# 13 Theoretical Probability

# learn by watching

The links below will launch the video lessons in YouTube

- Theoretical Probability Vocabulary ( 9 minutes 57 seconds)
- Finding Probabilities from Sample Spaces ( 11 minutes 58 seconds)

**Try Exercises #1-8.**

- Theoretical Probabilities Using AND and OR ( 9 minutes and 5 seconds)
- Probabilities with Replacement: Using AND and OR ( 8 minutes and 13 seconds)

**Try Exercises #9-10.**

- Probabilities without Replacement: Conditional Probabilities ( 8 minutes and 6 seconds)
- Probabilities with Cards ( 7 minutes and 12 seconds)

**Try Exercises #11-13.**

- Probability from Statistics: EXACTLY vs AT LEAST ( 12 minutes and 13 seconds)

**Try Exercises #14-15.**

# learn by reading

In the last section, we looked at calculating probabilities based on observed and collected data. Sometimes, however, observed data doesn’t end up making sense to use when reporting a probability. Consider, for example rolling a 6-sided die a total of 100 times. Using a simulator, I came up with the following results:

Dice Roll | 1 | 2 | 3 | 4 | 5 | 6 |

# of Times | 18 | 14 | 12 | 15 | 24 | 17 |

This is definitely a possible result (I did this for real, I promise!). However, it doesn’t seem like what I would *expect* to happen. I would expect to have an equal chance of rolling each number, yet if I use this data to calculate my probabilities, notice that it shows that it is twice as likely to roll a 5 as a 3!

In situations like this, where we have events that should be EQUALLY-LIKELY to occur, it is more accurate to report out a theoretical probability instead of an experimental one.

## theoretical probabilities using sample spaces

Theoretical Probability Definitions

Example 1: Suppose you are rolling a single 6-sided dice.

- Write out the sample space
- Find P(rolling a “2”)
- Find P(rolling a “number less than 10”)
- Find P(rolling “at least a 2”)

- The sample space is a list of all possible outcomes. In this case, we want all possible outcomes from rolling a dice. There are a total of six possibilities: {1,2,3,4,5,6}. NOTE: this total will be the bottom of all of our probabilities.
- P(rolling a “2”): There is only a single “2” in the sample space. So the number of favorable outcomes is 1. And our probability is 1/6. Just like in the last section, you should report your probabilities in both unsimplified fractions as well as in percentage form.

P(rolling a “2”) = 1/6 = 0.167 = 16.7% - P(rolling a “number less than 10”): As it turns, every number in our sample space is a number less than 10. So the number of favorable outcomes is 6. And our probability ends up as:

P(rolling a “number less than 10”) =6/6 = 1 = 100% - P(rolling “at least a 2”): The word “at least” means it could be that many or more. So, looking at our sample space, I could roll a 2,3,4,5,or 6 and get a favorable outcome. That is a total of 5 things on my list of 6 possible options that will work.

P(rolling “at least a 2”) = 5/6 = 0.833 = 83.3%

When figuring out probability and equally-likely outcomes, the order DOES matter.

Example 2: Write out the sample spaces for:

- Flipping one coin.
- Flipping two coins.
- Flipping three coins.
- Flipping four coins.

- When flipping a single coin, there are exactly 2 options: heads or tails. To simplify future calculations, I’m going to abbreviate heads (H) and tails (T). My sample space for a single coin flip can be listed as {H,T}
- When flipping two coins, at first it may seem like there are only 3 options: both heads, both tails, or one of each. However, when you think about equally-likely outcomes, it is actually more likely to end up with one of each. Why? Because the order presented matters. I actually have 4 options that are equally likely, and can write out my sample space as {HH, HT, TH, TT}. These correspond to the following outcomes:
- Flip heads then heads.
- Flip heads then tails.
- Flip tails then heads.
- Flip tails then tails

- When flipping three coins, we actually end up with 8 possible outcomes. Our first two coins could give us any of the 4 options we just mentioned. Then our third coin flip could be either heads or tails. It helps to try to keep a pattern as you are writing the sample space so you don’t miss any: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
- Keeping this same method of thinking, you’d expect to 16 possible outcomes this time. We could get heads or tails after each of the eight options from flipping three coins. {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

