Probability - AP Statistics
Card 0 of 328
Which of the following is NOT a discrete random variable?
Which of the following is NOT a discrete random variable?
By definition, a discrete random variable is a random variable whose values can be "counted" one by one. A continuous random variable is a random variable that can take any value on a certain interval. Of these choices, the number of lip products, the amount of money, and the number of midterms taken are all discrete random variables, as the respective values can be counted; however, the time taken to watch the first four seasons of a TV show is a continuous random variable, as not everyone will take the same amount of time to watch all those episodes (i.e. some might fastf-orward/replay parts of episodes).
By definition, a discrete random variable is a random variable whose values can be "counted" one by one. A continuous random variable is a random variable that can take any value on a certain interval. Of these choices, the number of lip products, the amount of money, and the number of midterms taken are all discrete random variables, as the respective values can be counted; however, the time taken to watch the first four seasons of a TV show is a continuous random variable, as not everyone will take the same amount of time to watch all those episodes (i.e. some might fastf-orward/replay parts of episodes).
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Given a fair coin, what is the probability of obtaining 5 heads and 3 tails from 8 tosses?
Given a fair coin, what is the probability of obtaining 5 heads and 3 tails from 8 tosses?
First, there are 8 trials and either choose 5 or 3 for heads or tails, respectively. Using this knowledge: 
. Next, the chance for either heads or tails is 0.5 and there are 5 heads and 3 tails. Thus: 
. Multiply: 
and 
and obtain 0.2188.
First, there are 8 trials and either choose 5 or 3 for heads or tails, respectively. Using this knowledge:
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There are 52 total cards in a full deck of playing cards. If a card dealer chooses 4 cards from the deck at random and without replacement, what is the chance that the dealer draws four kings as the first four cards?
There are 52 total cards in a full deck of playing cards. If a card dealer chooses 4 cards from the deck at random and without replacement, what is the chance that the dealer draws four kings as the first four cards?
In a normal deck of playing cards, there are 4 kings. Thus, when the dealer draws the first card, the chance of the dealer obtaining a king is 4 out of 52. Because this card has been picked and is not replaced, the chance that the next card chosen is a king is 3 out of 52. The chance the third card is a king is 2 out of 52 and the fourth card is 1 out of 52. Each of these events is multiplied together, thus obtaining the correct answer, 0.0000037.
In a normal deck of playing cards, there are 4 kings. Thus, when the dealer draws the first card, the chance of the dealer obtaining a king is 4 out of 52. Because this card has been picked and is not replaced, the chance that the next card chosen is a king is 3 out of 52. The chance the third card is a king is 2 out of 52 and the fourth card is 1 out of 52. Each of these events is multiplied together, thus obtaining the correct answer, 0.0000037.
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The probability that it will rain today is 0.35. What is the probability that it will not rain?
The probability that it will rain today is 0.35. What is the probability that it will not rain?
The answer is 0.65 because Pr(~Rain) is the complement of Pr(Rain) and both events are mutually exclusive.
When two events are mutually exclusive, 
. Since probabilities must sum up to 1, this implies that 
.
The answer is 0.65 because Pr(~Rain) is the complement of Pr(Rain) and both events are mutually exclusive.
When two events are mutually exclusive,
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Students collected 150 cans for a food drive. There were 23 cans of corn, 48 cans of beans, and 12 cans of tomato sauce. If a student randomly selects one can to give away, what is the probability that the can will be either tomato sauce or beans?
Students collected 150 cans for a food drive. There were 23 cans of corn, 48 cans of beans, and 12 cans of tomato sauce. If a student randomly selects one can to give away, what is the probability that the can will be either tomato sauce or beans?
In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. The possible outcomes are mutually exclusive because one can of food could not be both beans and tomato sauce. To determine the probability of the two possible outcomes, add them together and then find the least common denominator.
In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. The possible outcomes are mutually exclusive because one can of food could not be both beans and tomato sauce. To determine the probability of the two possible outcomes, add them together and then find the least common denominator.
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150 students are athletes at the school. 65 play baseball, 15 play basketball, and 10 play both basketball and baseball.
What is the probability that a randomly selected athlete will play either baseball or basketball or both sports?
150 students are athletes at the school. 65 play baseball, 15 play basketball, and 10 play both basketball and baseball.
What is the probability that a randomly selected athlete will play either baseball or basketball or both sports?
We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive. Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator.
We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive. Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator.
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Assume there is an election involving three parties: D, R, and I. The probability of D winning is .11, R winning is .78, and I winning is .11. What is the probability of D or R winning?
Assume there is an election involving three parties: D, R, and I. The probability of D winning is .11, R winning is .78, and I winning is .11. What is the probability of D or R winning?
Since all of the events are mutually exclusive (one of the parties must win), you can get the probability of either D or R winning by adding their probabilities.
Since the probability of D winning is .11 and R winning is .78, the probability of D or R winning is .89.
Since all of the events are mutually exclusive (one of the parties must win), you can get the probability of either D or R winning by adding their probabilities.
Since the probability of D winning is .11 and R winning is .78, the probability of D or R winning is .89.
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If 1 card is chosen at random from a deck of cards, what is the probability that it will be a heart or a king?
If 1 card is chosen at random from a deck of cards, what is the probability that it will be a heart or a king?
In a deck of cards, there are 52 total cards, 13 hearts, 4 kings, and 1 king that is a heart.
So, 
In a deck of cards, there are 52 total cards, 13 hearts, 4 kings, and 1 king that is a heart.
So,
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Using a standard deck of cards, what is the probability of choosing a single face card?
