MTH120 College Algebra Chapter 9.7

 9.7 Probability

Constructing probability models involves defining the components of a probabilistic system, specifying the possible outcomes, and assigning probabilities to those outcomes. Probability models are used to represent and analyze uncertain situations. Here's a step-by-step guide on how to construct probability models:

1. Identify the Random Experiment: Start by identifying the random experiment or situation you want to model. This could be anything from rolling a die to predicting stock market trends.

2. Define the Sample Space (S): The sample space is the set of all possible outcomes of the random experiment. It represents the entire range of possible results. For example, if you're rolling a fair six-sided die, the sample space is $\{1,2,3,4,5,6\}$.

3. Define Events: Events are specific outcomes or combinations of outcomes from the sample space. Events are represented as subsets of the sample space. For example, if you want to model the event of rolling an even number, the event could be $�=\{2,4,6\}$.

4. Assign Probabilities: Assign probabilities to each event. The probabilities should reflect the likelihood of each event occurring. In many cases, for simple models, you may assume that all outcomes are equally likely (equiprobable). For example, for a fair six-sided die, each outcome has a probability of $\frac{1}{6}$.

5. Verify the Probabilities: Ensure that the probabilities assigned to events add up to 1. In other words, the sum of the probabilities of all possible events in the sample space should be equal to 1.

6. Use Mathematical Notation: Use mathematical notation to represent the sample space, events, and probabilities. For example, you can use set notation and conditional probability notation. If $�$ and $�$ are events, $�(�)$ represents the probability of event $�$, and $�(�\mathrm{\mid }�)$ represents the conditional probability of event $�$ given that event $�$ has occurred.

7. Perform Calculations: Once you've constructed the probability model, you can perform various calculations, such as finding the probability of specific events, using conditional probability to analyze situations, and applying the rules of probability, such as the addition and multiplication rules.

8. Interpret the Results: Interpret the results in the context of the problem you're modeling. Probability models allow you to make informed decisions, assess risk, and understand the likelihood of different outcomes.

Probability models can range from simple, discrete models like coin tossing and dice rolling to more complex, continuous models like modeling stock prices or weather patterns. The key is to accurately define the sample space and assign appropriate probabilities to events based on your understanding of the system being modeled.

Computing probabilities of equally likely outcomes is relatively straightforward when each outcome in the sample space has the same probability of occurring. In such cases, you can calculate the probability of an event by dividing the number of favorable outcomes by the total number of possible outcomes. Here are some examples:

Example 1: Tossing a Fair Coin

In this example, when you toss a fair coin, there are two equally likely outcomes: heads (H) and tails (T). The sample space is $\{�,�\}$, and the probability of each outcome is $\frac{1}{2}$ because both outcomes are equally likely.

1. What is the probability of getting heads (H) when tossing the coin?

   • Probability of heads (H) = $\frac{1}{2}$ or 50%.

2. What is the probability of getting tails (T) when tossing the coin?

   • Probability of tails (T) = $\frac{1}{2}$ or 50%.

Example 2: Rolling a Fair Six-Sided Die

In this example, when you roll a fair six-sided die, there are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. The sample space is $\{1,2,3,4,5,6\}$, and the probability of each outcome is $\frac{1}{6}$ because all six outcomes are equally likely.

1. What is the probability of rolling a 3?

   • Probability of rolling a 3 = $\frac{1}{6}$.

2. What is the probability of rolling an even number (2, 4, or 6)?

   • Probability of rolling an even number = Probability of rolling a 2 or a 4 or a 6 = $\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$ or 50%.

3. What is the probability of rolling a number greater than 4 (5 or 6)?

   • Probability of rolling a number greater than 4 = Probability of rolling a 5 or a 6 = $\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$ or approximately 33.33%.

In both of these examples, the probabilities are straightforward to calculate because all outcomes are equally likely. You can extend this approach to more complex situations by ensuring that you correctly identify the sample space and the number of favorable outcomes for the event of interest.

To compute the probability of an event with equally likely outcomes, you can use the following formula:

$�(�)=\frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

Here's how to use this formula with examples:

Example 1: Tossing a Fair Six-Sided Die

In this example, let's calculate the probability of rolling a 4 when tossing a fair six-sided die. All six outcomes (1, 2, 3, 4, 5, 6) are equally likely.

