1. The Basics of Event Probability
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Chapter 6
1.
The Basics of Event Probability
This lesson delves into the basics of event probability, offering insights into sample space, favorable outcomes, and experimental probability. It aims to introduce these concepts in a manner that connects them to real-life situations. For instance, if one is participating in a game show and have to pick a door with a prize behind it, understanding sample space and favorable outcomes can increase your chances of winning. Similarly, experimental probability is discussed in the context of rolling dice or drawing marbles from a bag. This knowledge is not just theoretical; it has practical applications in various fields like statistics, game theory, and even in everyday decision-making.
Show less Show more expand_more| Student Learning Objectives: |
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| | 13 Theory slides |
| | 14 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The Basics of Event Probability
Slide of 13
In this lesson, some basics concepts of probability will be introduced and connected to real-life situations.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Explore
Possible Outcomes of Target Shooting
Magdalena is in a target shooting competition. For her next test, the target has three zones and she can shoot three arrows.
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Challenge
Probability of Winning a Prize
On a late-night game show, there are three closed doors. Behind one of the doors is a car, and there are sheep behind the other two. The host invites Tearrik to win the car by choosing only one door.
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b
What are the chances of choosing a door with a sheep? Round the answer to two decimal places.
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c
If there were 70 doors and 20 prizes, what is the chance that Tearrik picks a door with a sheep behind it? Round the answer to two decimal places.
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Discussion
Defining Experiments in Probability
Below, some basic definitions of probability are examined.
Concept
Experiment - Probability
An experiment is a process used to determine the probability of an event occurring in the future. In finding probability, an experiment is an action that can be repeated infinitely many times and has a variety of results called outcomes. Even rolling a die can be considered an experiment.
Concept
Outcome
An outcome is a possible result of a probability experiment. For example, when rolling a six-sided die, getting a 3 is one possible outcome.
Concept
Event
An event is a combination of one or more specific outcomes. For example, when playing cards, an event might be drawing a spade or a heart. For this event, one possible outcome is drawing the A♠ or drawing the 7♡.
However, these are not the only outcomes of this event. All the possible outcomes that satisfy the event are listed below.
Outcomes: 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♡Concept
Sample Space
The sample space of an experiment is the set of all possible outcomes. For example, when flipping a coin, there are two possible outcomes: heads, H, or tails, T. The sample space of flipping a coin is {H, T} .
Here, the sample space is shown in a tree diagram. Each row represents the possible outcomes of a toss. When the coin is flipped another time, the tree diagram gets another row with the possible outcomes.
Here, the sample space has four possible outcomes.
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Example
Sample Space of Rolling Dice
For each of the following experiments, list the possible outcomes in the sample space and count the total number of outcomes.
a
Roll one die.
b
Roll two dice at the same time.
Answer
a
Number of Outcomes: 6
Outcomes:
{1, 2, 3, 4, 5, 6}
b
Number of Outcomes: 36
Outcomes:
Hint
a
What are the possible outcomes when a die is rolled?
b
For each outcome of the first die, there are six different outcomes for the second die.
Solution
a
When a die is rolled, there are 6 different possible outcomes.
Consequently, the sample space is the set {1, 2, 3, 4, 5, 6}.
b
As was found in Part A, each possible outcome of rolling one die is just one number. However, when rolling two dice, one possible outcome is getting a 2 on the first die and a 5 on the second die.
{2,5}
To list all the possible outcomes, create all possible combinations by determining one outcome for the first die and then varying the outcome of the second die. Then, change the outcome of the first die and repeat the process.
Since each die has 6 possible outcomes, the total number of possible outcomes for this experiment is 6* 6 = 36.
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Example
Sample Space: Drawing Marbles From a Bag
Paulina bought two white and three black marbles, all of different sizes, and put them in a bag. When she got home, her little brother Diego and sister Emily asked her to give them two marbles. Paulina agreed but told them to draw one marble each without looking inside the bag. Diego drew the first marble, then Emily.
a
Assuming all the marbles are distinguishable, write down the sample space for the event of Diego and Emily randomly drawing two marbles from the bag. What is the total number of outcomes?
