displaystyle P(b) = \tfrac{1}{2}$, $\displaystyle P(c) = \tfrac{1}{4}$ is a valid prob­a­bil­ity assign­ment on $S = \{a, b, c\}$.

Under­stand­ing the prob­lem#

An assign­ment is valid only if it obeys the axioms: every prob­a­bil­ity is $\ge 0$, and the prob­a­bil­i­ties of all the sam­ple points add up to $P(S) = 1$.

The idea#

Check both con­di­tions. The first is easy to see; the sec­ond needs a sum of frac­tions.

Step-by-step solu­tion#

Step 1. Non-neg­a­tiv­ity: $\displaystyle \tfrac{1}{3}$, $\displaystyle \tfrac{1}{2}$ and $\displaystyle \tfrac{1}{4}$ are all pos­i­tive. This con­di­tion holds.

Step 2. Total. Use the com­mon denom­i­na­tor $12$.

$$\displaystyle \frac{1}{3} + \frac{1}{2} + \frac{1}{4} = \frac{4}{12} + \frac{6}{12} + \frac{3}{12} = \frac{13}{12}$$

Step 3. Com­pare with $1$: $\displaystyle \tfrac{13}{12} > 1$, so $P(S) \ne 1$. The sec­ond axiom fails.

Check­ing the answer#

In dec­i­mals: $0.333 + 0.5 + 0.25 = 1.083$, which is more than $1$.

Answer#

Not valid: the prob­a­bil­i­ties add up to $\displaystyle \tfrac{13}{12}$, not $1$.

Ques­tion 7: A biased coin#

The prob­lem#

A coin is biased so that heads is three times as likely as tails. Find $P(H)$ and $P(T)$.

Under­stand­ing the prob­lem#

The two out­comes are not equally likely, so you can­not use $\displaystyle \tfrac{1}{2}$. You are told a ratio: $P(H) = 3 \times P(T)$. The sam­ple space is $\{H, T\}$.

The idea#

Call the smaller prob­a­bil­ity $p$, express the other in terms of $p$, and use the axiom that the prob­a­bil­i­ties of all out­comes add to $1$.

Step-by-step solu­tion#

Step 1. Let $P(T) = p$. Then $P(H) = 3p$.

Step 2. The two prob­a­bil­i­ties must add to $1$.

$$\displaystyle 3p + p = 1 \quad\Longrightarrow\quad 4p = 1 \quad\Longrightarrow\quad p = \frac{1}{4}$$

Step 3. Sub­sti­tute back.

$$\displaystyle P(T) = \frac{1}{4}, \qquad P(H) = 3 \times \frac{1}{4} = \frac{3}{4}$$

Check­ing the answer#

$\displaystyle \tfrac{3}{4} + \tfrac{1}{4} = 1$, and $\displaystyle \tfrac{3}{4}$ is indeed three times $\displaystyle \tfrac{1}{4}$.

Answer#

$\displaystyle P(H) = \frac{3}{4}$ and $\displaystyle P(T) = \frac{1}{4}$.

Ques­tion 8: A prime num­ber on the loaded die#

The prob­lem#

For the loaded die of Exam­ple 3, in which each even num­ber is twice as likely as each odd num­ber, find $P(\text{prime})$.

Under­stand­ing the prob­lem#

You need the prob­a­bil­ity of each face first. Then pick out the prime faces. On a die the primes are $2$, $3$ and $5$ (remem­ber $1$ is not prime).

The idea#

As in Exam­ple 3, give each odd face prob­a­bil­ity $p$ and each even face $2p$, use the fact that all six add to $1$, and then add the prob­a­bil­i­ties of the prime faces.

Step-by-step solu­tion#

Step 1. Odd faces $1, 3, 5$ each have prob­a­bil­ity $p$; even faces $2, 4, 6$ each have $2p$. The total must be $1$.

$$\displaystyle 3 \times p + 3 \times 2p = 9p = 1 \quad\Longrightarrow\quad p = \frac{1}{9}$$

Step 2. So each odd face has prob­a­bil­ity $\displaystyle \tfrac{1}{9}$ and each even face $\displaystyle \tfrac{2}{9}$.