Example 3: Suppose you flip a coin three times. Find the following probabilities:

- P(all flips get the same result)
- P(at least 2 heads)

- Highlighting all the times I got the same result for all three flips {
**HHH**, HHT, HTH, HTT, THH, THT, TTH,**TTT**}, I find that 2 out of the 8 options have favorable outcomes. So P(all flips get the same result) = 2/8 = 0.25 = 25% - Highlighting all the times I got at least 2 heads: {
**HHH**,**HHT**,**HTH**, HTT,**THH**, THT, TTH, TTT}, I find that 4 out of the 8 options have favorable outcomes. So P(at least 2 heads) = 4/8 = 0.5 = 50%

Example 4: Suppose you roll two dice. Find the following probabilities:

- P(rolling the same number both times)
- P(rolling a bigger number on the second roll)
- P(a sum of 6 on the two dice)

Rolling a 1 first | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |

Rolling a 2 first | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |

Rolling a 3 first | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |

Rolling a 4 first | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |

Rolling a 5 first | (5,1) | (5,2) | (5,3) | (5,4) | (5.5) | (5,6) |

Rolling a 6 first | (6,1) | (6,2) | (6,3) | (6.4) | (6,5) | (6,6) |

Notice that there are a total of 36 different equally-likely things that could happen. So the bottom number of all the probability calculations will be 36.

- Rolling the same number both times:

**(1,1)**(1,2) (1,3) (1,4) (1,5) (1,6) (2,1) **(2,2)**(2,3) (2,4) (2,5) (2,6) (3,1) (3,2) **(3,3)**(3,4) (3,5) (3,6) (4,1) (4,2) (4,3) **(4,4)**(4,5) (4,6) (5,1) (5,2) (5,3) (5,4) **(5.5)**(5,6) (6,1) (6,2) (6,3) (6.4) (6,5) **(6,6)**P(rolling the same number) = 6/36 = 0.167 = 16.7%

- Rolling a bigger number on the second roll:

(1,1) **(1,2)****(1,3)****(1,4)**( **1,5)****(1,6)**(2,1) (2,2) **(2,3)****(2,4)****(2,5)****(2,6)**(3,1) (3,2) (3,3) **(3,4)****(3,5)****(3,6)**(4,1) (4,2) (4,3) (4,4) **(4,5)****(4,6)**(5,1) (5,2) (5,3) (5,4) (5.5) **(5,6)**(6,1) (6,2) (6,3) (6.4) (6,5) (6,6) Counting up the highlighted values, I find that there are 15 different ways that my second dice roll will be bigger than my first. So P(rolling a bigger number on the second roll) = 15/36= 0.417 = 41.7%

- A sum of 6 on the two dice. Again, we will highlight all the ways this can happen. Recall that a “sum” is when you add numbers together. So, for example, a dice roll of (1,1) has a sum of 1+1 = 2.

(1,1) (1,2) (1,3) (1,4) **(1,5)**(1,6) (2,1) (2,2) (2,3) **(2,4)**(2,5) (2,6) (3,1) (3,2) **(3,3)**(3,4) (3,5) (3,6) (4,1) **(4,2)**(4,3) (4,4) (4,5) (4,6) **(5,1)**(5,2) (5,3) (5,4) (5.5) (5,6) (6,1) (6,2) (6,3) (6.4) (6,5) (6,6) Counting up the highlighted values, I find that there are 5 different ways that I can get two different dice to add up to 6. So P(rolling a sum of 6) = 5/36 = 0.139 = 13.9%

At some point, listing out the sample spaces just is not practical. For example, just rolling 3rd dice will end up with 216 total possibilities (each of the 36 options in our last list ALSO gets 6 options for each).

Exercises: Try These!

1) Write out the sample space for rolling a single standard die.

