Chapter 7: ProbabilityContent check — Contemporary Math, Chapter 7: Probability

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 $n\times m$ 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.

$4\times 15=60$

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 $n\times m$ 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.

$4\times 15\times 3\times 2=360$

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 $n\times m$ 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.

$3\times 10\times 10\times 10\times 10\times 10\times 26\times 26\times 26=5,272,800,000$

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 $n\times m$ 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.

$18\times 30=540$

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 $n\times m$ 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.

$2\times 7=14$

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 $n\times m$ 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.

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=1,024$

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 $n\times m$ 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.

$5\times 8\times 4\times 5=800$

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 $n\times m$ 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.

$15\times 18\times 12\times 8=25,920$

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 $n\times m$ 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 $5\times 4\times 3\times 2\times 1=120$.

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.

$6!=6\times 5\times 4\times 3\times 2\times 1=720$

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.

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

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

$\frac{12!}{10!}=\frac{12\times 11\times 10!}{10!}=\frac{12\times 11\times 10!}{10!}=132$

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.

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

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.

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

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.

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

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.

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

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.

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

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

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.

$5!=5\times 4\times 3\times 2\times 1=120$

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.

$\frac{10!}{7!3!}=\frac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}=\frac{720}{6}=120$

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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 $n\times m$ 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 $13\text{C}3\times 13\text{C}2\times 13\text{C}1$.

Evaluate each combination separately.

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

$13C2=\frac{n!}{r!\left(n-r\right)!}=\frac{13!}{2!\left(13-2\right)!}=\frac{13!}{2!11!}=\frac{13\times 1211!}{2\times 1\times 11!}=\frac{156}{2}=78$

$13C1=\frac{n!}{r!\left(n-r\right)!}=\frac{13!}{1!\left(13-1\right)!}=\frac{13!}{1!12!}=\frac{1312!}{1\times 12!}=13$

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

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.

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

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.

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

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 $nCr=\frac{n!}{r!\left(n-r\right)!}$.

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

There are 20 ways to pick three other people.

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 $n\times m$ 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 $8\text{C}3\times 6\text{C}3$.

Evaluate each combination separately.

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

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

Now you can evaluate $8\text{C}3\times 5\text{C}3=56\times 10=560$.

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

The sample space is the set of all possible outcomes.

{hearts, spades, clubs, diamonds}

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}

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

{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}.

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 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.

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) }

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$\mathrm{♡}$)}

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}

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.

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

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.

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

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.

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

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) = $\frac{2}{6}=\frac{1}{3}$.

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) $=\frac{3}{6}=\frac{1}{2}$.

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) = $\frac{4}{6}=\frac{2}{3}$.

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) $=\frac{13}{1,000}=1.3\mathrm{\%}$.

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.

$P(\mathrm{w}\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{i}\mathrm{n}\mathrm{g})=\frac{1}{500}=0.002=0.2\mathrm{\%}$

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.

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

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

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

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

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

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 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:

$13\text{C}2\times 13\text{C}3$

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

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 $\text{13: 3}$.

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 $\text{3:1}$.

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

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

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

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

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

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

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

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

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 $1:2$.

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 $5:1$.

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 $1:1$.

Odds for:

If the probability is $\frac{3}{8}$ then the odds for the event are $\frac{3}{8}:\left(1-\frac{3}{8}\right)$.

$\frac{3}{8}:\frac{5}{8}$

Multiply both numbers by 8 to clear fractions.

$\text{3:5}$

Odds for:

If the probability is $\frac{2}{13}$ then the odds for the event are $\frac{2}{13}:\left(1-\frac{2}{13}\right)$.

$\frac{2}{13}:\frac{11}{13}$

Multiply both numbers by 13 to clear fractions.

$\text{2:11}$

Reverse the ratio to find the odds against the event.

$\text{11: 2}$

The odds in favor of $E$ are $\text{4:9}$.

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

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

$P(E)=\frac{A}{A+B}=\frac{4}{4+9}=\frac{4}{13}$

The odds in favor of $E$ are $\text{5:7}$.

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

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

$P(E)=\frac{A}{A+B}=\frac{5}{5+7}=\frac{5}{12}$

$E$ = {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.

They have the Ace of hearts in common.

They are not mutually exclusive.

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) = $\frac{4}{10}+\frac{2}{10}=\frac{6}{10}=\frac{3}{5}$.

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.

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) = $\frac{2}{6}+\frac{2}{6}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$.

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) = $\frac{2}{6}+\frac{2}{6}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$.

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) = $\frac{2}{6}+\frac{2}{6}=\frac{4}{6}=\frac{2}{3}$.

Sample Space:

These are mutually exclusive.

There are 10 cards.

You can use the Addition Rule for Mutually Exclusive Events:

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

There are 4 Aces and 2 kings.

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

Sample Space:

These are mutually exclusive.

There are 10 cards.

You can use the Addition Rule for Mutually Exclusive Events:

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

There are 4 spades and 2 clubs.

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

Sample Space:

Ace or heart are not mutually exclusive.

Aces: A heart, A spade, A club, A diamond.

Hearts: Q heart, J heart.

In common: Ace of hearts.

There are not mutually exclusive events.

There are 10 cards.

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

$\text{P}\left(E\text{or}F\right)=\text{ P}\left(E\right)-\text{ P}\left(F\right)$($E$ and $F$).

P(Ace or heart) = P(A) + P(heart) – P(A and heart) = $\frac{4}{10}+\frac{3}{10}-\frac{1}{10}=\frac{6}{10}=\frac{3}{5}$.