Find probabilities - HiSET
Card 0 of 36
If I have 
quarters, what is the probability that I will get no more than 
heads?
If I have 
Tap to see back →
Step 1: If we have 3 quarters and each quarter has two outcomes, then there are 
total outcomes that can come out of all of the rolls of the quarters.
Those rolls are:
HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.
Step 2: Determine how many outcomes have no more than 
heads....
We cannot count HHH because it has 
heads.
We also cannot count TTT because it has no heads at all.
Step 3: We had 
options and we can't take 
options..so we have a total of 
options left.
Step 1: If we have 3 quarters and each quarter has two outcomes, then there are 
Those rolls are:
HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.
Step 2: Determine how many outcomes have no more than 
We cannot count HHH because it has 
We also cannot count TTT because it has no heads at all.
Step 3: We had 
In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?
In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?
Tap to see back →
Step 1: Find the probability of getting a green marble...
The probability of a green marble is 
.
Step 2: Find the probability of getting a blue marble (without replacement)
When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is 
less.
Probability of a blue marble is 
.
Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.
So, 
Step 4: Simplify...
is the answer.
Step 1: Find the probability of getting a green marble...
The probability of a green marble is 
Step 2: Find the probability of getting a blue marble (without replacement)
When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is 
Probability of a blue marble is 
Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.
So,
Step 4: Simplify...
What is the probability that we will roll a one on a six sided die and flip a coin to reveal "heads"?
What is the probability that we will roll a one on a six sided die and flip a coin to reveal "heads"?
Tap to see back →
In order to solve this standard we need to understand two primary components: probabilities and the property of independent events. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
An intersection is the likelihood that one event will occur with another.
We can calculate that the probability of rolling a one on a six-sided die is:
Now, let's determine the probability of flipping a coin to reveal the "heads" side. A coin has two sides: heads or tails. The probability of rolling heads is as follows:
An intersection is the probability of two events occurring simultaneously; therefore, we need to multiply the probabilities.
Convert to a percentage.
In order to solve this standard we need to understand two primary components: probabilities and the property of independent events. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
An intersection is the likelihood that one event will occur with another.
We can calculate that the probability of rolling a one on a six-sided die is:
Now, let's determine the probability of flipping a coin to reveal the "heads" side. A coin has two sides: heads or tails. The probability of rolling heads is as follows:
An intersection is the probability of two events occurring simultaneously; therefore, we need to multiply the probabilities.
Convert to a percentage.
The spades and the aces are removed from a standard deck of 52. Give the probability that a card drawn at random will be a red three.
The spades and the aces are removed from a standard deck of 52. Give the probability that a card drawn at random will be a red three.
Tap to see back →
In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of
cards, to leave a modified deck of
cards.
None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is
.
In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of
cards, to leave a modified deck of
cards.
None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is
How many ways can eight people be seated at two tables if four people should be at each table, and it only matters at which table each person sits (ass opposed to which seat)?
How many ways can eight people be seated at two tables if four people should be at each table, and it only matters at which table each person sits (ass opposed to which seat)?
Tap to see back →
The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is 
.
Since
,
then, setting 
,
,
the correct number of arrangements.
The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is
Since
then, setting
the correct number of arrangements.
A standardized test comprises five examinations: reading, writing, math, science, and social studies, and these examinations may be taken in any order; however, in one state, there is one restriction: neither the writing nor the math test may be taken last.
In how many different orders can the five tests be taken so that this restriction is met?
A standardized test comprises five examinations: reading, writing, math, science, and social studies, and these examinations may be taken in any order; however, in one state, there is one restriction: neither the writing nor the math test may be taken last.
In how many different orders can the five tests be taken so that this restriction is met?
Tap to see back →
Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:
First, the test to be taken last must be one of three tests: reading, science, or social studies;
Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.
By the multiplication principle, this makes
possible schedules.
Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:
First, the test to be taken last must be one of three tests: reading, science, or social studies;
Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.
By the multiplication principle, this makes
possible schedules.
What is the probability that a person will roll an even number on a six-sided fair die?
What is the probability that a person will roll an even number on a six-sided fair die?
Tap to see back →
In order to solve this problem, we need to discuss probabilities. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
The die has six sides with the following values: one tow, three, four, five, and six. Of these values the die has three that are even: two, four, and six. We can write the following probability.
Simplify.
Now, let's convert this into a percentage:
In order to solve this problem, we need to discuss probabilities. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
The die has six sides with the following values: one tow, three, four, five, and six. Of these values the die has three that are even: two, four, and six. We can write the following probability.
Simplify.
Now, let's convert this into a percentage:
Two fair dice are tossed. You are not shown the dice but you are told that exactly one die shows an even number. Give the probability that a total of seven was thrown.
Two fair dice are tossed. You are not shown the dice but you are told that exactly one die shows an even number. Give the probability that a total of seven was thrown.
