Chapter 7

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

$4\times <!--\; \times \; -->15=60$

$4\times <!--\; \times \; -->15=60$

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

$4\times <!--\; \times \; -->15\times <!--\; \times \; -->3\times <!--\; \times \; -->2=360$

$4\times <!--\; \times \; -->15\times <!--\; \times \; -->3\times <!--\; \times \; -->2=360$

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

You have 3 choices for letters, 10 choices each time you choose a digit, and 26 choices for each time you choose a letter.

$3\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->26\times <!--\; \times \; -->26\times <!--\; \times \; -->26=5,272,800,000$

$3\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->10\times <!--\; \times \; -->26\times <!--\; \times \; -->26\times <!--\; \times \; -->26=\text{5,272,800,000}$

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

The number of ways you can make your choices is $5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1=120$.

120

A factorial for any positive whole number n (denoted n!) is the product of every whole number less than or equal to n. We define 0! to be equal to one.

$6!=6\times <!--\; \times \; -->5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1=720$

720

A factorial for any positive whole number n (denoted n!) is the product of every whole number less than or equal to n. We define 0! to be equal to one.

$\frac{12!}{10!}=\frac{12\times <!--\; \times \; -->11\times <!--\; \times \; -->10\times <!--\; \times \; -->9\times <!--\; \times \; -->8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}{10\times <!--\; \times \; -->9\times <!--\; \times \; -->8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{12\times <!--\; \times \; -->11\times <!--\; \times \; -->\cancel{10\times <!--\; \times \; -->9\times <!--\; \times \; -->8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}}{\cancel{10\times <!--\; \times \; -->9\times <!--\; \times \; -->8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}}=132$

Instead of writing out all the factors, you could have written this expression an easier way.

$\frac{12!}{10!}=\frac{12\times <!--\; \times \; -->11\times <!--\; \times \; -->10!}{10!}=\frac{12\times <!--\; \times \; -->11\times <!--\; \times \; -->\cancel{10!}}{\cancel{10!}}=132$

132

A factorial for any positive whole number n (denoted n!) is the product of every whole number less than or equal to n. We define 0! to be equal to one.

$\frac{8!}{4!4!}=\frac{8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->5\times <!--\; \times \; -->\cancel{4!}}{\cancel{4!}\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{1,680}{24}=70$

70

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$6\mathrm{P}2=\frac{n!}{\left(n-<!--\; -\; -->r\right)!}=\frac{6!}{\left(6-<!--\; -\; -->2\right)!}=\frac{6!}{4!}=\frac{6\times <!--\; \times \; -->5\times <!--\; \times \; -->4!}{4!}=30$

30

24,024

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$14\mathrm{P}4=\frac{n!}{\left(n-<!--\; -\; -->r\right)!}=\frac{14!}{\left(14-<!--\; -\; -->4\right)!}=\frac{14!}{10!}=\frac{14\times <!--\; \times \; -->13\times <!--\; \times \; -->12\times <!--\; \times \; -->11\times <!--\; \times \; -->10!}{10!}=24,024$

5,814

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$19\mathrm{P}3=\frac{n!}{\left(n-<!--\; -\; -->r\right)!}=\frac{19!}{\left(19-<!--\; -\; -->3\right)!}=\frac{19!}{16!}=\frac{19\times <!--\; \times \; -->18\times <!--\; \times \; -->17\times <!--\; \times \; -->16!}{16!}=5,814$

${}_{15}{P}_3=\mathrm{2,730}$

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$15\mathrm{P}3=\frac{15!}{\left(15-<!--\; -\; -->3\right)!}=\frac{15!}{12!}=\frac{15\times <!--\; \times \; -->14\times <!--\; \times \; -->13\times <!--\; \times \; -->12!}{12!}=2,730$

There are 2,730 ways to pick the top three finishers.

Every kid gets a piece of candy. It does not matter who gets their candy first. This is a combination.

combination

It does not matter in what order the instructor names the students to go in the first vehicle. They are all going in the same vehicle, no matter how the names are called. This is a combination.

combination

The number of combinations of n objects taken r at a time is $nCr=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}$.

The answer will always be an integer, so check your math if you get a fraction.

$6C4=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{6!}{4!\left(6-<!--\; -\; -->4\right)!}=\frac{6!}{4!2!}=\frac{6\times <!--\; \times \; -->5\times <!--\; \times \; -->\cancel{4!}}{\cancel{4!}\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{30}{2}=15$

15

The number of combinations of n objects taken r at a time is $nCr=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}$.

The answer will always be an integer, so check your math if you get a fraction.

$10C8=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{10!}{8!\left(10-<!--\; -\; -->8\right)!}=\frac{10!}{8!2!}=\frac{10\times <!--\; \times \; -->9\times <!--\; \times \; -->\cancel{8!}}{\cancel{8!}\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{90}{2}=45$

45

The number of combinations of n objects taken r at a time is $nCr=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}$.