2) Find the following probabilities when rolling a single standard die (write your answers in both fraction AND percentage form for credit):

a) The probability of rolling a 3?

b) The probability of rolling an even number?

c) The probability of rolling at least a 3?

d) The probability of rolling a number less than 3?

e) The probability of rolling an 8?

f) The probability of rolling an a number less than 10?

g) The probability of not rolling a 6?

3) Write out the sample space for flipping a standard coin twice?

4) Find the following probabilities when flipping a standard coin twice? (write your answers in both fraction AND percentage form for credit):

a) The probability of getting both heads?

b) The probability of getting heads followed by tails?

c) The probability of getting heads on the first flip?

d) The probability of getting 1 heads and 1 tails?

e) The probability of getting at least one tails?

f) The probability of not getting any heads?

5) Write out the sample space for flipping a standard coin three times?

6) Find the following probabilities when flipping a standard coin three times (write your answers in both fraction AND percentage form for credit):

a) The probability of getting all heads?

b) The probability of getting heads first?

c) The probability of getting at least 1 tails?

d) The probability of getting 1 heads and 2 tails?

e) The probability of not getting any heads?

7) Write out the sample space for rolling 2 standard dice?

8) Find the following probabilities when rolling a standard dice two times (write your answers in both fraction AND percentage form for credit):

a) The probability of getting doubles (the same result on each die)?

b) The probability of getting a sum of 7 on the two dice?

c) The probability of getting a sum of 7 or 11 on the two dice?

d) The probability of getting at least one even number?

e) The probability of not rolling any 5’s?

f) The probability that the first die roll is a smaller number than the second die roll?

g) The probability that the second die roll is a 6?

## theoretical probabilities using multiple events

To find the probability of two events both happening together, one after the other, we can multiply the individual probability of each event together. The order of occurrence DOES make a difference when you use this multiplication rule.

Multiplication Rule for Probabilities

Example 5: Suppose you are going to roll a dice 3 times in a row. What is the probability that you will roll at least a 5 each time?

This is a situation where we really don’t want to write out the full sample space. Instead we can calculate each roll’s probability, and multiply them together.

1st roll: There is 2/6 probability of rolling at least a 5 (rolling a “5” or a “6” would both be favorable outcomes).

2nd roll: There is 2/6 probability of rolling at least a 5 (rolling a “5” or a “6” would both be favorable outcomes).

3rd roll: There is 2/6 probability of rolling at least a 5 (rolling a “5” or a “6” would both be favorable outcomes).

So the probability of rolling at least a 5, when rolling a dice three times, would be 2/6 * 2/6 * 2/6 = .037 = 3.7%

We multiply probabilities when we want multiple things to occur. What if we only want one option or a completely different option? In this case, we will add the probabilities together (similar to the OR probabilities that we used in contingency tables).

Addition Rule for Probability

The probability of one outcome OR another completely different outcome occurring can be found by adding the probabilities together.

P(A or B) = P(A) + P(B)

Example 6: Suppose you have a bag of marbles. There are 10 clear marbles, 8 red marbles, 1 white marble, and 1 black marble. You pick a single marble from the bag.

- What is the probability of picking a red marble?
- What is the probability of picking a white marble or a black marble.

- P (red). There are 8 red marbles that would be “favorable” outcomes. So P(red) = 8/20 = 0.4 = 40%
- P (white or black). The probability of picking a white marble is 1/20. The probability of picking a black marble is 1/20. These are two completely different outcomes, and we want one or the other to apply. So we can find the P(white or black) = P(white) + P(black) = 1/20 + 1/20 = 0.1 = 10%

Complement of an Outcome

When looking at all possible different outcomes, the total probabilities must add up 100% .

The “complement” of an outcome can be described as “all the rest of the possible outcomes”.

P(A) = 1 – P(complement of A)

Note: if using percentages, the equivalent definition would be: P(A) = 100% – P(complement of A)

Example 7: Consider the bag of marbles in the last example, containing 10 clear marbles, 8 red marbles, 1 white marble, and 1 black marble. You pick a single marble from the bag. What is the probability of NOT picking a red marble.

There are 2 ways you can approach this problem.