1. Number of Favorable Outcomes: There is one favorable outcome, which is rolling a 4.
2. Total Number of Possible Outcomes: There are six possible outcomes (the numbers 1 through 6) on the die.

Now, use the formula to calculate the probability:

$�(\text{Rolling a 4})=\frac{1}{6}$

So, the probability of rolling a 4 is $\frac{1}{6}$, which is approximately 16.67%.

Example 2: Flipping a Fair Coin

In this example, let's calculate the probability of getting heads (H) when flipping a fair coin. There are two equally likely outcomes: heads (H) and tails (T).

1. Number of Favorable Outcomes: There is one favorable outcome, which is getting heads (H).
2. Total Number of Possible Outcomes: There are two possible outcomes (heads and tails) when flipping a coin.

Now, use the formula to calculate the probability:

$�(\text{Getting Heads})=\frac{1}{2}$

So, the probability of getting heads is $\frac{1}{2}$, which is 50%.

Example 3: Drawing a Card from a Standard Deck

In this example, let's calculate the probability of drawing a red card (hearts or diamonds) from a standard deck of 52 cards. There are 26 red cards (13 hearts and 13 diamonds) and 52 total cards in the deck.

1. Number of Favorable Outcomes: There are 26 favorable outcomes (red cards) in the deck.
2. Total Number of Possible Outcomes: There are 52 possible outcomes (total number of cards) in the deck.

Now, use the formula to calculate the probability:

$�(\text{Drawing a Red Card})=\frac{26}{52}=\frac{1}{2}$

So, the probability of drawing a red card is $\frac{1}{2}$, which is 50%.

In each of these examples, the probability is calculated by dividing the number of favorable outcomes (outcomes that result in the event of interest) by the total number of possible outcomes. This approach works well when all outcomes are equally likely.

To compute the probability of the union of two events (the probability that either one of the two events occurs), you can use the Addition Rule for Probability. The formula for the probability of the union of two events, $�(�\cup �)$, is:

$�(�\cup �)=�(�)+�(�)-�(�\cap �)$

Here are some examples of how to use this formula:

Example 1: Rolling a Fair Six-Sided Die

In this example, let's calculate the probability of rolling either an even number or a number greater than 4 when rolling a fair six-sided die. The events are:

• Event $�$: Rolling an even number (2, 4, or 6).
• Event $�$: Rolling a number greater than 4 (5 or 6).

1. Probability of Event $�$:

   • The probability of rolling an even number is $\frac{3}{6}$ because there are three favorable outcomes (2, 4, 6) out of six possible outcomes.
   • $�(�)=\frac{1}{2}$ or 50%.

2. Probability of Event $�$:

   • The probability of rolling a number greater than 4 is $\frac{2}{6}$ because there are two favorable outcomes (5, 6) out of six possible outcomes.
   • $�(�)=\frac{1}{3}$ or approximately 33.33%.

3. Probability of the Intersection of Events $�$ and $�$ (rolling an even number and a number greater than 4):

   • The intersection of $�$ and $�$ is the outcome 6, which is the only outcome that satisfies both events.
   • $�(�\cap �)=\frac{1}{6}$.

Now, use the Addition Rule to calculate the probability of either event $�$ or event $�$ occurring:

$�(�\cup �)=�(�)+�(�)-�(�\cap �)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{3}{6}+\frac{2}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$

So, the probability of rolling either an even number or a number greater than 4 is $\frac{2}{3}$ or approximately 66.67%.

Example 2: Drawing Cards from a Deck

In this example, let's calculate the probability of drawing either a red card (hearts or diamonds) or a face card (jack, queen, or king) from a standard deck of 52 cards.

1. Probability of Event $�$: Drawing a red card (hearts or diamonds).

   • There are 26 red cards and 52 total cards, so $�(�)=\frac{26}{52}=\frac{1}{2}$.

2. Probability of Event $�$: Drawing a face card (jack, queen, or king).

   • There are 12 face cards (3 in each of the four suits), so $�(�)=\frac{12}{52}=\frac{3}{13}$.