The answer will always be an integer, so check your math if you get a fraction.

$14C5=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{14!}{5!\left(14-<!--\; -\; -->5\right)!}=\frac{14!}{5!9!}=\frac{14\times <!--\; \times \; -->13\times <!--\; \times \; -->12\times <!--\; \times \; -->11\times <!--\; \times \; -->10\times <!--\; \times \; -->\cancel{9!}}{5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{9!}}=\frac{240,240}{120}=2,002$

2,002

Since they all receive the same prize, it does not matter what order they call the names. This is a combination, choosing 3 names from 58.

$58C3=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{58!}{3!\left(58-<!--\; -\; -->3\right)!}=\frac{58!}{3!55!}=\frac{58\times <!--\; \times \; -->57\times <!--\; \times \; -->56\times <!--\; \times \; -->\cancel{55!}}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{55!}}=\frac{185,136}{6}=30,856$

30,856

Since they all have equal responsibility, it does not matter in which order you name the members. This is a combination, choosing 4 members from 42 members.

$42C4=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{42!}{4!\left(42-<!--\; -\; -->4\right)!}=\frac{42!}{4!38!}=\frac{42\times <!--\; \times \; -->41\times <!--\; \times \; -->40\times <!--\; \times \; -->39\times <!--\; \times \; -->\cancel{38!}}{4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{38!}}=\frac{2,686,320}{24}=111,930$

111,930

It does not matter in which order the cards are dealt to you. For each requirement, it is a combination. There are 13 cards in each suit.

Then, use the Multiplication Rule of Counting. The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

The number of ways to have the required hand is $13\text{C}3\times <!--\; \times \; -->13\text{C}2\times <!--\; \times \; -->13\text{C}1$.

Evaluate each combination separately.

$13C3=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{13!}{3!\left(13-<!--\; -\; -->3\right)!}=\frac{13!}{3!10!}=\frac{13\times <!--\; \times \; -->12\times <!--\; \times \; -->11\times <!--\; \times \; -->\cancel{10!}}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{10!}}=\frac{1,716}{6}=286$

$13C2=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{13!}{2!\left(13-<!--\; -\; -->2\right)!}=\frac{13!}{2!11!}=\frac{13\times <!--\; \times \; -->12\cancel{11!}}{2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{11!}}=\frac{156}{2}=78$

$13C1=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{13!}{1!\left(13-<!--\; -\; -->1\right)!}=\frac{13!}{1!12!}=\frac{13\cancel{12!}}{1\times <!--\; \times \; -->\cancel{12!}}=13$

Now you can evaluate $13\text{C}3\times <!--\; \times \; -->13\text{C}2\times <!--\; \times \; -->13\text{C}1=286\times <!--\; \times \; -->78\times <!--\; \times \; -->13=290,004$.

290,004

The sample space is the set of all possible outcomes.

{hearts, spades, clubs, diamonds}

$\{\mathrm{♡<!--\; ♡\; -->},\mathrm{♠<!--\; ♠\; -->},\mathrm{♣<!--\; ♣\; -->},\mathrm{♢<!--\; ♢\; -->}\}$

The sample space is the set of all possible outcomes.

A represents an Ace, J for Jack, Q for Queen, and K for King.

{A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}

{A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K}

The sample space is the set of all possible outcomes. There are four possible outcomes.

{1, 2, 3, 4}

{1, 2, 3, 4}

The sample space is the set of all possible outcomes. The possible outcomes are the possible sums from adding the choice from {1, 2, 3, 4} on three dice.

Every number between 3 and 12 is possible, too.

Sample space: {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

{3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Independent stages do not affect each other. Dependent stages do affect each other.

If you do not care which colors go together, the choices of blouse and skirt are independent.

Independent

Independent stages do not affect each other. Dependent stages do affect each other.

Once you decide that if you choose the blue shirt, you will not wear the orange skirt (or something like that), the choices of blouse and skirt are dependent.

Dependent

The sample space: {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) }

{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Due to the limitations of graphics in the solution area, the “tree” will be simulated with a table.

Sample space: {((J$\mathrm{♡<!--\; ♡\; -->}$, Q$\mathrm{♡<!--\; ♡\; -->}$), (J$\mathrm{♡<!--\; ♡\; -->}$, K$\mathrm{♡<!--\; ♡\; -->}$),(Q$\mathrm{♡<!--\; ♡\; -->}$, J$\mathrm{♡<!--\; ♡\; -->}$),(Q$\mathrm{♡<!--\; ♡\; -->}$, K$\mathrm{♡<!--\; ♡\; -->}$),(K$\mathrm{♡<!--\; ♡\; -->}$, J$\mathrm{♡<!--\; ♡\; -->}$),(K$\mathrm{♡<!--\; ♡\; -->}$, Q$\mathrm{♡<!--\; ♡\; -->}$)}

