Probability - GED Math
Card 0 of 20
When rolling a 
-sided die, what is the probability of rolling 
or greater?
When rolling a
When rolling a die, the following outcomes are possible:
Of the 
outcomes, 
outcomes are 
or greater. Therefore,
When rolling a die, the following outcomes are possible:
Of the
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The two red queens are removed from a standard deck of 52 cards. What is the probability that a randomly dealt card from this altered deck will be red?
The two red queens are removed from a standard deck of 52 cards. What is the probability that a randomly dealt card from this altered deck will be red?
After the removal of two red cards from a deck of 52, there are now fifty cards, of which twenty-four are red. The probability of a random card being red is therefore 
, which as a percent is
After the removal of two red cards from a deck of 52, there are now fifty cards, of which twenty-four are red. The probability of a random card being red is therefore
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What is the probability of drawing a red jack from a deck of standard playing cards?
What is the probability of drawing a red jack from a deck of standard playing cards?
A standard deck of play cards has 
cards. There are 
jacks, 
of which are red. Therefore, the probability is:
A standard deck of play cards has
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A die is altered so that it comes up a "6" with probability 
. The other five outcomes are equally likely. If this die and a fair die are rolled, what is the probability that the outcome will be a total of "2"?
A die is altered so that it comes up a "6" with probability
For a "2" to be rolled with two dice, both dice must show a "1". In the fair die, this is one of six equally likely outcomes, so it will happen with probability 
.
In the loaded die, since a "6" will come up with probability 
, for the other five rolls to be equally likely, each, including "1", must come up with probability
.
A double "1" will appear with probability
.
For a "2" to be rolled with two dice, both dice must show a "1". In the fair die, this is one of six equally likely outcomes, so it will happen with probability
In the loaded die, since a "6" will come up with probability
A double "1" will appear with probability
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Given the data set 
, which of the following quantities are equal to each other/one another?
I: The mean
II: The median
III: The mode
Given the data set
I: The mean
II: The median
III: The mode
Since this data set is arranged in ascending order and has an even number of elements, the median of the data set is the arithmetic mean of its middle two elements. Both elements are 6, so this is the median.
6 is the mode, since it occurs most frequently.
The mean is the sum of the elements divided by the number of elements, which is 8:
The median and the mode are equal to each other, but not to the mean, so the correct answer is "II and III only".
Since this data set is arranged in ascending order and has an even number of elements, the median of the data set is the arithmetic mean of its middle two elements. Both elements are 6, so this is the median.
6 is the mode, since it occurs most frequently.
The mean is the sum of the elements divided by the number of elements, which is 8:
The median and the mode are equal to each other, but not to the mean, so the correct answer is "II and III only".
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Which of the following elements can be added to the data set
so that its mode(s) remain unchanged?
I: 
II: 
III: 
Which of the following elements can be added to the data set
so that its mode(s) remain unchanged?
I:
II:
III:
The mode of the data set is the most frequently occurring element. In the set given, 4 occurs three times, 5 occurs three times, 6 occurs two times, and 3 and 7 occur one time each. Therefore, the set has two modes, 4 and 5, and we want to preserve this condition.
If 3 is added to this set, it becomes
and 4 and 5 are still tied for the most frequently occurring element. The same happens if 7 is added to yield
.
If 5 is added to this set, it becomes
and 5 appears more frequently than 4 or any other element. This changes the data set to one with only one mode, 5.
The correct response is therefore "I and III only".
The mode of the data set is the most frequently occurring element. In the set given, 4 occurs three times, 5 occurs three times, 6 occurs two times, and 3 and 7 occur one time each. Therefore, the set has two modes, 4 and 5, and we want to preserve this condition.
If 3 is added to this set, it becomes
and 4 and 5 are still tied for the most frequently occurring element. The same happens if 7 is added to yield
If 5 is added to this set, it becomes
and 5 appears more frequently than 4 or any other element. This changes the data set to one with only one mode, 5.
The correct response is therefore "I and III only".
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A red die is altered so that it comes up a "6" with probability . The other five outcomes are equally likely. A blue die is altered similarly. If these two dice are rolled, what is the probability that the outcome will be a total of "2"?
A red die is altered so that it comes up a "6" with probability
For a "2" to be rolled with two dice, both dice must show a "1".