Method 1: If we don’t want to pick a red marble, that means we can pick a clear marble or a white marble or a black marble.

P(not red) = P(clear or white or black) = P(clear) + P(white) + P(black) = 10/20 + 1/20 + 1/20 = 0.6 = 60%

Method 2: If we don’t want to pick a red marble, that means we can pick ANYTHING else. So we are finding the complement of picking a red marble.

P(not red) = 1 – P(red) = 1 – 8/20 = 0.6 = 60%

Sometimes, using this “backwards”-style complement method can make calculations a little easier.

Let’s put these concepts altogether to find probabilities. Keep in mind that when you are doing events one after the other, sometimes the second probability will be affected by what you picked first.

Example 8a: Consider the bag of marbles in the last example, containing 10 clear marbles, 8 red marbles, 1 white marble, and 1 black marble. This time, let’s pick two marbles from the bag. Suppose after you pick the first marble, you put it back into the bag and shake it up before picking the second marble. (We call this selection WITH replacement).

- Find the probability of picking both clear marbles.
- Find the probability of picking both white marbles.
- Find the probability of picking a red marble followed by a black marble.
- Find the probability of picking a red and a black marble.
- Find the probability of getting at least 1 red marble.

- P(both clear) = P(1st pick clear) * P(2nd pick clear). Since there are 10 clear marbles in the bag, this becomes 10/20 * 10/20 = 0.25 = 25% chance of picking both clear marbles.
- P(both white) = P(1st pick white) * P(2nd pick white). Since there is only 1 white marble in the bag, this becomes 1/20 * 1/20 = 0.0025 = 0.25%
- P(red then black) = P(1st pick red) * P(2nd pick black). There are 8/20 marbles to get red on the first pick. Then there is only 1/20 marbles to get black on the second pick. So P(red then black) = 8/20 * 1/20 = .02 = 2% chance of picking red then black.
- P(red and black). This may sound a lot like the last question, but it is fundamentally different. In this problem, I only care that I end up picking a red marble and a black marble. I don’t care whether I pick it 1st or 2nd. However, with probabilities the order matters. So there are two different ways this can happen. P(red then black)
**OR**P(black then red). Since we want one or the other, we can add these options together.P(red then black) + P(black then red) = (8/20 * 1/20) + (1/20 * 8/20) = .04 = 4% chance of having a red and a black pick - P(at least 1 red). There are two ways we can think about this
- There are lots of ways we can have at least one red when picking 2 marbles. You can find each of these probabilities separately and add them together.
- P(red then clear)
- P(clear then red)
- P(red then red)
- P(red then white)
- P(white then red)
- P(red then black)
- P(black then red)

- Alternatively, there is only ONE way that this can NOT happen. The word “at least 1” means 1 or more red marbles. The only way that can NOT happen is if we pick NO red marbles. Keep in mind that since there were 8 red marbles, then the other 12 marbles were NOT red. So P(at least one red) = 1 – P(no red marbles) = 1 – (12/20 * 12/20) = 1 – 0.36 = 0.64 = 64% chance of picking at least one red marble. [note: this is the same answer you’d get if you used method 1. Try it and see!]

- There are lots of ways we can have at least one red when picking 2 marbles. You can find each of these probabilities separately and add them together.

Example 8b: Consider the bag of marbles in the last example, containing 10 clear marbles, 8 red marbles, 1 white marble, and 1 black marble. This time, let’s pick two marbles from the bag. Suppose after you pick the first marble, you set it aside, then make your second pick from the marbles still remaining in the bag. (We call this selection WITHOUT replacement).

- Find the probability of picking both clear marbles.
- Find the probability of picking both white marbles.
- Find the probability of picking a red marble followed by a black marble.
- Find the probability of picking a red and a black marble.
- Find the probability of getting at least 1 red marble.