{J$\mathrm{♡<!--\; ♡\; -->}$ Q$\mathrm{♡<!--\; ♡\; -->}$, J$\mathrm{♡<!--\; ♡\; -->}$ K$\mathrm{♡<!--\; ♡\; -->}$, Q$\mathrm{♡<!--\; ♡\; -->}$ J$\mathrm{♡<!--\; ♡\; -->}$, Q$\mathrm{♡<!--\; ♡\; -->}$ K$\mathrm{♡<!--\; ♡\; -->}$, K$\mathrm{♡<!--\; ♡\; -->}$ J$\mathrm{♡<!--\; ♡\; -->}$, K$\mathrm{♡<!--\; ♡\; -->}$ Q$\mathrm{♡<!--\; ♡\; -->}$}

Jorge is flipping a coin only 4 times. He cannot have more than 4 heads, so the probability of less than 5 heads is equal to 1.

Jorge is flipping a coin only 4 times. He cannot have more than 4 heads, so the probability of more than 5 heads is equal to zero.

$P(\text{H}\ge 5)=0$

$\frac{1}{6}$

For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is $P(E)=\frac{n(E)}{n(S)}$, where n(E) and n(S) denote the number of outcomes in the event and in the sample space, respectively.

$P(4)=\frac{1}{6}$

$\frac{2}{3}$

For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is $P(E)=\frac{n(E)}{n(S)}$, where n(E) and n(S) denote the number of outcomes in the event and in the sample space, respectively.

$P(\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{n}2)=P\left(3,4,5,\mathrm{o}\mathrm{r}6\right)=\frac{4}{6}=\frac{2}{3}$

$\frac{1}{2}$

For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is $P(E)=\frac{n(E)}{n(S)}$, where n(E) and n(S) denote the number of outcomes in the event and in the sample space, respectively.

$P\left(odd\right)=P\left(1,3,\mathrm{o}\mathrm{r}5\right)=\frac{3}{6}=\frac{1}{2}$

$\frac{5}{16}$

Of the 16 possible sums, five of them are 6 or 7.

For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is $P(E)=\frac{n(E)}{n(S)}$, where n(E) and n(S) denote the number of outcomes in the event and in the sample space, respectively.

$P\left(6\mathrm{o}\mathrm{r}7\right)=\frac{5}{16}$

The “tree” as a table: These are all hearts, so only the number of the card is in the table.

There are six possible results.

P(the first card is a 3 heart) = $\frac{2}{6}=\frac{1}{3}$.

$\frac{1}{3}$

The “tree” as a table: These are all hearts, so only the number of the card is in the table.

There are six possible results. The ones that satisfy the requirement are 34, 35, and 45.

P(the first card drawn has a lower number than the second card) $=\frac{3}{6}=\frac{1}{2}$.

$\frac{1}{2}$

The “tree” as a table: These are all hearts, so only the number of the card is in the table.

There are six possible results. The ones that satisfy the requirement are 34, 43, 45, and 54.

P(one of the cards drawn is the 4 hearts) = $\frac{4}{6}=\frac{2}{3}$.

$\frac{2}{3}$

The empirical probability is $\frac{\text{number of times the event occurred in the previous replications}}{\text{total number of previous replications}}$.

P(bad placement) $=\frac{13}{1,000}=1.3\mathrm{\%<!--\; \%\; -->}$.

$\frac{13}{1,000}=1.3\mathrm{\%<!--\; \%\; -->}$

theoretical

This is theoretical. You know that 500 people have tickets.

For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is $P(E)=\frac{n(E)}{n(S)}$, where n(E) and n(S) denote the number of outcomes in the event and in the sample space, respectively.

$P(\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{g})=\frac{1}{500}=0.002=0.2\mathrm{\%<!--\; \%\; -->}$

subjective

When you see “you believe,” you know this is not yet based on facts. It is the first day of school. The person has not yet seen the bus arrive and is guessing when it will arrive. This is subjective, which is based on instinct.

empirical

This is the last day of school. The person is basing the probability on previous replications of the experiment, having seen the bus arrive repeatedly between those times. This is empirical.

Of the 16 possible sums, 13 of them are greater than 3.

For an experiment whose sample space S consists of equally likely outcomes, the theoretical probability of the event E is $P(E)=\frac{n(E)}{n(S)}$, where n(E) and n(S) denote the number of outcomes in the event and in the sample space, respectively.

$P\left(\mathrm{g}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{n}3\right)=\frac{13}{16}$

$\frac{13}{16}$

The number of possible five-card hands out of a 52-card deck is $\text{52C5 = 2,598,960}$.

There are 13 cards in each suit.

The number of ways to draw 2 spades out of 13 spades is $\text{13C2 = 78}$.

The number of ways to draw 3 hearts out of 13 hearts is $\text{13C3 = 286}$.