For each die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "1", must come up with probability
.
The probability of a double "1" showing up will be
.
For a "2" to be rolled with two dice, both dice must show a "1".
For each die, since a "6" will come up with probability
The probability of a double "1" showing up will be
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A red die is altered so that it comes up a "6" with probability . The other five numbers are equally likely outcomes to one another. A blue die is altered similarly. If these dice are rolled, what is the probability that the outcome will be a total of "11"?
A red die is altered so that it comes up a "6" with probability
For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". For each die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "5", must come up with probability
.
The probability of rolling a red "5" and a blue "6" will be
which is also the probability of rolling a blue "5" and a red "6". Add:
For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". For each die, since a "6" will come up with probability
The probability of rolling a red "5" and a blue "6" will be
which is also the probability of rolling a blue "5" and a red "6". Add:
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A die is altered so that it comes up a "6" with probability . The other five outcomes are equally likely. If this die and a fair die are rolled, what is the probability that the outcome will be a total of "11"?
A die is altered so that it comes up a "6" with probability
For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". In the fair die, "5" and "6" are two of six equally likely outcomes, so each will happen with probability .
In the loaded die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "5", must come up with probability
.
The probability of rolling a "5" on the loaded die and a "6" on the fair die is
.
The probability of rolling a "6" on the loaded die and a "5" on the fair die is
.
Add these probabilities:
For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". In the fair die, "5" and "6" are two of six equally likely outcomes, so each will happen with probability
In the loaded die, since a "6" will come up with probability
The probability of rolling a "5" on the loaded die and a "6" on the fair die is
The probability of rolling a "6" on the loaded die and a "5" on the fair die is
Add these probabilities:
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A penny and a nickel are altered so that the penny comes up heads 65% of time when tossed and the nickel comes up heads 60% of the time when tossed. Both coins are tossed; what is the probability that at least one coin will come up heads?
A penny and a nickel are altered so that the penny comes up heads 65% of time when tossed and the nickel comes up heads 60% of the time when tossed. Both coins are tossed; what is the probability that at least one coin will come up heads?
The easiest way to determine the probability of at least one head is to determine the complement of this event, which is the probability of two tails. The penny will come up tails 35% of the time (0.35) and the nickel will come up tails 40% of the time (0.40). These are independent events, so the probability of both happening is the product, or
.
The probability of at least one head - the complement of this event - is this probability subtracted from one, so
, or 86%.
The easiest way to determine the probability of at least one head is to determine the complement of this event, which is the probability of two tails. The penny will come up tails 35% of the time (0.35) and the nickel will come up tails 40% of the time (0.40). These are independent events, so the probability of both happening is the product, or
The probability of at least one head - the complement of this event - is this probability subtracted from one, so
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Roll a fair dice two times. What is the probability of NOT rolling a six on both trials?
Roll a fair dice two times. What is the probability of NOT rolling a six on both trials?
A fair dice has 6 faces. Therefore, there is a sixth chance to land on any number between 1 through 6. Rolling the die the first time is independent from the second time, which means that we must account for the probabilities for each trial.
Probability of rolling a 6 is: 1/6
Since probability adds up to 1, the odds of rolling anything other than a 6 for one trial is:
1 - 1/6 = 5/6
On two trials:
A fair dice has 6 faces. Therefore, there is a sixth chance to land on any number between 1 through 6. Rolling the die the first time is independent from the second time, which means that we must account for the probabilities for each trial.
Probability of rolling a 6 is: 1/6
Since probability adds up to 1, the odds of rolling anything other than a 6 for one trial is:
1 - 1/6 = 5/6
On two trials:
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Roll a fair dice. Then flip a coin. What's the probability of rolling a 3 or greater, and then flipping a heads?
Roll a fair dice. Then flip a coin. What's the probability of rolling a 3 or greater, and then flipping a heads?
There are 4 possibilities that a 3 or higher can be rolled on a die.
The probability of rolling a 3 or higher on a fair die is:
Rolling a heads on a coin has a probability of one half. Therefore,
There are 4 possibilities that a 3 or higher can be rolled on a die.
The probability of rolling a 3 or higher on a fair die is:
Rolling a heads on a coin has a probability of one half. Therefore,
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A penny is altered so that the odds are 5 to 4 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed, what are the odds of there being two heads or two tails?