- P(both clear) = P(1st pick clear) * P(2nd pick clear). We have 10 clear marbles to choose from during our first draw. But since we are keeping this marble out, when we go to pick the second marble, there are only 9 clear marbles left. So, P(both clear) = 10/20 * 9/19 = 0.237 = 23.7% chance of picking both clear marbles.
- P(both white) = P(1st pick white) * P(2nd pick white). Since there is only 1 white marble in the bag, when we go to pick the second marble, there are not any marbles remaining. So P(both white) = 1/20 * 0/20 = 0.0 = 0%. Not surprising! It’s impossible to pick 2 white marbles out when there is only one in the bag!
- P(red then black) = P(1st pick red) * P(2nd pick black). There are 8/20 marbles to get red on the first pick. When we go to make the second pick there is the 1 black marble out of only 19 remaining marbles. So P(red then black) = 8/20 * 1/19 = 0.021 = 2.1% chance of picking red then black.
- P(red and black). Again, in this problem, I only care that I end up picking a red marble and a black marble. I don’t care whether I pick it 1st or 2nd. However, with probabilities the order matters. So there are two different ways this can happen. P(red then black)
**OR**P(black then red). Since we want one or the other, we can add these options together. Remember that there are only 19 marbles left when we go back to do the second draw.

P(red then black) + P(black then red) = (8/20 * 1/19) + (1/20 * 8/19) = .042 = 4.2% chance of having a red and a black pick. - P(at least 1 red). There are two ways we can think about this
- There are lots of ways we can have at least one red when picking 2 marbles. You can find each of these probabilities separately and add them together.
- P(red then clear)
- P(clear then red)
- P(red then red)
- P(red then white)
- P(white then red)
- P(red then black)
- P(black then red)

- Alternatively, there is only ONE way that this can NOT happen. The word “at least 1” means 1 or more red marbles. The only way that can NOT happen is if we pick NO red marbles. Keep in mind that since there were 8 red marbles, then the other 12 marbles were NOT red. On the first pick, I will have all 12 “not red” marbles to choose from. On the second pick, there are only 11 “not red marbles left. So my probability looks like this: P(at least one red) = 1 – P(no red marbles) = 1 – (12/20 * 11/19) = 1 – 0.347 = 0.653 = 65.3% chance of picking at least one red marble. [note: this is the same answer you’d get if you used method 1. Try it and see!]

- There are lots of ways we can have at least one red when picking 2 marbles. You can find each of these probabilities separately and add them together.

Exercises: Try These!

9) Consider a jar containing 6 blue marbles, 8 red marbles, and 5 green marbles. You pick one marble from the jar.

a) What is the probability of picking a blue marble?

b) What is the probability of picking a blue or a green marble?

c) What is the probability of picking a blue and a green marble?

d) What is the probability of not picking a red marble?

10) Consider a jar containing 6 blue marbles, 8 red marbles, and 5 green marbles. You pick one marble from the jar, then put it back in the jar. Then you draw a second marble. (This is selection with replacement)

a) What is the probability that both selections are red?

b) What is the probability that you pick a red marble followed by a green marble?

c) What is the probability that you pick a red and a green marble?

d) What is the probability of not picking a green marble?

11) Consider a jar containing 6 blue marbles, 8 red marbles, and 5 green marbles. You pick one marble from the jar and set it aside. Then you draw a second marble. (This is selection without replacement)

a) What is the probability that both selections are red?

b) What is the probability that you pick a red marble followed by a green marble?

c) What is the probability that you pick a red and a green marble?

d) What is the probability of not picking a green marble?

For the following questions, consider a complete deck of playing cards with no jokers. If you are not familiar with cards, I’ve put an image of each of the 52 cards in a deck at the end of these exercises for your reference.

12) Find the following probabilities of drawing a single card from the deck:

a) Picking a 10?

b) Picking the Queen of hearts?

c) Picking a Diamond or a 3?

d) Picking a red card?

e) Picking a black Ace?

f) Not picking a black Ace?

13) Find the following probabilities of drawing two cards from the deck (without replacement):

a) Picking two cards with numbers on them?

b) Picking two Jacks?

c) Picking two red cards?

d) Picking a Queen followed by a King?

e) Picking a Queen and a King?

## CALCULATING PROBABILITIES FROM STATISTICS

Sometimes we may read about statistics that are true about a population in general. In these cases, we can still apply our multiplication, addition, and complement rules to find probabilities of multiple events.