Use the Multiplication Rule of Counting to count the ways to draw 2 spades and 3 hearts:

$13\text{C}2\times <!--\; \times \; -->13\text{C}3$

P(drawing 2 spades and 3 hearts) = $\frac{\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{d}\mathrm{r}\mathrm{a}\mathrm{w}2\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{d}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{n}\mathrm{d}3\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{s}}{\mathrm{w}\mathrm{a}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{o}\mathrm{d}\mathrm{r}\mathrm{a}\mathrm{w}5\mathrm{c}\mathrm{a}\mathrm{r}\mathrm{d}\mathrm{s}}=\frac{78\times <!--\; \times \; -->286}{2,598,960}\approx <!--\; \approx \; -->0.086\mathrm{\%<!--\; \%\; -->}$.

$\frac{{}_{13}{C}_2\times <!--\; \times \; -->{}_{13}{C}_3}{{}_{52}{C}_5}\approx <!--\; \approx \; -->0.86\mathrm{\%<!--\; \%\; -->}$

The sample space:

The total number of outcomes: 16.

The outcomes greater than three: 13.

The number of outcomes not in the desired event is $16-<!--\; -\; -->13=3$.

The ratio of the number of equally likely outcomes in an event $E$ to the number of equally likely outcomes not in the event is called the odds for (or odds in favor of) the event.

The odds in favor of your event are $\text{13: 3}$.

$13:3$

The sample space:

The total number of outcomes: 16.

The outcomes with both dice giving the same number: 4.

The number of outcomes not in the desired event is $16-<!--\; -\; -->4=12$.

The ratio of the number of equally likely outcomes in an event $E$ to the number of equally likely outcomes not in the event is called the odds for (or odds in favor of) the event.

The odds against your event are $\text{12:4}$ or $\text{3:1}$.

$12:4=3:1$

Odds for:

If the probability is 80%, then the odds for the event are $0.8:\left(1-<!--\; -\; -->0.8\right)$.

$0.8:0.2$

You can simplify the ratio by dividing both numbers by 0.8.

$1:0.25$

You can multiply again by 4 to get rid of decimals.

$4:1$

Odds against: (reverse the numbers).

$1:4$

The odds for $E$ are $4:1$ and the odds against $E$ are $1:4$.

The odds in favor of $E$ are $\text{1: 15}$.

If the odds in favor of $E$ are $A:B$, then

$P(E)=\frac{A}{A+B}$.

$P(E)=\frac{A}{A+B}=\frac{1}{15+1}=\frac{1}{16}$

$P(E)=\frac{1}{16}$

The odds in favor of $E$ are $2.5:1$.

If the odds in favor of $E$ are $A:B$, then

$P(E)=\frac{A}{A+B}$.

$P(E)=\frac{A}{A+B}=\frac{2.5}{2.5+1}=\frac{2.5}{3.5}=\frac{25}{35}=\frac{5}{7}\approx <!--\; \approx \; -->0.714$

$P(E)=\frac{2.5}{3.5}\approx <!--\; \approx \; -->0.714$

$E$ = {A hearts, A spades, A clubs, A diamonds}

$F$ = {K hearts, K spades, K clubs, K diamonds}

There are no outcomes in common.

These are mutually exclusive.

Mutually exclusive

They have the Ace of hearts in common.

They are not mutually exclusive.

Not mutually exclusive

$\frac{4}{10}+\frac{2}{10}=\frac{3}{5}$

Sample Space:

There are 10 cards, 4 of which are Aces and 2 are kings.

The Addition Rule for Mutually Exclusive Events:

If $E$ and $F$ are mutually exclusive events, then $\text{P}\left(E\text{or}F\right)=\text{ P}\left(E\right)+\text{ P}\left(F\right)$.

P(Ace or King) = $\frac{4}{10}+\frac{2}{10}=\frac{6}{10}=\frac{3}{5}$.

Not appropriate; the events are not mutually exclusive.

Sample Space:

There are 10 cards.

These are not mutually exclusive events. There is only one card that is both a heart and an Ace, and that is the Ace of hearts. It is not appropriate to use the Addition Rule, so we do not continue this exercise.

$\frac{3}{10}+\frac{2}{10}=\frac{1}{2}$

Sample Space:

There are 10 cards, three of which are hearts and two of which are kings.

(Note, the kings are a spade and a club.)

The Addition Rule for Mutually Exclusive Events:

If $E$ and $F$ are mutually exclusive events, then $\text{P}\left(E\text{or}F\right)=\text{ P}\left(E\right)+\text{ P}\left(F\right)$.

P(heart or king) = $\frac{3}{10}+\frac{2}{10}=\frac{5}{10}=\frac{1}{2}$.

Sample Space:

Orange: Orange 1, Orange 2. This is 2 outcomes.

Even: Orange 2, Orange 4. This is 2 outcomes.

Outcomes in common: Orange 2. This is 1 outcome in common.

You can use the Inclusion/Exclusion Principle. If $E$ and $F$ are not mutually exclusive, then

$\text{P}\left(E\text{or}F\right)=\text{ P}\left(E\right)+\text{ P}\left(F\right)-<!--\; -\; -->\text{ P}$ ($E$ and $F$).