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b
How many of the outcomes in the sample space represent the case where Diego and Emily both drew white marbles?
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c
How many of the outcomes in the sample space represent the case where Diego and Emily both drew black marbles?
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d
How many of the outcomes in the sample space represent the case where Diego and Emily drew marbles of different colors?
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e
In a few words, write a different event for the situation presented.
Answer
a
Number of Outcomes: 20
Outcomes:
{W_1,W_2},{W_1,B_1},{W_1,B_2},{W_1,B_3} {W_2,W_1},{W_2,B_1},{W_2,B_2},{W_2,B_3} {B_1,W_1},{B_1,W_2},{B_1,B_2},{B_1,B_3} {B_2,W_1},{B_2,W_2},{B_2,B_1},{B_2,B_3} {B_3,W_1},{B_3,W_2},{B_3,B_1},{B_3,B_2}
Here, W_1 and W_2 represent the white marbles, and B_1, B_2, and B_3 represent the black marbles.
b
Number of Outcomes: 2
Outcomes: {W_1,W_2} and {W_2,W_1}
c
Number of Outcomes: 6
Outcomes:
{B_1,B_2}, {B_1,B_3}, {B_2,B_1}, {B_2,B_3}, {B_3,B_1}, {B_3,B_2}
d
Number of Outcomes: 12
Outcomes:
{W_1,B_1}, {W_1,B_2}, {W_1,B_3}, {W_2,B_1}, {W_2,B_2}, {W_2,B_3}, {B_1,W_1}, {B_1,W_2}, {B_2,W_1}, {B_2,W_2}, {B_3,W_1}, {B_3,W_2}
e
Example Event: Diego and Emily ask Paulina to give them two marbles each.
Hint
a
Give each marble a label. How many outcomes does the first drawn marble have? How many outcomes does the second marble have?
b
Look for the outcomes containing only the labels assigned to the white marbles.
c
Look for the outcomes containing only the labels assigned to the black marbles.
d
Look for the outcomes containing one label assigned to the white marbles and one label assigned to the black marbles.
e
If Diego and Emily ask for a different number of marbles, this is a different event.
Solution
a
Despite having only two colors, all five marbles are said to be distinguishable. That is why starting by labeling each of the marbles is a good option.
Since both marbles are randomly drawn, each outcome of the event will consist of two labels. For the first marble, there are 5 possible outcomes. For the second marble, there are 4 possible outcomes because one marble has already been removed from the bag.
Since the Diego draws a marble first and Emily draws a marble after him, the order in which the marbles are drawn matters. Therefore, the outcomes {W_1,W_2} and {W_2,W_1} are different. The following table lists all the possible outcomes for the event of drawing two marbles from the bag. {W_1,W_2},{W_1,B_1},{W_1,B_2},{W_1,B_3} {W_2,W_1},{W_2,B_1},{W_2,B_2},{W_2,B_3} {B_1,W_1},{B_1,W_2},{B_1,B_2},{B_1,B_3} {B_2,W_1},{B_2,W_2},{B_2,B_1},{B_2,B_3} {B_3,W_1},{B_3,W_2},{B_3,B_1},{B_3,B_2} Consequently, there are a total of 20 possible outcomes in the sample space.
b
The case in which two white marbles are drawn is represented by the outcomes containing only W_1 and W_2. From Part A, it can be seen that there are only two such outcomes — {W_1,W_2} and {W_2,W_1}.
c
The case in which two black marbles are drawn is represented by the outcomes containing neither W_1 nor W_2. Analyzing the sample space from Part A, six such outcomes can be found.
&{B_1,B_2}, {B_1,B_3}, &{B_2,B_1}, {B_2,B_3}, &{B_3,B_1}, {B_3,B_2}
d
The case in which one white and one black marble are drawn is represented by the outcomes containing one W-label and one B-label. From Part A, there are twelve such outcomes.