Example 9: Suppose the unemployment rate in a town is currently 4%.

- If you talk to 2 people, what is the probability that both people are unemployed?
- If you talk to 3 people, what is the chance that they are all employed?
- If you talk to 3 people what is the chance that exactly 1 person is unemployed?
- If you talk to 3 people, what is the probability that at least one person is unemployed?

Keep in mind, again, that probabilities imply an order. So when we want multiple outcomes to occur, we can multiply each probability – in order – to find the total probability. It is important to always convert percentage probabilities into their decimal form before doing any calculations.

- P(both people are unemployed) = P(first person is unemployed) * P(second person is unemployed). We can apply the 4% unemployment rate to both people. 4% = 4/100 = 0.04 in decimal form. So P(both unemployed) = (0.04)*(0.04) = 0.0016 = 0.16%. So there is a less than one percent chance that you will randomly talk to 2 people that are both unemployed.
- P(all 3 people are employed). This time we are talking to 3 people, so we will need to multiply the probabilities for each person together to find the chance that all 3 are employed. P(first person is employed) * P(second person is employed) * P(third person is employed).We are not told the probability of a person being employed. We are only told that there is a 4% unemployment rate. However, if a person is
*not*unemployed, that means they*are*employed. So if 4% are unemployed, then 100%-4% = 96% are employed. Again, don’t forget to change probabilities to decimal form before doing any calculations.P(all 3 people are employed) = (0.96)*(0.96)*(0.96) = 0.9216 = 92.6%. So there is a 92.6% chance that all three people I talk to are employed. - P(exactly 1 of 3 people is unemployed). This one gets a little bit tricky. We know we will need to multiply 1 unemployed rate (4%) by 2 employed rates (96%). The problem is — we don’t know which of the three people is the one that is unemployed! Remember that order is implied in your probability calculation. So there are 3 ways we can have exactly 1 person unemployed:P(only the 1st of 3 people is unemployed) = (0.04)*(0.96)*(0.96) = 0.037

P(only the 2nd of 3 people is unemployed) = (0.96)*(0.04)*(0.96) =0.037

P(only the 3rd of 3 people is unemployed) = (0.96)*(0.96)*(0.04) = 0.037Since we want either the 1st or 2nd or 3rd case to apply, and these are all distinct and different outcomes, we can add the probabilities together: P(exactly 1 of of 3 people is unemployed) = 0.037 + 0.037 + 0.037 = 0.111 = 11.1% that exactly one of the people I talk to is unemployed. - (at least 1 of 3 people is unemployed). This is actually even trickier than the last one. To calculate this directly, we would need all the cases where exactly one person was unemployed, plus all the cases where exactly 2 people were unemployed, plus all the cases where all 3 people were unemployed.However, the only way to NOT have at least one person unemployed would be to have ALL the people employed. So we can use the complement rule to calculate this result more easily.P(at least one unemployed) = 1 – P(all employed) = 1 – (0.96*0.96*0.96) = 1 – .9216 = 0.0784 = 7.84%

So if I randomly chose three people to talk to, there would be a 7.84% chance that at least one of them would be unemployed.

Exercises: Try These!

14) In the 2020 election, 66.8% of citizens aged 18 and older voted.

a) If you randomly talk to 3 citizens, what is the probability that all 3 of them voted in the 2020 election?

b) If you randomly talk to 3 citizens, what is the probability that none of them voted in the 2020 election?

c) If you randomly talk to 3 citizens, what is the probability that at least 1 of them voted in the 2020 election?

d) If you randomly talk to 3 citizens, what is the probability that exactly 1 person voted in the 2020 election?

15) During the pandemic, 78% of pet owners acquired a new pet.

a) If you randomly talk to 4 pet owners, what is the probability that all 4 of them got new pets during the pandemic?

b) If you randomly talk to 4 pet owners, what is the probability that none of them got new pets during the pandemic?

c) If you randomly talk to 4 pet owners, what is the probability that at least one of them got new pets during the pandemic?