P(orange or even) = P(orange) + P(even) – P(orange and even) = $\frac{2}{6}+\frac{2}{6}-<!--\; -\; -->\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$.

$\frac{1}{2}$

Sample Space:

Green: Green A, Green B. This is 2 outcomes.

A: Blue A, Green A. This is 2 outcomes.

Common: Green A. This is 1 outcome in common.

They are not mutually exclusive.

You can use the Inclusion/Exclusion Principle. If $E$ and $F$ are not mutually exclusive, then

$\text{P}\left(E\text{or}F\right)=\text{ P}\left(E\right)+\text{ P}\left(F\right)-<!--\; -\; -->\text{ P}$ ($E$ and $F$).

P(green or A) = P(green) + P(A) – P(green and A) = $\frac{2}{6}+\frac{2}{6}-<!--\; -\; -->\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$.

$\frac{1}{2}$

$\frac{2}{3}$

Sample Space:

Even: Orange 2, Orange 4, which makes 2 outcomes.

Green: Green A, Green B, which makes 2 outcomes.

They are mutually exclusive.

You can use the Addition Rule for Mutually Exclusive Events:

If $E$ and $F$ are mutually exclusive events, then $\text{P}\left(E\text{or}F\right)=\text{ P}\left(E\right)+\text{ P}\left(F\right)$.

P(even or green) = $\frac{2}{6}+\frac{2}{6}=\frac{4}{6}=\frac{2}{3}$.

$\frac{1}{3}$

Sample space:

What is the probability of a vowel given that an orange has been rolled?

An orange has been rolled, which means that one of these sides is the result.

Vowels: E.

From these choices, $\text{P}\left(\text{letter us a vowel | the number is orange}\right)=\frac{1}{3}$.

$1$

Sample space:

What is the probability of a 1 given that a red has been rolled?

A red has been rolled, meaning that one of these sides is the result.

Both sides have a 1.

From these choices, $\text{P}\left(\text{the number is a 1 | the number is red}\right)=\frac{1}{1}=1$.

$\frac{2}{3}$

Sample space:

What is the probability of a red given that a 1 has been rolled?

A red has been rolled, which means that one of these sides is the result.

Two of these sides are red.

From these choices, $\text{P}\left(\text{the number is red | the number is a 1}\right)=\frac{2}{3}$.

Sample space:

The Multiplication Rule for Probability: If $E$ and $F$ are events associated with the first and second stages of an experiment, then $\text{P}\left(E\text{and}F\right)=\text{ P}\left(E\right)\times <!--\; \times \; -->\text{ P}\left(F|E\right)$.

There are 3 hearts to be drawn first out of 8 cards.

There are two spades to be drawn second out of 7 cards.

$\text{P(heart, then spade) = }\frac{3}{8}\left(\frac{2}{7}\right)=\frac{6}{56}=\frac{3}{28}$.

$\frac{3}{28}$

Sample space:

The Multiplication Rule for Probability: If $E$ and $F$ are events associated with the first and second stages of an experiment, then $\text{P}\left(E\text{and}F\right)=\text{ P}\left(E\right)\times <!--\; \times \; -->\text{ P}\left(F|E\right)$.

There are 3 hearts to be drawn first out of 8 cards.

For the second draw, there are 2 remaining hearts to be drawn out of the remaining 7 cards.

$\text{P(9, then a 6) = }\frac{3}{8}\left(\frac{1}{7}\right)=\frac{3}{56}$.

$\frac{3}{56}$

Sample space:

The Multiplication Rule for Probability: If $E$ and $F$ are events associated with the first and second stages of an experiment, then $\text{P}\left(E\text{and}F\right)=\text{ P}\left(E\right)\times <!--\; \times \; -->\text{ P}\left(F|E\right)$.

There are 3 nines to be drawn first out of 8 cards.

There is 1 six be drawn second out of 7 cards.

$\text{P(a heart, then another heart) = }\frac{3}{8}\left(\frac{2}{7}\right)=\frac{6}{56}=\frac{3}{28}$.

$\frac{3}{28}$

A table will be used to represent the tree.

There are 12 total outcomes. The probabilities are in the last column of the table.

A binomial trial has two outcomes. A binomial experiment is an experiment with a fixed number of repeated binomial trials, where each trial has the same probability of success.

This is not binomial because it has more than two outcomes.

Not binomial (more than two outcomes)

A binomial trial has two outcomes. A binomial experiment is an experiment with a fixed number of repeated binomial trials, where each trial has the same probability of success.

This is not binomial because there are more than two outcomes.

Not binomial (not independent)

A binomial trial has two outcomes. A binomial experiment is an experiment with a fixed number of repeated binomial trials, where each trial has the same probability of success.

This is binomial because there are two outcomes (club or not club) and a fixed number of trials.

Binomial

A binomial trial has two outcomes. A binomial experiment is an experiment with a fixed number of repeated binomial trials, where each trial has the same probability of success.