{W_1,B_1}, {W_1,B_2}, {W_1,B_3}, {W_2,B_1}, {W_2,B_2}, {W_2,B_3}, {B_1,W_1}, {B_1,W_2}, {B_2,W_1}, {B_2,W_2}, {B_3,W_1}, {B_3,W_2}
e
There are an infinite number events that can be considered. For example, Diego and Emily could ask Paulina to give them two marbles each. In that case, the event would be the drawing of four marbles.
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Discussion
Union, Intersection, and Complement of Events
Sometimes more than one event can be involved in an experiment. In such cases, it is important to know how to determine the union or intersection of the events.
Concept
Union - Probability
The union of two events A and B is the set of all the outcomes that are in A or are in B or in both A and B. The union of A and B is usually written as A or B
or A⋃ B.
Concept
Intersection - Probability
The intersection of two events A and B is the set of all the outcomes that satisfy both events A and B simultaneously. The intersection of A and B is usually written as AandB
or A⋂ B.
Also, there may be situations where it is easier to determine which outcomes do not satisfy an event rather than determining which outcomes do. For such cases, the following concept will be useful.
Concept
Complementary Events
Complementary events are pairs of events where only one can occur at a time. The complement of an event A is written as A', A^c, or A. It represents all possible outcomes that are not part of event A. For instance, when rolling a fair six-sided die, rolling an even number and rolling an odd number are complementary events.
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Example
Operating With Events
Let U be the set of all integers from 1 to 9.
- Let A be the event of picking a prime number.
- Let B be the event of picking an odd number.
- Let C be the event of picking a multiple of 3.
a
Write the complement of each event in words.
b
List the outcomes of each event and its complement.
c
Write the event A⋂ B in words and list its outcomes.
d
Write the event B⋃ C' in words and list its outcomes.
e
Make a Venn diagram illustrating the set U and the events A, B, and C.
Answer
a
Complements:
- A' is the event of picking a non-prime number.
- B' is the event of picking an even number.
- C' is the event of picking a number that is not a multiple of 3.
b
Outcomes of the Events:
A &= {2,3,5,7} [0.15cm] B &= {1,3,5,7,9} [0.15cm] C &= {3,6,9} [0.15cm] Outcomes of the Complements: A' &= {1,4,6,8,9} [0.15cm] B' &= {2,4,6,8} [0.15cm] C' &= {1,2,4,5,7,8}
c
Intersection of Events: A⋂ B is the event of picking a prime, odd number.
Outcomes: A⋂ B = {3,5,7}
d
Union of Events: B⋃ C' is the event of picking either an odd number or a number that is not a multiple of 3.
Outcomes: B⋃ C' = {1,2,3,4,5,7,8,9}
e
Venn Diagram:
Hint
a
To write the complement of an event, consider the opposite event.
b
The complement of an event is the set of all the possible outcomes that are not in that event.
c
Remember how the intersection of events is defined.
d
Use the definition of a union of events.
e
First, look for the numbers that are common for the three events. Then, look for the numbers that are in two events but not in the third one. Finally, look for the numbers from the sample space that do not occur in any of the events.
Solution
a
By definition, the complement of an event is the set of all possible outcomes that are not in that event. Therefore, the complement of picking a prime number is picking a non-prime number. The complement of events B and C can be similarly written.
| Event | Complement |
|---|---|
| A: picking a prime number | A': picking a non-prime number |
| B: picking an odd number | B': picking an even number |
| C: picking a multiple of 3 | C': picking a number that is not a multiple of 3 |
b
The outcomes of the event A are all numbers in U that are prime.