Step 3. The prime faces are $2$ (even), $3$ (odd) and $5$ (odd).

$$\displaystyle P(\text{prime}) = P(2) + P(3) + P(5) = \frac{2}{9} + \frac{1}{9} + \frac{1}{9} = \frac{4}{9}$$

Check­ing the answer#

The non-prime faces are $1, 4, 6$ with total $\displaystyle \tfrac{1}{9} + \tfrac{2}{9} + \tfrac{2}{9} = \tfrac{5}{9}$, and $\displaystyle \tfrac{4}{9} + \tfrac{5}{9} = 1$.

Answer#

$\displaystyle P(\text{prime}) = \frac{4}{9}$.

Com­mon mis­take to avoid#

Count­ing $1$ as a prime, which would wrongly give $\displaystyle \tfrac{5}{9}$.

Ques­tion 9: Mutu­ally exclu­sive but not exhaus­tive#

The prob­lem#

Give an exam­ple of three events that are mutu­ally exclu­sive but not exhaus­tive.

Under­stand­ing the prob­lem#

You need three events such that no two can hap­pen together (mutu­ally exclu­sive: every pair­wise inter­sec­tion is empty), yet together they do not cover the whole sam­ple space (not exhaus­tive: their union is not $S$).

The idea#

Choose a sim­ple exper­i­ment and pick three sep­a­rate out­comes, leav­ing some out­comes unused.

Step-by-step solu­tion#

Step 1. Take the roll of one die: $S = \{1, 2, 3, 4, 5, 6\}$.

Step 2. Let $E_1 = \{1\}$, $E_2 = \{2\}$, $E_3 = \{3\}$.

Step 3. Mutu­ally exclu­sive: $E_1 \cap E_2 = E_1 \cap E_3 = E_2 \cap E_3 = \varnothing$, since one roll can­not show two dif­fer­ent num­bers.

Step 4. Not exhaus­tive: $E_1 \cup E_2 \cup E_3 = \{1, 2, 3\} \ne S$. A roll of $4$, $5$ or $6$ makes none of them hap­pen.

Check­ing the answer#

Both con­di­tions are shown by direct list­ing, so the exam­ple works. (Many other exam­ples are cor­rect too.)

Answer#

On one roll of a die, the events $\{1\}$, $\{2\}$ and $\{3\}$ are mutu­ally exclu­sive but not exhaus­tive.

Ques­tion 10: An event and its com­ple­ment#

The prob­lem#

Show that $A$ and $A'$ are always mutu­ally exclu­sive and exhaus­tive.

Under­stand­ing the prob­lem#

This is a proof. You must show two facts for any event $A$ in any sam­ple space $S$: (i) $A \cap A' = \varnothing$, and (ii) $A \cup A' = S$. Recall $A' = S - A$, the out­comes of $S$ that are not in $A$.

The idea#

Take an arbi­trary out­come $\omega$ of $S$ and ask where it can lie. It is either in $A$ or not in $A$, never both and never nei­ther.

Step-by-step solu­tion#

Step 1 (mutu­ally exclu­sive). Sup­pose some out­come $\omega$ were in $A \cap A'$. Then $\omega \in A$, and also $\omega \in A'$, which means $\omega \notin A$. An out­come can­not both be in $A$ and not be in $A$, so no such $\omega$ exists.

$$A \cap A' = \varnothing$$

Step 2 (exhaus­tive). Take any out­come $\omega \in S$. Either $\omega \in A$, or $\omega \notin A$, in which case $\omega \in A'$ by the def­i­n­i­tion of $A'$. Either way $\omega \in A \cup A'$. So every out­come of $S$ is in $A \cup A'$, and $A \cup A'$ con­tains noth­ing out­side $S$.

$$A \cup A' = S$$

Step 3. Hav­ing shown both facts, $A$ and $A'$ are mutu­ally exclu­sive and exhaus­tive.

Check­ing the answer#

Test with Exam­ple 1: $A = \{2, 4, 6\}$ gives $A' = \{1, 3, 5\}$; these share noth­ing and together make $\{1, \ldots, 6\}$.

Answer#

$A \cap A' = \varnothing$ (they can­not occur together) and $A \cup A' = S$ (one of them must occur), so $A$ and $A'$ are always mutu­ally exclu­sive and exhaus­tive.