This is not binomial because the number of trials was not determined ahead of time.

Not binomial (number of trials isn’t fixed)

Binomial formula: For a binomial experiment with n trials and p probability of success in each trial, then P(number of a successes) = $nCa\times <!--\; \times \; -->p^a\times <!--\; \times \; -->{\left(1-<!--\; -\; -->p\right)}^{n-<!--\; -\; -->a}$.

There is one 2 out of 4 sides, so p is $\frac{1}{4}$.

P(exactly four 2s in 10 rolls) =$\left(10C4\right){\left(\frac{1}{4}\right)}^4{\left(1-<!--\; -\; -->\frac{1}{4}\right)}^{10-<!--\; -\; -->4}$.

$\left(210\right){\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^6=\frac{153,090}{1,048,576}\approx <!--\; \approx \; -->0.146$

0.146

Binomial formula: For a binomial experiment with n trials and p probability of success in each trial, then P(number of a successes) = $nCa\times <!--\; \times \; -->p^a\times <!--\; \times \; -->{\left(1-<!--\; -\; -->p\right)}^{n-<!--\; -\; -->a}$.

There is one 2 out of 4 sides, so p is $\frac{1}{4}$.

P(exactly four 2s in 20 rolls) =$\left(20C4\right){\left(\frac{1}{4}\right)}^4{\left(1-<!--\; -\; -->\frac{1}{4}\right)}^{20-<!--\; -\; -->4}$.

$\left(4,845\right){\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^{16}\approx <!--\; \approx \; -->0.190$

0.190

Binomial formula: For a binomial experiment with n trials and p probability of success in each trial, then P(number of a successes) = $nCa\times <!--\; \times \; -->p^a\times <!--\; \times \; -->{\left(1-<!--\; -\; -->p\right)}^{n-<!--\; -\; -->a}$.

There is one 2 out of 4 sides, so p is $\frac{1}{4}$.

P(exactly four 2s in 20 rolls) =$\left(30C4\right){\left(\frac{1}{4}\right)}^4{\left(1-<!--\; -\; -->\frac{1}{4}\right)}^{30-<!--\; -\; -->4}$.

$\left(27,405\right){\left(\frac{1}{4}\right)}^4{\left(\frac{3}{4}\right)}^{26}\approx <!--\; \approx \; -->0.060$

0.060

Google Sheets can be used to compute cumulative binomial distributions. The event is sides less than or equal to 5.

There are two sides greater than 4: 5 and 6. The probability is $\frac{2}{6}=\frac{1}{3}\approx <!--\; \approx \; -->0.333333$.

In Google Sheets, enter “=BINOMDIST(5, 20, 0.33333, TRUE)” to get your answer: 0.02972.

0.2972

This uses the wrong inequality. Instead of S < 8, you want S ≤ 7.

There are two sides greater than 4: 5 and 6. The probability is $\frac{2}{6}=\frac{1}{3}\approx <!--\; \approx \; -->0.333333$.

In Google Sheets, enter “=BINOMDIST(7, 20, 0.33333, TRUE)” to get your answer: 0.6615.

0.6615

This goes in the wrong direction, so you will use complements. First find S ≤ 6.

In Google Sheets, enter “=BINOMDIST(6, 20, 0.33333, TRUE)” to get the answer.0.4793554…

Your answer is $1-<!--\; -\; -->0.4793=0.5207$.

0.5207

You are looking for the outcome {6,7}.

To do this, you will find S ≤ 7 and subtract S ≤ 5.

In Google Sheets, enter “=BINOMDIST(7, 20, 0.33333, TRUE)” to get the answer 0.66148…

In Google Sheets, enter “=BINOMDIST(5, 20, 0.33333, TRUE)” to get the answer 0.2972248…

Subtract the two answers to find the chance of 5 < S < 8 is approximately 0.3643.

0.3643

$\frac{10}{3}$

If O represents an outcome of an experiment and n(O) the value of that outcome, then the expected value of the experiment is $\sum <!--\; \sum \; -->n\left(O\right)\times <!--\; \times \; -->P\left(O\right)$.

The sides of the die each has a probability of $\frac{1}{6}$. Multiply the value showing on each face times the probability of each side. Do not simplify the fractions because it makes it easier to add the fractions to keep this way.

The sum of the numbers in the last column is the expected value = $\frac{20}{6}=\frac{10}{3}$.

The expected value is $\frac{10}{3}$.

$\frac{3}{2}$

The sum of the numbers in the last column is the expected value = $\frac{12}{8}=\frac{3}{2}$.

$\frac{25}{4}$ = $6.25

The sum of the numbers in the last column is the expected value = $\frac{50}{8}=\frac{25}{4}$= $6.25.

From the previous exercise, you know the expected value is $\frac{10}{3}$, which is approximately 3.33. If you roll the die repeatedly, you expect the mean of the numbers showing would be around 3.33.