A = {2,3,5,7} In a similar way, the outcomes of the remaining events can be written. B &= {1,3,5,7,9} [0.15cm] C &= {3,6,9} [0.15cm] Finally, the outcomes of A' are the outcomes in U that are not in A. Similarly for the outcomes of B' and C'. A' &= {1,4,6,8,9} [0.15cm] B' &= {2,4,6,8} [0.15cm] C' &= {1,2,4,5,7,8}
c
By definition, the intersection of two events is the set of all outcomes that satisfy both events simultaneously. To write A⋂ B, start by recalling what events A and B represent.
A - event of picking a prime number
B - event of picking an odd number
Therefore, A⋂ B is the event of picking a prime and odd number.
| Number | Is it prime? | Is it odd? |
|---|---|---|
| 1 | No | Yes |
| 2 | Yes | No |
| 3 | Yes | Yes |
| 4 | No | No |
| 5 | Yes | Yes |
| 6 | No | No |
| 7 | Yes | Yes |
| 8 | No | No |
| 9 | No | Yes |
Consequently, A⋂ B = {3,5,7}.
d
By definition, the union of two events is the set of all outcomes that satisfy either of the events. Before writing what B⋃ C' is, begin by remembering what each of these events represents.
B - event of picking an odd number
C' - event of picking a number that is not a multiple of 3
Therefore, B⋃ C' is the event of picking either an odd number or a number that is not multiple of 3.
| Number | Is it odd? | Is not a multiple of 3 |
|---|---|---|
| 1 | Yes | Yes |
| 2 | No | Yes |
| 3 | Yes | No |
| 4 | No | Yes |
| 5 | Yes | Yes |
| 6 | No | No |
| 7 | Yes | Yes |
| 8 | No | Yes |
| 9 | Yes | No |
Consequently, B⋃ C' = {1,2,3,4,5,7,8,9}. Notice that 6 is the only element of U that is not in B⋃ C' since it satisfies neither B nor C'.
e
In order to make a Venn diagram, start by considering the sample space U, which is a set of integers from 1 to 9.
Next, draw three sets representing the events A, B, and C and write down the outcomes of each event inside the corresponding set.
Comparing the outcomes of the three events, some conclusions can be drawn.
- The number 3 is common for the three events. Thus, the three sets intersect each other.
- The numbers 5 and 7 are common for events A and B, but they are not in C.
- The number 9 is common for events B and C, but it is not in A.
- The events A and C have nothing in common other than the number 3.
- The numbers 4 and 8 do not belong to any of the events.
With this information, the Venn diagram can be drawn.
In the following diagram, the events and their complements can be appreciated separately.
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Discussion
Theoretical and Experimental Probability
Since an event is a combination of possible outcomes of an experiment, in some cases the event happens rarely, while in others it happens frequently. This frequency depends on the experiment and the event itself.
Concept
Probability
Probability measures the likelihood that something will occur. It can be any value from 0 to 1 or from 0 % to 100 %, inclusive. When it is certain that the situation will not occur, the probability is 0. Likewise, when it is certain that the situation will occur, the probability is 1.
The probability of an event occurring can be determined both theoretically and experimentally. Theoretical probability expresses the expected probability when all outcomes in a sample space are equally likely. In comparison, experimental probability is expressed by analyzing data collected from repeated trials of an experiment.
Concept
Theoretical Probability
When all outcomes in a sample space are equally likely to occur, the theoretical probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.
P(event)=Number of favorable outcomes/Number of possible outcomes
Consider the roll of a standard six-sided die. There are six equally likely outcomes in the sample space. For the event of rolling an even number, there are three favorable outcomes. This can be called A.
When an experiment is performed, the results may be a little different from what was expected. In other words, slightly different results may be obtained from what the theoretical probability predicted.
Concept
Experimental Probability
Experimental probability is the probability of an event occurring based on data collected from repeated trials of a probability experiment. For each trial, the outcome is noted. When all trials are performed, the experimental probability of an event is calculated by dividing the number of times the event occurs, or its frequency, by the number of trials.