A penny is altered so that the odds are 5 to 4 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed, what are the odds of there being two heads or two tails?
5 to 4 odds in favor of heads is equal to a probability of 
, which is the probability that the penny will come up heads. The probability that the penny will come up tails is 
.
Similarly, 4 to 3 odds in favor of heads is equal to a probability of 
, which is the probability that the nickel will come up heads. The probablity that the nickel will come up tails is 
.
The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.
The probability of the penny and the nickel both coming up heads is 
.
The probability of the penny and the nickel both coming up tails is 
.
The probabilities are added:
Since 
, this translates to 32 to 31 odds in favor of this event.
5 to 4 odds in favor of heads is equal to a probability of
Similarly, 4 to 3 odds in favor of heads is equal to a probability of
The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.
The probability of the penny and the nickel both coming up heads is
The probability of the penny and the nickel both coming up tails is
The probabilities are added:
Since
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Six boys and four girls are finalists in a drawing. The ten names are placed into a hat, and two are drawn at random. What is the probability that both names drawn will be girls?
Six boys and four girls are finalists in a drawing. The ten names are placed into a hat, and two are drawn at random. What is the probability that both names drawn will be girls?
Since order is irrelevant here, we are dealing with combinations.
There are
ways to choose two out of ten students.
There are
ways to choose two out of four girls.
The probability that two girls will be chosen is therefore
.
Since order is irrelevant here, we are dealing with combinations.
There are
ways to choose two out of ten students.
There are
ways to choose two out of four girls.
The probability that two girls will be chosen is therefore
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Seven boys and three girls are finalists in a drawing. The ten names are placed into a hat, and two are drawn at random. What is the probability that the names of one boy and one girl will be drawn?
Seven boys and three girls are finalists in a drawing. The ten names are placed into a hat, and two are drawn at random. What is the probability that the names of one boy and one girl will be drawn?
Since order is irrelevant here, we are dealing with combinations.
There are
ways to choose two out of ten students.
There are 
ways to choose one boy and one girl.
The probability that one boy and one girl will be chosen is therefore
.
Since order is irrelevant here, we are dealing with combinations.
There are
ways to choose two out of ten students.
There are
The probability that one boy and one girl will be chosen is therefore
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Find the probability of drawing a red card from a deck of cards.
Find the probability of drawing a red card from a deck of cards.
To find the probability of an event, we will use the following formula:
Given the event of drawing a red card from a deck of cards, we can calculate the following:
because there are 26 red cards in a deck
- 13 hearts
- 13 diamonds
Now, we can calculate the following:
because there are 52 total cards we could potentially draw.
So, we can substitute. We get
Therefore, the probability of drawing a red card from a deck of cards is 
.
To find the probability of an event, we will use the following formula:
Given the event of drawing a red card from a deck of cards, we can calculate the following:
because there are 26 red cards in a deck
- 13 hearts
- 13 diamonds
Now, we can calculate the following:
because there are 52 total cards we could potentially draw.
So, we can substitute. We get
Therefore, the probability of drawing a red card from a deck of cards is
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A parking lot contains the following:
- 3 blue cars
- 2 white cars
- 4 black cars
Find the probability the next car that leaves is a blue car.
A parking lot contains the following:
- 3 blue cars
- 2 white cars
- 4 black cars
Find the probability the next car that leaves is a blue car.
To find the probability of an event, we will use the following formula:
Now, given the event of a blue car leaves next, we can calculate the following:
because there are 3 blue cars in the lot.
We can also calculate the following:
because there are 9 total cars in the lot that could potentially leave:
- 3 blue cars
- 2 white cars
- 4 black cars
Now, we can substitute. We get
To find the probability of an event, we will use the following formula:
Now, given the event of a blue car leaves next, we can calculate the following:
because there are 3 blue cars in the lot.
We can also calculate the following:
because there are 9 total cars in the lot that could potentially leave:
- 3 blue cars
- 2 white cars
- 4 black cars
Now, we can substitute. We get
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Find the probability of drawing a black card from a deck of cards.
Find the probability of drawing a black card from a deck of cards.