If you roll the special die many times, the mean of the numbers showing will be around 3.33.

From the previous exercise, you know the expected value is $\frac{3}{2}$, or 1.5. If you repeatedly flip a coin three times, you can expect the average number of heads to be 1.5 heads.

If you repeat the coin-flipping experiment many times, the mean of the number of heads you get will be around 1.5.

Possible sums when rolling two dice:

There are 36 possible outcomes. Six of those outcomes have a sum of 7.

30 outcomes do not have a sum of 7.

The expected value of a bet on 7 is $-<!--\; -\; -->\frac{1}{6}$or a loss of $0.17.

If the player bets on 7, the expected value is $-<!--\; -\; -->\mathrm{\$<!--\; \$\; -->}0.17$.

Possible sums when rolling two dice:

There are 36 possible outcomes. Only one of those outcomes has a sum of 12.

35 outcomes do not have a sum of 12.

The expected value of a bet on 12 is $-<!--\; -\; -->\frac{5}{36}$or a loss of $0.14.

If the player bets on 12, the expected value is $-<!--\; -\; -->\mathrm{\$<!--\; \$\; -->}0.14$.

Possible sums when rolling two dice:

There are 36 possible outcomes. Four of those outcomes have a sum of 2, 3, or 12.

32 outcomes do not have a sum of 12.

The expected value of a bet on 2, 3, or 12 is $-<!--\; -\; -->\frac{1}{9}$or a loss of $0.11.

With all the bets, you lose money over the long run. You are better off not betting at all. If you must bet, you lose money at the slowest rate with the last bet. The “any craps” bet loses $0.11 on average per bet. Regarding the casino’s interests, the casino is better off with the bet on 7, since players lose the most with that bet.

If the player bets on any craps, the expected value is $-<!--\; -\; -->\mathrm{\$<!--\; \$\; -->}0.11$.
The best bet for the player is any craps; the best bet for the casino is the bet on 7.

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

$18\times <!--\; \times \; -->30=540$

540

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

You have two choices of how many chicken strips you get (three or five).

You have 7 choices of side dishes.

$2\times <!--\; \times \; -->7=14$

14

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

Each time you flip, you have two choices.

You flip ten times, so you have two choices each flip.

$2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2\times <!--\; \times \; -->2=1,024$

1,024

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

$5\times <!--\; \times \; -->8\times <!--\; \times \; -->4\times <!--\; \times \; -->5=800$

800

The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

Multiply the choices during each class period.

$15\times <!--\; \times \; -->18\times <!--\; \times \; -->12\times <!--\; \times \; -->8=25,920$

25,920

A factorial for any positive whole number n (denoted n!) is the product of every whole number less than or equal to n. We define 0! to be equal to one.

$5!=5\times <!--\; \times \; -->4\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1=120$

120

A factorial for any positive whole number n (denoted n!) is the product of every whole number less than or equal to n. We define 0! to be equal to one.

$\frac{10!}{7!3!}=\frac{10\times <!--\; \times \; -->9\times <!--\; \times \; -->8\times <!--\; \times \; -->\cancel{7!}}{\cancel{7!}\times <!--\; \times \; -->3\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{720}{6}=120$

120

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$12\mathrm{P}3=\frac{n!}{\left(n-<!--\; -\; -->r\right)!}=\frac{12!}{\left(12-<!--\; -\; -->3\right)!}=\frac{12!}{9!}=\frac{12\times <!--\; \times \; -->11\times <!--\; \times \; -->10\times <!--\; \times \; -->\cancel{9!}}{\cancel{9!}}=1,320$

1,320

1,680

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$8\mathrm{P}4=\frac{8!}{\left(8-<!--\; -\; -->4\right)!}=\frac{8!}{4!}=\frac{8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->5\times <!--\; \times \; -->\cancel{4!}}{\cancel{4!}}=1,680$

The permutation of n objects taken r at a time is $\mathrm{n}\mathrm{P}\mathrm{r}=\frac{n!}{(n-<!--\; -\; -->r)!}$where 1 ≤ r ≤ n.

$15\mathrm{P}4=\frac{n!}{\left(n-<!--\; -\; -->r\right)!}=\frac{15!}{\left(15-<!--\; -\; -->4\right)!}=\frac{15!}{11!}=\frac{15\times <!--\; \times \; -->14\times <!--\; \times \; -->13\times <!--\; \times \; -->12\times <!--\; \times \; -->\cancel{11!}}{\cancel{11!}}=32,760$

${}_{15}{P}_4=\mathrm{32,760}$

Anytime you see the word “arrange,” you know that order matters. This is a permutation.

permutations

The only thing that matters is which toys go on the trip. This is a combination.

combinations

The number of combinations of n objects taken r at a time is $nCr=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}$.

The answer will always be an integer, so check your math if you get a fraction.