P(event)=Number of times the event occurs/Number of trials [0.5em] ⇕ [0.5em] P(event)=Frequency of the event/Number of trials
By repeating an experiment many times, the result will trend toward the theoretical probability of the event. For example, consider the flip of a fair coin and the event that tails is the outcome. Use the applet to simulate the outcomes and calculate the experimental probability.
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Example
Theoretical Probability vs. Experimental Probability
Ramsha and Mark conducted an experiment consisting of rolling two dice and adding their results. The following diagram shows the numbers obtained in each roll.
Consider the event of getting a result greater than or equal to 8.
a
What is the theoretical probability of this event? Round the answer to two decimal places.
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b
Find the experimental probability of the given event according to Ramsha's data. Round the answer to two decimal places.
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c
Find the experimental probability of the given event according to Mark's data. Round the answer to two decimal places.
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Hint
a
Determine all the possible outputs of rolling two dice. Then, calculate the sum of the outputs and count how many of them are greater than or equal to 8.
b
Start by finding the number of successes, the number of times the event was fulfilled.
c
Divide the number of successes by the number of trials.
Solution
a
To find the theoretical probability, the number of favorable outcomes should be divided by the total number of possible outcomes.
P=Number of favorable outcomes/Number of possible outcomes Since the experiment consists of rolling two dice and each die has 6 possible outcomes, there is a total of 36 possible combinations. Calculate the sum of the outputs for each combination.
The 5 in the second row and third column represents the event of rolling a 2 on the first die and a 3 on the second die. The other outcomes can be calculated in the same fashion. Next, highlight the outcomes satisfying the given event, that the sum of the dice is greater than or equal to 8.
There is a total of 15 favorable outcomes. With this information, the theoretical probability can be calculated.
P(Sum≥ 8) = Favorable outcomes/Total outcomes
▼
Substitute values and simplify
SubstituteII
Favorable outcomes= 15, Total outcomes= 36
P(Sum≥ 8) = 15/36
ReduceFrac
a/b=.a /3./.b /3.
P(Sum≥ 8) = 5/12
CalcQuot
Calculate quotient
P(Sum≥ 8) = 0.416666...
RoundDec
Round to 2 decimal place(s)
P(Sum≥ 8) ≈ 0.42
Consequently, the probability of getting a sum greater than or equal to 8 when two dice are rolled is about 0.42, or 42 %. Notice that the sample space of the experiment contains only 11 outcomes. {2,3,4,5,6,7,8,9,10,11,12} However, the outcomes are not equally likely. From the above table, there is greater chance of rolling a 7 than rolling a 2.
b
The experimental probability of an event equals the number of successes obtained divided by the number of trials.
P(Event) = Number of Successes/Number of Trials Therefore, to find the experimental probability obtained by Ramsha, divide the number of successes she got by the number of trials she conducted. These two numbers can be deduced from the given diagram.
As can be seen, Ramsha got 2 successes in 7 trials. With this information, the experimental probability can be found.
P(Sum≥ 8) = Number of Successes/Number of Trials
▼
Substitute values and simplify
SubstituteII
Number of Successes= 2, Number of Trials= 7
P(Sum≥ 8) = 2/7
CalcQuot
Calculate quotient
P(Sum≥ 8) = 0.285714...
RoundDec
Round to 2 decimal place(s)
P(Sum≥ 8) ≈ 0.29
Consequently, for the given event, Ramsha got an experimental probability of 0.29, or 29 %.
c
Similar to Part B, to find the experimental probability obtained by Mark, divide the number of successes he got by the number of trials he conducted. As before, this information can be found in the given diagram.
Mark got 6 successes in 9 trials. With this information, the experimental probability can be found.
P(Sum≥ 8) = Number of Successes/Number of Trials
▼
Substitute values and simplify
SubstituteII
Number of Successes= 6, Number of Trials= 9
P(Sum≥ 8) = 6/9
ReduceFrac
a/b=.a /3./.b /3.