3. Probability of the Intersection of Events $�$ and $�$ (drawing a red face card):

   • There are 6 red face cards (2 in each of the two red suits), so $�(�\cap �)=\frac{6}{52}=\frac{3}{26}$.

Now, use the Addition Rule to calculate the probability of either event $�$ or event $�$ occurring:

$�(�\cup �)=�(�)+�(�)-�(�\cap �)=\frac{1}{2}+\frac{3}{13}-\frac{3}{26}$

You can simplify this expression to get the final probability.

In both of these examples, the Addition Rule for Probability is used to calculate the probability of the union of two events. It involves finding the probabilities of each event, the probability of their intersection, and then applying the formula.

To compute the probability of mutually exclusive events (events that cannot occur simultaneously), you can simply add the individual probabilities of each event. This is because the probability of both events occurring at the same time is zero for mutually exclusive events. Here are some examples:

Example 1: Tossing a Fair Coin

In this example, let's calculate the probability of either getting heads (H) or tails (T) when flipping a fair coin. These two events, getting heads and getting tails, are mutually exclusive because you can't get both at the same time.

1. Probability of Getting Heads: $�(\text{H})=\frac{1}{2}$ (50%)
2. Probability of Getting Tails: $�(\text{T})=\frac{1}{2}$ (50%)

Since the events are mutually exclusive, you can simply add their probabilities:

$�(\text{H or T})=�(\text{H})+�(\text{T})=\frac{1}{2}+\frac{1}{2}=1$

So, the probability of getting either heads or tails when flipping a coin is 1, which means it's certain (100%).

Example 2: Rolling a Fair Six-Sided Die

In this example, let's calculate the probability of either rolling an even number (2, 4, or 6) or an odd number (1, 3, or 5) when rolling a fair six-sided die. These two events are mutually exclusive because a single roll cannot result in both an even and an odd number simultaneously.

1. Probability of Rolling an Even Number: $�(\text{Even})=\frac{3}{6}=\frac{1}{2}$ (50%)
2. Probability of Rolling an Odd Number: $�(\text{Odd})=\frac{3}{6}=\frac{1}{2}$ (50%)

Since the events are mutually exclusive, you can add their probabilities:

$�(\text{Even or Odd})=�(\text{Even})+�(\text{Odd})=\frac{1}{2}+\frac{1}{2}=1$

So, the probability of rolling either an even number or an odd number when rolling a six-sided die is 1, which means it's certain (100%).

In both of these examples, the probabilities of mutually exclusive events are added together because they cannot occur simultaneously. The sum of their probabilities will always be 1, indicating certainty that at least one of the events will occur.

The probability of the union of mutually exclusive events can be calculated by simply adding the individual probabilities of the events. Mutually exclusive events are events that cannot occur simultaneously, so there is no need to subtract the probability of their intersection. Here's the formula:

$�(�\cup �)=�(�)+�(�)$

Where:

• $�(�\cup �)$ is the probability of the union of events $�$ and (B).
• $�(�)$ is the probability of event (A).
• $�(�)$ is the probability of event (B).

Here are a couple of examples to illustrate this:

Example 1: Tossing a Fair Coin

In this example, let's calculate the probability of either getting heads (H) or tails (T) when flipping a fair coin. These two events, getting heads and getting tails, are mutually exclusive.

1. Probability of Getting Heads: $�(\text{H})=\frac{1}{2}$ (50%)
2. Probability of Getting Tails: $�(\text{T})=\frac{1}{2}$ (50%)

Now, use the formula to find the probability of either event occurring:

$�(\text{H or T})=�(\text{H})+�(\text{T})=\frac{1}{2}+\frac{1}{2}=1$

So, the probability of getting either heads or tails when flipping a coin is 1, which means it's certain (100%).

Example 2: Rolling a Fair Six-Sided Die

In this example, let's calculate the probability of either rolling an even number (2, 4, or 6) or an odd number (1, 3, or 5) when rolling a fair six-sided die. These two events are mutually exclusive.