To find the probability of an event, we will use the following formula:
So, given the event of drawing a black card from a deck of cards, we can calculate the following:
because there are 26 black cards in a deck
- 13 clubs
- 13 spades
Now, we can calculate the following:
because there are 52 total cards in a deck we could potentially draw. So, we can substitute. We get
Therefore, the probability of drawing a black card from a deck of cards is 
.
To find the probability of an event, we will use the following formula:
So, given the event of drawing a black card from a deck of cards, we can calculate the following:
because there are 26 black cards in a deck
- 13 clubs
- 13 spades
Now, we can calculate the following:
because there are 52 total cards in a deck we could potentially draw. So, we can substitute. We get
Therefore, the probability of drawing a black card from a deck of cards is
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A bookshelf at the library contains the following:
- 10 chapter books
- 5 picture books
- 3 history books
Find the probability the next visitor selects a history book.
A bookshelf at the library contains the following:
- 10 chapter books
- 5 picture books
- 3 history books
Find the probability the next visitor selects a history book.
To find the probability of an event, we will use the following formula:
Now, given the event of selecting a history book, we can calculate the following:
because there are 3 history books on the shelf.
We can also calculate the following:
because there are 18 total books on the shelf we could potentially choose from:
- 10 chapter books
- 5 non-fiction books
- 3 history books
Now, we can substitute. We get
Therefore, the probability of selecting a history book is 
.
To find the probability of an event, we will use the following formula:
Now, given the event of selecting a history book, we can calculate the following:
because there are 3 history books on the shelf.
We can also calculate the following:
because there are 18 total books on the shelf we could potentially choose from:
- 10 chapter books
- 5 non-fiction books
- 3 history books
Now, we can substitute. We get
Therefore, the probability of selecting a history book is
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Find the probability of drawing a King, Queen, or Jack from a deck of cards.
Find the probability of drawing a King, Queen, or Jack from a deck of cards.
To find the probability of an event, we will use the following formula:
Now, we will look at finding the probability of drawing a King, Queen, or Jack from a deck of cards. We will consider these as 3 separate events. We will find the probability of each, then add them together.
So, we will find the probability of drawing a King from a deck of cards. We get
because there are 4 Kings in a deck of cards:
- King of Hearts
- King of Diamonds
- King of Clubs
- King of Spades
We also get
because there are 52 cards in a deck that we would potentially draw.
So, we get
Now, we will do the same for finding the probability of drawing a Queen from a deck of cards.
because there are 4 Queens in a deck of cards:
- Queen of Hearts
- Queen of Diamonds
- Queen of Clubs
- Queen of Spades
We also get
because there are 52 cards in a deck that we would potentially draw.
So, we get
And lastly, we will find the probability of drawing a Jack. We get
because there are 4 Jacks in a deck of cards:
- Jack of Hearts
- Jack of Diamonds
- Jack of Clubs
- Jack of Spades
We also get
because there are 52 cards in a deck that we would potentially draw.
So, we get
Now, to find the probability of drawing a King, Queen, or Jack, we will add them all together. We get
Therefore, the probability of drawing a King, Queen, or Jack from a deck of cards is 
.
To find the probability of an event, we will use the following formula:
Now, we will look at finding the probability of drawing a King, Queen, or Jack from a deck of cards. We will consider these as 3 separate events. We will find the probability of each, then add them together.
So, we will find the probability of drawing a King from a deck of cards. We get
because there are 4 Kings in a deck of cards:
- King of Hearts
- King of Diamonds
- King of Clubs
- King of Spades
We also get
because there are 52 cards in a deck that we would potentially draw.
So, we get
Now, we will do the same for finding the probability of drawing a Queen from a deck of cards.
because there are 4 Queens in a deck of cards:
- Queen of Hearts
- Queen of Diamonds
- Queen of Clubs
- Queen of Spades
We also get
because there are 52 cards in a deck that we would potentially draw.
So, we get
And lastly, we will find the probability of drawing a Jack. We get
because there are 4 Jacks in a deck of cards:
- Jack of Hearts
- Jack of Diamonds
- Jack of Clubs
- Jack of Spades
We also get
because there are 52 cards in a deck that we would potentially draw.
So, we get
Now, to find the probability of drawing a King, Queen, or Jack, we will add them all together. We get
Therefore, the probability of drawing a King, Queen, or Jack from a deck of cards is
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