P(Sum≥ 8) = 2/3
CalcQuot
Calculate quotient
P(Sum≥ 8) = 0.66666...
RoundDec
Round to 2 decimal place(s)
P(Sum≥ 8) ≈ 0.67
Therefore, for the given event, Mark got an experimental probability of 0.67, or 67 %.
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Discussion
Probability of the Complement of an Event
The probability of drawing a club from a standard deck of cards is 0.25. Knowing this, what is the probability of drawing a spade, heart, or diamond if a card is drawn randomly?
To figure it out, instead of counting the favorable outcomes, the complement rule can be used.
Rule
Complement Rule - Probability
The sum of the probability of an event and the probability of its complement is equal to 1.
P(A) + P(A') = 1
This formula is useful when calculating the probability of the complement of an event is easier than calculating the probability of the event itself. Then, the probability of the event is calculated as follows. P(A) = 1 - P(A')
Proof
Complement RuleLet A be an event, A' be its complement, and S be the sample space. By the definition of the complement, the union of an event and its complement is equal to the entire sample space.
A⋃ A' = S
Because A⋃ A' and S represent the same event, their probabilities are equal.
P(A⋃ A') = P(S)
Since the complement of A consists of the outcomes that are not in A, events A and A' are disjoint. By the Addition Rule of Probability, the probability of the union is the sum of the individual probabilities of each of the events.
P(A⋃ A') = P(A) + P(A')
Now the Transitive Property of Equality can be applied to the equalities.
P(A⋃ A') = P(A) + P(A') P(A⋃ A') = P(S) ⇓ P(A) + P(A') = P(S)
Additionally, the probability of the entire sample space P(S) is equal to 1. By applying the Transitive Property of Equality once more, the rule is proven.
P(A) + P(A') = 1
Using the Subtraction Property of Equality, the formula for the probability of A is obtained. P(A) + P(A') = 1 ⇕ P(A) = 1 - P(A')
Applying this formula, the probability of drawing a spade, heart, or diamond can be computed.
P(♠, ♡, or◊) &= 1 - P(♣) &= 1 - 0.25 &= 0.75
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Example
Counting Not Multiples of 5 vs. Counting Multiples of 5
Dylan cut out 100 squares of paper and wrote a number from 1 to 100 on each square. He then put the papers in a bag and asked his dad to choose a paper at random.
What is the probability that Dylan's father picks a number that is not a multiple of 5?
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Hint
What is easier to count, the numbers that are not multiples of 5 or the numbers that are multiples of 5? Note that these sets of numbers are the complements of each other.
Solution
Start by remembering what the probability of an event is.
P=Number of favorable outcomes/Number of possible outcomes
Let A be the given event, picking a number that is not a multiple of 5. However, counting the numbers from 1 to 100 that are not multiples of 5 can be tedious. In this case, it is worth considering the complement of A.
| Event | Complement |
|---|---|
| Picking a number that is not a multiple of 5. | Picking a number that is a multiple of 5. |
Counting how many numbers are multiples of 5 is an easier task. 5,10,15,20,25,30,35,40,45,50, 55,60,65,70,75,80,85,90,95,100 There is a total of 20 favorable outcomes for A'. Knowing this, the probability of A' can be calculated. P(A') = 20/100 = 0.2 The probability of picking a multiple of 5 is 0.2. Applying the Complement Rule, the probability of A can be computed.
Consequently, the probability that Dylan's father picks a number that is not a multiple of 5 is 0.8.
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Closure
Chances of Picking the Right Door
At the beginning of this lesson, Tearrik was invited by the host of a late-night game show to pick one door out of thee to win a car. However, behind two of the doors are sheep.
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b
What are the chances of choosing a door with a sheep? Round the answer to two decimal places.
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c
If there were 70 doors and 20 prizes, what is the chance that Tearrik picks a door with a sheep behind it? Round the answer to two decimal places.
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Hint
a
Divide the number of favorable outcomes by the number of possible outcomes.
b
Use the Complement Rule.
c
How many of the doors have sheep behind them?