1. Probability of Rolling an Even Number: $�(\text{Even})=\frac{3}{6}=\frac{1}{2}$ (50%)
2. Probability of Rolling an Odd Number: $�(\text{Odd})=\frac{3}{6}=\frac{1}{2}$ (50%)

Use the formula to find the probability of either event occurring:

$�(\text{Even or Odd})=�(\text{Even})+�(\text{Odd})=\frac{1}{2}+\frac{1}{2}=1$

So, the probability of rolling either an even number or an odd number when rolling a six-sided die is 1, which means it's certain (100%).

In both of these examples, the probabilities of mutually exclusive events are added together to find the probability of their union. This is because the events cannot occur simultaneously, and the sum of their probabilities will always be 1, indicating certainty that at least one of the events will occur.

The Complement Rule is a fundamental principle in probability that allows you to calculate the probability of an event not happening (the complement of an event). It states that the probability of the complement of an event (${�}^{\mathrm{\prime }}$ or $\overline{�}$) is equal to 1 minus the probability of the event itself ($�$).

The Complement Rule can be expressed as follows:

$�({�}^{\mathrm{\prime }})=1-�(�)$

Here are some examples of how to use the Complement Rule to compute probabilities:

Example 1: Rolling a Fair Six-Sided Die

In this example, let's calculate the probability of not rolling a 1 when rolling a fair six-sided die. The event $�$ is rolling a 1.

1. Probability of Rolling a 1: $�(�)=\frac{1}{6}$ (There is one favorable outcome, 1, out of six possible outcomes.)

Now, use the Complement Rule to find the probability of not rolling a 1:

$�({�}^{\mathrm{\prime }})=1-�(�)=1-\frac{1}{6}=\frac{5}{6}$

So, the probability of not rolling a 1 when rolling a six-sided die is $\frac{5}{6}$, which is approximately 83.33%.

Example 2: Drawing Cards from a Standard Deck

In this example, let's calculate the probability of not drawing a spade when drawing a card from a standard deck of 52 cards. The event $�$ is drawing a spade.

1. Probability of Drawing a Spade: $�(�)=\frac{13}{52}$ (There are 13 spades out of 52 total cards in the deck.)

Now, use the Complement Rule to find the probability of not drawing a spade:

$�({�}^{\mathrm{\prime }})=1-�(�)=1-\frac{13}{52}=1-\frac{1}{4}=\frac{3}{4}$

So, the probability of not drawing a spade when drawing a card from a standard deck is $\frac{3}{4}$, which is 75%.

In both of these examples, the Complement Rule is used to find the probability of the complement of an event (not rolling a 1 or not drawing a spade) by subtracting the probability of the event itself from 1. This provides a straightforward way to calculate the probability of events that are defined by what does not happen.

Probability can be computed using counting theory, specifically the principles of combinatorics. Combinatorics helps calculate the number of possible outcomes, and by dividing the number of favorable outcomes by the total number of possible outcomes, you can find the probability of an event. Here are examples of using counting theory to compute probabilities:

Example 1: Probability of Drawing Cards

Suppose you want to find the probability of drawing two hearts in a row from a standard deck of 52 cards without replacement.

1. Total Number of Possible Outcomes: In a standard deck, there are 52 cards, and you draw one card at a time. So, the total number of possible outcomes for the first card is 52.

2. Number of Favorable Outcomes for the First Card: There are 13 hearts in the deck, so the number of favorable outcomes for the first card is 13.

3. After drawing the first heart, there are now 51 cards left in the deck, and 12 hearts remaining. So, for the second card, the number of favorable outcomes is 12.

Now, use combinatorics to calculate the probability:

• Probability of drawing a heart on the first draw = $\frac{13}{52}$
• Probability of drawing a heart on the second draw after drawing one heart = $\frac{12}{51}$

Since the events are independent (your first draw does not affect the second draw), you can multiply the probabilities:

$�(\text{Two Hearts in a Row})=\frac{13}{52}\cdot \frac{12}{51}=\frac{13}{68}$

Example 2: Probability of Arranging Letters

Suppose you want to find the probability of randomly arranging the letters of the word "PROBABILITY."

1. Total Number of Possible Arrangements: There are 11 letters in the word "PROBABILITY," so there are 11 possible positions for the first letter.