$12C10=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{12!}{10!\left(12-<!--\; -\; -->10\right)!}=\frac{12!}{10!2!}=\frac{12\times <!--\; \times \; -->11\times <!--\; \times \; -->\cancel{10!}}{\cancel{10!}\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{132}{2}=66$

66

The number of combinations of n objects taken r at a time is $nCr=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}$.

The answer will always be an integer, so check your math if you get a fraction.

$16C3=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{16!}{3!\left(16-<!--\; -\; -->3\right)!}=\frac{13!}{3!13!}=\frac{16\times <!--\; \times \; -->15\times <!--\; \times \; -->14\times <!--\; \times \; -->\cancel{13!}}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{13!}}=\frac{3,360}{6}=560$

560

This is a combination. It doesn’t matter which order you choose the 3 friends out of 6 friends.

The number of combinations of n objects taken r at a time is $nCr=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}$.

$6C3=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{6!}{3!\left(6-<!--\; -\; -->3\right)!}=\frac{6!}{3!3!}=\frac{6\times <!--\; \times \; -->5\times <!--\; \times \; -->4\times <!--\; \times \; -->\cancel{3!}}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{3!}}=\frac{120}{6}=20$

There are 20 ways to pick three other people.

20

It does not matter in which order you pack the shirts and skirts. They will all be with you on the trip.

Then, use the Multiplication Rule of Counting. The Multiplication Rule of Counting says that if there are n ways to accomplish one task and m ways to accomplish a second task, then there are $n\times <!--\; \times \; -->m$ ways to accomplish both tasks. You can tack on additional tasks by multiplying the number of ways to accomplish those tasks to your previous product.

The number of ways to have the required hand is $8\text{C}3\times <!--\; \times \; -->6\text{C}3$.

Evaluate each combination separately.

$8C3=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{8!}{3!\left(8-<!--\; -\; -->5\right)!}=\frac{8!}{3!5!}=\frac{8\times <!--\; \times \; -->7\times <!--\; \times \; -->6\times <!--\; \times \; -->\cancel{5!}}{3\times <!--\; \times \; -->2\times <!--\; \times \; -->1\times <!--\; \times \; -->\cancel{5!}}=\frac{336}{6}=56$

$5C3=\frac{n!}{r!\left(n-<!--\; -\; -->r\right)!}=\frac{5!}{3!\left(5-<!--\; -\; -->3\right)!}=\frac{5!}{3!2!}=\frac{5\times <!--\; \times \; -->4\times <!--\; \times \; -->\cancel{3!}}{\cancel{3!}\times <!--\; \times \; -->2\times <!--\; \times \; -->1}=\frac{20}{2}=10$

Now you can evaluate $8\text{C}3\times <!--\; \times \; -->5\text{C}3=56\times <!--\; \times \; -->10=560$.

There are 560 ways to choose 3 shirts and 3 skirts.

560

The minimum number of heads is none. The maximum number of heads is 6. Every number between 0 and 6 is possible.

The sample space: {0, 1, 2, 3, 4, 5, 6}

{0, 1, 2, 3, 4, 5, 6}

$P(\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{d}\mathrm{s})=P\left(\mathrm{H}\mathrm{H}\right)=\frac{1}{4}$

$\frac{1}{4}$

$P(\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{d}\mathrm{s})=P\left(\mathrm{H}\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{H}\mathrm{T}\right)=\frac{2}{4}=\frac{1}{2}$

$\frac{1}{2}$

$P\left(\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}\right)=P\left(\mathrm{H}\mathrm{T}\mathrm{o}\mathrm{r}\mathrm{T}\mathrm{H}\right)=\frac{2}{4}=\frac{1}{2}$

$\frac{1}{2}$

$P(\mathrm{t}\mathrm{h}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{s})=0$ There are only two tosses.

0

This is theoretical because you know the composition of a deck and can calculate the chance of drawing a particular card.

theoretically

Subjective. There is no good way to make measurements or collect data for this probability. A subjective probability is an assignment of a probability to an event using only one’s instincts.

subjectively

Empirical. They are using data from replications to calculate from experience, not from theoretical expected values.

empirically

$\text{number of heads}>20$

The complement of E is the collection of all the outcomes that are not in E.

E is 20 or fewer heads.

The complement of E is the number of heads is > 20.

The P(complement of E) = 1 – P(E) = 1 – 10.1 = 89.9%.

89.9%

The total number of outcomes: 6.

The outcomes with a green face: 2.

The number of outcomes not in the desired event is $6-<!--\; -\; -->2=4$.

The odds in favor of rolling a green face are $2:4$ or $1:2$.

$1:2$

The total number of outcomes is 6.

The number of outcomes with a blue face is 1.

The number of outcomes not in the desired event is $6-<!--\; -\; -->1=5$.

The odds against rolling a blue face are $5:1$.

$5:1$

The total number of outcomes is 6.

The number of outcomes with an orange face is 3.

The number of outcomes not in the desired event is $6-<!--\; -\; -->3=3$.

The odds in favor of rolling an orange face is $3:3$ or $1:1$.