Tap to see back →
Two fair dice can be thrown with 36 equally probable rolls. 18 rolls comprise one even number and one odd number. They are as follows - with the rolls that total seven in boldface:
6 of the 18 possible rolls have total 7, making the probability of a roll of 7, given that exactly one doe shows an even number, equal to 
.
Two fair dice can be thrown with 36 equally probable rolls. 18 rolls comprise one even number and one odd number. They are as follows - with the rolls that total seven in boldface:
6 of the 18 possible rolls have total 7, making the probability of a roll of 7, given that exactly one doe shows an even number, equal to 
Two fair dice are tossed. You are not shown the dice but you are told that at least one die shows an even number. Give the probability that a total of seven was thrown.
Two fair dice are tossed. You are not shown the dice but you are told that at least one die shows an even number. Give the probability that a total of seven was thrown.
Tap to see back →
Two fair dice can be thrown with 36 equally probable rolls.9 rolls can be thrown that show two odd numbers: they are
This leaves 27 rolls with at least one even number, 6 of which result in a sum of 7:
This makes the probability of rolling a sum of 7
,
the correct choice.
Two fair dice can be thrown with 36 equally probable rolls.9 rolls can be thrown that show two odd numbers: they are
This leaves 27 rolls with at least one even number, 6 of which result in a sum of 7:
This makes the probability of rolling a sum of 7
the correct choice.
If I have quarters, what is the probability that I will get no more than heads?
If I have
Tap to see back →
Step 1: If we have 3 quarters and each quarter has two outcomes, then there are total outcomes that can come out of all of the rolls of the quarters.
Those rolls are:
HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.
Step 2: Determine how many outcomes have no more than heads....
We cannot count HHH because it has heads.
We also cannot count TTT because it has no heads at all.
Step 3: We had options and we can't take options..so we have a total of options left.
Step 1: If we have 3 quarters and each quarter has two outcomes, then there are
Those rolls are:
HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.
Step 2: Determine how many outcomes have no more than
We cannot count HHH because it has
We also cannot count TTT because it has no heads at all.
Step 3: We had
In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?
In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?
Tap to see back →
Step 1: Find the probability of getting a green marble...
The probability of a green marble is .
Step 2: Find the probability of getting a blue marble (without replacement)
When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is less.
Probability of a blue marble is .
Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.
So,
Step 4: Simplify...
is the answer.
Step 1: Find the probability of getting a green marble...
The probability of a green marble is
Step 2: Find the probability of getting a blue marble (without replacement)
When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is
Probability of a blue marble is
Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.
So,
Step 4: Simplify...
What is the probability that we will roll a one on a six sided die and flip a coin to reveal "heads"?
What is the probability that we will roll a one on a six sided die and flip a coin to reveal "heads"?
Tap to see back →
In order to solve this standard we need to understand two primary components: probabilities and the property of independent events. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
An intersection is the likelihood that one event will occur with another.
We can calculate that the probability of rolling a one on a six-sided die is:
Now, let's determine the probability of flipping a coin to reveal the "heads" side. A coin has two sides: heads or tails. The probability of rolling heads is as follows:
An intersection is the probability of two events occurring simultaneously; therefore, we need to multiply the probabilities.
Convert to a percentage.
In order to solve this standard we need to understand two primary components: probabilities and the property of independent events. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
An intersection is the likelihood that one event will occur with another.
We can calculate that the probability of rolling a one on a six-sided die is:
Now, let's determine the probability of flipping a coin to reveal the "heads" side. A coin has two sides: heads or tails. The probability of rolling heads is as follows:
An intersection is the probability of two events occurring simultaneously; therefore, we need to multiply the probabilities.
Convert to a percentage.
The spades and the aces are removed from a standard deck of 52. Give the probability that a card drawn at random will be a red three.
The spades and the aces are removed from a standard deck of 52. Give the probability that a card drawn at random will be a red three.
Tap to see back →
In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of
cards, to leave a modified deck of
cards.
None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is
.
In a deck of 52 cards, one-fourth of the cards, or 13 cards, are spades. Also, there are four aces, one of which is a spade and three of which are not spades. It follows that the removal of the spades and the aces results in the removal of
cards, to leave a modified deck of
cards.
None of the two red threes (the three of clubs, the three of diamonds) were removed, so the probability of a random draw of a red three is
How many ways can eight people be seated at two tables if four people should be at each table, and it only matters at which table each person sits (ass opposed to which seat)?
How many ways can eight people be seated at two tables if four people should be at each table, and it only matters at which table each person sits (ass opposed to which seat)?
Tap to see back →
The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is .
Since
,
then, setting ,
,
the correct number of arrangements.
The number of ways to choose four people to sit at the first table without regard to the arrangement at either table is the number of combinations of four people from a set of eight; once the people who will sit at the first table are chosen, the people who will sit at the second are automatically chosen. Therefore, the correct number of such arrangements is
Since
then, setting
the correct number of arrangements.