Solution
a
The probability of an event is the quotient of the number of favorable outcomes to the number of possible outcomes.
P=Number of favorable outcomes/Number of possible outcomes Since there are three doors, there are 3 total possible outcomes, of which only 1 is favorable. Knowing this, the chance that Tearrik will choose the wining door can be calculated. P(winning door) = 1/3 ≈ 0.33
b
Note that the event of picking a door with a sheep is the complement of picking the winning door. Therefore, the Complement Rule can be used.
P(A') = 1 - P(A) In Part A, the probability of picking the winning door was found to be about 0.33. Substituting this value into the previous equation, the probability of picking a door with a sheep will be obtained. P(A') = 1 - 0.33 ≈ 0.67
c
In this case, there are 70 doors and 20 prizes. This time finding a sheep is the favorable outcome.
Since there are 20 prizes, there are 50 doors with sheep. With this information, the required probability can be found.
P(Sheep) = Favorable outcomes/Total outcomes
SubstituteII
Favorable outcomes= 50, Total outcomes= 70
P(Sheep) = 50/70
ReduceFrac
a/b=.a /10./.b /10.
P(Sheep) = 5/7
CalcQuot
Calculate quotient
P(Sheep) = 0.71428...
RoundDec
Round to 2 decimal place(s)
P(Sheep) ≈ 0.71
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The Basics of Event Probability
Exercises
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Let's draw all the possible combinations when rolling two dice.
There are multiple combinations that add up to the same number. Remember, we cannot count an outcome multiple times. Therefore, the sample space is as follows. {2,3,4,5,6,7,8,9,10,11,12} The sample space has 11 outcomes.
To determine the probability of rolling an even number, we will highlight all even combinations in the table made earlier. Remember, an even number is divisible by 2.
We see that 18 combinations are even. Since the sample space is 6 by 6, the total number of possible combinations is 6* 6= 36. Now we can calculate the probability by dividing the number of favorable combinations by the number of possible combinations. P=18/36= 1/2 The probability of rolling an even number is 12.
Let's list all of the dice combinations that give a difference of 3. & Die 1-Die 2 [-1.1em] &D1 - D2 = 4-1 = 3 &D1 - D2 = 5-2 = 3 &D1 - D2 = 6-3 = 3 [1em] & Die 2-Die 1 [-1.1em] &D2 -D1 = 4-1 = 3 &D2 -D1 = 5-2 = 3 &D2 -D1 = 6-3 = 3 Let's highlight all of these combinations in the table we made earlier.
Of the 36 possible combinations, 6 of them give a favorable outcome. Now we can calculate the probability that the difference between the dice is 3. P=6/36= 1/6
The most likely outcome is the number in the sample space that occurs most frequently. If we again look at the table with all the possible combinations for the dice, we see that 7 is the most frequent outcome.
There are 6 different combinations that give a seven. With this information, we can calculate the probability of rolling this outcome. P=6/36= 1/6
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From the exercise, we know that of the 2000 computers tested, 8 have defects. If we divide the number of defective computers by the total number of tested computers, we get the probability of a computer having a defect.
Consequently, there is a 0.4 % chance that a computer picked at random has a defect.
To predict the number of computers with defects in a shipment of 100 000, we must multiply the probability found earlier by the number of computers in the shipment. 100 000(0.004)=400 About 400 computers in the shipment could be defective.
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A card is drawn from a standard deck of cards. |
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We flip a coin and draw a marble from a bag containing two purple marbles and one gold marble. |
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We draw three marbles, one at a time and without replacing them, from a bag containing three green marbles and four blue marbles. |
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A standard deck has 52 playing cards. Therefore, the number of possible outcomes when drawing a card from one is 52.
In this case, we have a coin and a bag of marbles. A coin has two outcomes, heads and tails. This produces two outcomes.