2. The first letter can be any one of the 11 letters.

3. After placing the first letter, there are 10 remaining letters for the second position.

Continue this process until you have placed all 11 letters.

Now, use combinatorics to calculate the probability:

• Probability of randomly arranging the letters = $\frac{1}{11}\cdot \frac{1}{10}\cdot \frac{1}{9}\cdot \dots \cdot \frac{1}{1}=\frac{1}{11!}$

In this example, you have a 1 in 11! (11 factorial) chance of arranging the letters in the exact order of "PROBABILITY."

These are examples of using combinatorics and counting principles to calculate probabilities in different scenarios. Counting theory is essential for finding the total number of outcomes and then determining the probability based on favorable outcomes.

College Algebra Chapter 9 Quiz

1. Write the first four terms of the sequence defined by the recursive formula a=–14,an=2+an–12. 2. Write the first four terms of the sequence defined by the explicit formula an=n2–n–1n!. 3. Is the sequence 0.3,1.2,2.1,3,… arithmetic? If so find the common difference. 4. An arithmetic sequence has the first term a1=−4 and common difference d=–43. What is the 6th term? 5. Write a recursive formula for the arithmetic sequence −2,−72,−5,−132,… and then find the 22nd term. 6. Write an explicit formula for the arithmetic sequence 15.6,15,14.4,13.8,… and then find the 32nd term. 7. Is the sequence −2,−1,−12,−14,… geometric? If so find the common ratio. If not, explain why. 8. What is the 11th term of the geometric sequence −1.5,−3,−6,−12,…? 9. Write a recursive formula for the geometric sequence 1,−12,14,−18,… 10. Write an explicit formula for the geometric sequence 4,−43,49,−427,… 11. Use summation notation to write the sum of terms 3k2−56k from k=−3 to k=15. 12. A community baseball stadium has 10 seats in the first row, 13 seats in the second row, 16 seats in the third row, and so on. There are 56 rows in all. What is the seating capacity of the stadium? 13. Use the formula for the sum of the first n terms of a geometric series to find ∑k=17−0.2⋅(−5)k−1. 14. Find the sum of the infinite geometric series ∑k=1∞13⋅(−15)k−1. 15. Rachael deposits $3,600 into a retirement fund each year. The fund earns 7.5% annual interest, compounded monthly. If she opened her account when she was 20 years old, how much will she have by the time she’s 55? How much of that amount was interest earned? 16. In a competition of 50 professional ballroom dancers, 22 compete in the fox-trot competition, 18 compete in the tango competition, and 6 compete in both the fox-trot and tango competitions. How many dancers compete in the fox-trot or tango competitions? 17. A buyer of a new sedan can custom order the car by choosing from 5 different exterior colors, 3 different interior colors, 2 sound systems, 3 motor designs, and either manual or automatic transmission. How many choices does the buyer have? 18. To allocate annual bonuses, a manager must choose his top four employees and rank them first to fourth. In how many ways can he create the “Top-Four” list out of the 32 employees? 19. A rock group needs to choose 3 songs to play at the annual Battle of the Bands. How many ways can they choose their set if have 15 songs to pick from? 20. A self-serve frozen yogurt shop has 8 candy toppings and 4 fruit toppings to choose from. How many ways are there to top a frozen yogurt? 21. How many distinct ways can the word EVANESCENCE be arranged if the anagram must end with the letter E? 22. Use the Binomial Theorem to expand (32x−12y)5. 23. Find the seventh term of (x2−12)13 without fully expanding the binomial.

1. The recursive formula is ${�}_1=-14$ and ${�}_{�}=2+{�}_{�-1}$. Let's find the first four terms:

   • ${�}_1=-14$
   • ${�}_2=2+{�}_1=2-14=-12$
   • ${�}_3=2+{�}_2=2-12=-10$
   • ${�}_4=2+{�}_3=2-10=-8$

2. The explicit formula is ${�}_{�}=\frac{{�}^2-�-1}{�!}$. Let's find the first four terms:

   • ${�}_1=\frac{1^2-1-1}{1!}=-1$
   • ${�}_2=\frac{2^2-2-1}{2!}=\frac{2}{2}=1$
   • ${�}_3=\frac{3^2-3-1}{3!}=\frac{5}{6}$
   • ${�}_4=\frac{4^2-4-1}{4!}=\frac{6}{24}=\frac{1}{4}$

3. The sequence $\{0.3,1.2,2.1,3,\dots \}$ is not arithmetic because the differences between consecutive terms are not constant.