A standardized test comprises five examinations: reading, writing, math, science, and social studies, and these examinations may be taken in any order; however, in one state, there is one restriction: neither the writing nor the math test may be taken last.
In how many different orders can the five tests be taken so that this restriction is met?
A standardized test comprises five examinations: reading, writing, math, science, and social studies, and these examinations may be taken in any order; however, in one state, there is one restriction: neither the writing nor the math test may be taken last.
In how many different orders can the five tests be taken so that this restriction is met?
Tap to see back →
Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:
First, the test to be taken last must be one of three tests: reading, science, or social studies;
Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.
By the multiplication principle, this makes
possible schedules.
Because the last test cannot be either math or writing, the selection of an order in which the tests are taken must be looked at as a sequence of events as follows:
First, the test to be taken last must be one of three tests: reading, science, or social studies;
Second, the remaining four tests must be arranged in a particular order. There are four ways to schedule the first test, three remaining ways to schedule the third, two for the third, and one for the fourth.
By the multiplication principle, this makes
possible schedules.
What is the probability that a person will roll an even number on a six-sided fair die?
What is the probability that a person will roll an even number on a six-sided fair die?
Tap to see back →
In order to solve this problem, we need to discuss probabilities. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
The die has six sides with the following values: one tow, three, four, five, and six. Of these values the die has three that are even: two, four, and six. We can write the following probability.
Simplify.
Now, let's convert this into a percentage:
In order to solve this problem, we need to discuss probabilities. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:
The die has six sides with the following values: one tow, three, four, five, and six. Of these values the die has three that are even: two, four, and six. We can write the following probability.
Simplify.
Now, let's convert this into a percentage:
Two fair dice are tossed. You are not shown the dice but you are told that exactly one die shows an even number. Give the probability that a total of seven was thrown.
Two fair dice are tossed. You are not shown the dice but you are told that exactly one die shows an even number. Give the probability that a total of seven was thrown.
Tap to see back →
Two fair dice can be thrown with 36 equally probable rolls. 18 rolls comprise one even number and one odd number. They are as follows - with the rolls that total seven in boldface:
6 of the 18 possible rolls have total 7, making the probability of a roll of 7, given that exactly one doe shows an even number, equal to .
Two fair dice can be thrown with 36 equally probable rolls. 18 rolls comprise one even number and one odd number. They are as follows - with the rolls that total seven in boldface:
6 of the 18 possible rolls have total 7, making the probability of a roll of 7, given that exactly one doe shows an even number, equal to
Two fair dice are tossed. You are not shown the dice but you are told that at least one die shows an even number. Give the probability that a total of seven was thrown.
Two fair dice are tossed. You are not shown the dice but you are told that at least one die shows an even number. Give the probability that a total of seven was thrown.
Tap to see back →
Two fair dice can be thrown with 36 equally probable rolls.9 rolls can be thrown that show two odd numbers: they are
This leaves 27 rolls with at least one even number, 6 of which result in a sum of 7:
This makes the probability of rolling a sum of 7
,
the correct choice.
Two fair dice can be thrown with 36 equally probable rolls.9 rolls can be thrown that show two odd numbers: they are
This leaves 27 rolls with at least one even number, 6 of which result in a sum of 7:
This makes the probability of rolling a sum of 7
the correct choice.
If I have quarters, what is the probability that I will get no more than heads?
If I have
Tap to see back →
Step 1: If we have 3 quarters and each quarter has two outcomes, then there are total outcomes that can come out of all of the rolls of the quarters.
Those rolls are:
HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.
Step 2: Determine how many outcomes have no more than heads....
We cannot count HHH because it has heads.
We also cannot count TTT because it has no heads at all.
Step 3: We had options and we can't take options..so we have a total of options left.
Step 1: If we have 3 quarters and each quarter has two outcomes, then there are
Those rolls are:
HHH, HHT, HTH, THH, THT, TTH, TTT, HTT.
Step 2: Determine how many outcomes have no more than
We cannot count HHH because it has
We also cannot count TTT because it has no heads at all.
Step 3: We had
In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?
In a sack, I have 20 red marbles, 15 blue marbles and 5 green marbles. What is the chance, without replacement, that I pick a green marble first and then pick a blue marble?
Tap to see back →
Step 1: Find the probability of getting a green marble...
The probability of a green marble is .
Step 2: Find the probability of getting a blue marble (without replacement)
When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is less.
Probability of a blue marble is .
Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.
So,
Step 4: Simplify...
is the answer.
Step 1: Find the probability of getting a green marble...
The probability of a green marble is
Step 2: Find the probability of getting a blue marble (without replacement)
When we have without replacement, we do not put the first marble we take out back into the sack. In this case, the total number of marbles in the sack is
Probability of a blue marble is
Step 3: To find the probability of both events happening at the same time, we need to multiply the probabilities in the first two steps.
So,
Step 4: Simplify...