Next up, we have to choose a marble. This can be purple or gold. Notice that we are not concerned with probabilities to figure out the possible outcomes. Only the number of possible outcomes is important, which means we will disregard the fact that there are more purple marbles than gold marbles in the bag.
Therefore, the number of possible outcomes in the sample space is 4. {H, P}, {H, G }, {T, P}, {T, G}
As in the previous situation, we are not concerned with probabilities at this point. Rather, we want to list all the possible outcomes. We can make another tree diagram to show all the possibilities.
As we can see, we have a total of 8 possible outcomes. {G, B, B}, {G, B, G}, {G, G, B}, {G, G, G}, {B, B, B}, {B, B, G}, {B, G, B}, {B, G, G}
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Calculate the probability of the following events happening. Write the answers as a fraction.
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To calculate the probability of an event, we have to divide the number of favorable outcomes by the number of possible outcomes. P=Number of favorable outcomes/Number of possible outcomes In this case, Jordan has a regular six-sided die and we want to find the probability of her rolling at least a 2. This means we cannot roll a 1.
We have 5 favorable outcomes of 6 possible outcomes. Now we can determine the probability. P(At least a2)=5/6
Let's find all the possible even outcomes on a die.
We have 3 favorable outcomes out of 6 possible outcomes. Now we can determine the probability. P(Even)=3/6=1/2
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Kriz pulls out a ball without looking. Give the probability as a fraction that the ball is...
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Probability is calculated by dividing the number of favorable outcomes by the number of possible outcomes. P=Number of favorable outcomes/Number of possible outcomes We have 4 blue balls, 3 red balls, and 5 yellow balls. blue balls:& && 4 red balls:& && 3 yellow balls:& && 5 [-1.2em] total:& &&12 We want to know the probability of picking a blue ball. With the given information, we can determine the probability of picking a blue ball. P( Blue) = 4/12=1/3
This time, we must divide the sum of the blue and yellow balls by the total number of balls. From the previous section we know that 4 balls are blue and 5 are yellow, for a total of 9 balls. Now we can determine the probability. P(BlueorYellow) = 9/12=3/4 The probability of picking a blue or a yellow ball is 34.
The event of picking a ball that is neither blue nor yellow is the complement of picking a blue or yellow ball. We already calculated the probability of picking a blue or yellow ball in the previous section. P(Blue or Yellow) = 3/4 Therefore, the probability of picking a ball that is neither blue nor yellow can be found using the complement rule.
The probability of picking a ball that is neither blue nor yellow is 14.
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Consider the following probabilities: 0, 0.02, 0.7, and 1 Pair these probabilities with the likelihood of the following statements. Note that the given probabilities are not necessarily representative of the statement. i:& A person with a high income went & to college. ii:& An ant can fly a jet. iii:& A card dealer deals the ace of spades. iv:& Flipping a coin gives either heads or tails.
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Let's think about the statements one at the time.
i. A Person With a High Income Went to College
A person does not have to have gone to college to have a high income. However, it still increases the possibility of securing a high-paying job if they have. The most appropriate probability would be 0.7. This is completely arbitrary, by the way — someone can definitely succeed without having gone to college!
ii. An Ant Can Fly a Jet
Ants are not able to fly a jet created for a human. The probability of this is therefore 0.
iii. A Card Dealer Deals the Ace of Spades
The ace of spades is one card out of 52 cards in a standard deck. To calculate the probability of being dealt this card, we divide 1 by 52. P=1/52≈ 0.02 The probability is about 0.02.
iv. Flipping a Coin Gives Either Heads or Tails
There are two possible outcomes when we flip a coin, either heads or tails. Therefore, the probability of either of them happening must be 1.
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In a lottery there are three types of tickets. No prize $10 prize $20 prize Let A represent the event that a person wins $10. What is the complement to this event?
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If the person does not win $10, they will have drawn either a no prize ticket or a $20 prize ticket. They therefore have the following complement. $20 prize and No prize
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The Basics of Event Probability
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