4. In an arithmetic sequence, ${�}_{�}={�}_1+(�-1)�$, where ${�}_1=-4$ and $�=-43$. To find the 6th term (${�}_6$): ${�}_6=-4+(6-1)\cdot (-43)=-4+5\cdot (-43)=-4-215=-219$

5. The recursive formula for the sequence $\{-2,-72,-5,-132,\dots \}$ is ${�}_1=-2$ and ${�}_{�}={�}_{�-1}-70$. To find the 22nd term: ${�}_2=-2-70=-72$ ${�}_3=-72-70=-142$ Continue this process until you reach the 22nd term.

6. The explicit formula for the arithmetic sequence $\{15.6,15,14.4,13.8,\dots \}$ is ${�}_{�}=15.6-0.6(�-1)$. To find the 32nd term: ${�}_{32}=15.6-0.6(32-1)=15.6-0.6\cdot 31=15.6-18.6=-3$

7. The sequence $\{-2,-1,-12,-14,\dots \}$ is not geometric because the ratios between consecutive terms are not constant.

8. The common ratio for the geometric sequence $\{-1.5,-3,-6,-12,\dots \}$ is 2. To find the 11th term: ${�}_1=-1.5$ $�=2$ ${�}_{11}={�}_1\cdot {�}^{(11-1)}=-1.5\cdot 2^{10}=-1.5\cdot 1024=-1536$

9. The recursive formula for the geometric sequence is ${�}_1=1$ and ${�}_{�}=-\frac{3}{2}{�}_{�-1}$. To find the 22nd term, use the formula or recursively calculate it.

10. The explicit formula for the geometric sequence $\{4,-43,49,-427,\dots \}$ is ${�}_{�}=4\cdot (-11{)}^{�-1}$. To find the 11th term: ${�}_{11}=4\cdot (-11{)}^{11-1}=4\cdot (-11{)}^{10}=4\cdot 11^{10}$

11. The sum of terms $3{�}^2-56�$ from $�=-3$ to $�=15$ using summation notation is: ${\sum }_{�=-3}^{15}(3{�}^2-56�)$

12. The seating capacity of the stadium is given by the sum of an arithmetic sequence: ${\sum }_{�=1}^{56}(10+3(�-1))$. Calculate this sum.

13. Use the formula for the sum of the first $�$ terms of a geometric series to find ${\sum }_{�=1}^7-0.2\cdot (-5{)}^{�-1}$.

14. Find the sum of the infinite geometric series ${\sum }_{�=1}^{\mathrm{\infty }}13\cdot (-15{)}^{�-1}$.

15. To calculate Rachael's retirement fund, you'll need to use the formula for compound interest. First, find the total number of compounding periods (months) and then use the formula to calculate the final amount.

16. To find the number of dancers who compete in the fox-trot or tango competitions, use the principle of inclusion-exclusion.

17. To find the number of choices the buyer has when custom ordering the car, use the multiplication principle to multiply the number of choices in each category (exterior color, interior color, sound system, motor design, and transmission).

18. To find the number of ways the manager can create the "Top-Four" list out of the 32 employees, you need to calculate the number of permutations.

19. To find the number of ways the rock group can choose their setlist from 15 songs, you need to calculate the number of combinations.

20. To find the number of ways to top a frozen yogurt with either candy or fruit, use the multiplication principle to multiply the number of choices for each type of topping.

21. To find the number of distinct ways the word "EVANESCENCE" can be arranged with the requirement of ending with the letter "E," you can treat the word as having 10 distinct characters (ignoring repeated letters). Calculate the number of arrangements and then consider the repeated letters.

22. Use the Binomial Theorem to expand $(32�-12�{)}^5$.

23. To find the seventh term of $({�}^2-12{)}^{13}$ without fully expanding the binomial, use the Binomial Theorem, specifically focusing on the 7th term of the expansion.