Using a standard deck of cards, what is the probability of choosing a single face card?
There are 
cards in a standard deck: 
cards in 
suits. There are 
face cards (King, Queen, Jack) in each suit, so there are 
total face cards.
Thus the probability of choosing a single face card is 
or 
.
There are
Thus the probability of choosing a single face card is
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A student randomly selected a highlighter from her desk. There were five highlighters on the desk, each of a different color--blue, green, yellow, red, and orange. What is the probability that the student selected either the red or the yellow highlighter?
A student randomly selected a highlighter from her desk. There were five highlighters on the desk, each of a different color--blue, green, yellow, red, and orange. What is the probability that the student selected either the red or the yellow highlighter?
In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. To determine the probability of the two possible outcomes, simply add them together. This is called the addition rule.
In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. To determine the probability of the two possible outcomes, simply add them together. This is called the addition rule.
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At a business conference, one participant was selected to give a presentation. There were 
total participants. 
had business degrees and 
did not. 
were entrepreneurs and 
were not. 
of those with business degrees were entrepreneurs. What is the probability that the person selected will have a business degree or be an entrepreneur?
At a business conference, one participant was selected to give a presentation. There were
We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive.
Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator. In our case, we are told that 2 of those who have a business degree are also entrepreneurs therefore we need to subtract to from the total to get the correct probability.
We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive.
Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator. In our case, we are told that 2 of those who have a business degree are also entrepreneurs therefore we need to subtract to from the total to get the correct probability.
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What is the probability of obtaining at most 
heads when tossing a fair coin 
times?
What is the probability of obtaining at most
First find the number of ways to get 0, 1, or 2 heads.
Remember, 
.Also, 
and 
.
heads: 
head: 
heads: 
Now combine these to find the probability of seeing at most 2 heads in 10 coin tosses:
First find the number of ways to get 0, 1, or 2 heads.
Remember,
Now combine these to find the probability of seeing at most 2 heads in 10 coin tosses:
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In a standard deck of 
playing cards, what is the probability that the first card drawn is a face card (a king, queen or jack) or a diamond?
In a standard deck of
Bear in mind how the addition rule applies here. This will be a lengthy application, but applied properly, it will see you through this problem. It helps to think of this as four events that are not mutually exclusive - drawing a queen, drawing a jack, drawing a king or drawing a diamond. We then realize that we have to avoid counting the king, queen and jack of diamonds twice.
Bear in mind how the addition rule applies here. This will be a lengthy application, but applied properly, it will see you through this problem. It helps to think of this as four events that are not mutually exclusive - drawing a queen, drawing a jack, drawing a king or drawing a diamond. We then realize that we have to avoid counting the king, queen and jack of diamonds twice.
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When rolling a 6 sided dice, what is the probability of rolling a 2 or a 4?
When rolling a 6 sided dice, what is the probability of rolling a 2 or a 4?
In the single roll of a dice, rolling a 2 is mutually exclusive of rolling a 4. When a question asks for the probability of event A "or" event B, the probability is the sum of each event.
First, find the probability of each individual event.
Because the problem asks for a 2 OR a 4, add the indivual probabilities together.
In the single roll of a dice, rolling a 2 is mutually exclusive of rolling a 4. When a question asks for the probability of event A "or" event B, the probability is the sum of each event.
First, find the probability of each individual event.
Because the problem asks for a 2 OR a 4, add the indivual probabilities together.
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A magician has a bag containing 13 black marbles, 1 red marble, 1 green marble, 1 blue marble, and 1 yellow marble. What is the probability that the magician draws a red or a green marble?
A magician has a bag containing 13 black marbles, 1 red marble, 1 green marble, 1 blue marble, and 1 yellow marble. What is the probability that the magician draws a red or a green marble?
In the draw of a marble, picking a red is mutually exclusive of picking a green marble. When a question asks for the probability of event A "or" event B, the probability is the sum of each event.
First, find the probability of each event occuring.
Because the problem asks for the probability of either red OR green, we add the two probabilities.
In the draw of a marble, picking a red is mutually exclusive of picking a green marble. When a question asks for the probability of event A "or" event B, the probability is the sum of each event.
First, find the probability of each event occuring.
Because the problem asks for the probability of either red OR green, we add the two probabilities.
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A person is drawing a single card from a regular deck of 52 cards. What is the probability that they draw a heart or a spade?
A person is drawing a single card from a regular deck of 52 cards. What is the probability that they draw a heart or a spade?
In a single draw of a card, drawing a heart is mutually exclusive of drawing a spade, so we can use the addition rule to find the probability of a heart of a spade.
First, you must find the probability of each seperate event.
To find the probability of a heart or a spade, just add the probability of each event occuring.
In a single draw of a card, drawing a heart is mutually exclusive of drawing a spade, so we can use the addition rule to find the probability of a heart of a spade.
First, you must find the probability of each seperate event.
To find the probability of a heart or a spade, just add the probability of each event occuring.
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A person draws a single card from a regular deck of 52 playing cards. What is the probability of drawing a heart or a jack?
A person draws a single card from a regular deck of 52 playing cards. What is the probability of drawing a heart or a jack?
This problem asks for the probability of one event or another, so we will be using the addition rule. Because the event drawing a heart is not mutually exclusive from the event drawing a Jack, we must subtract the probability of getting both. Otherwise, we will double-count the Jack of Hearts.
First, find the probability of each event occuring
Now, find the probability of drawing both a Jack and a Heart. There is one card that is both.
Now, solve the equation.
This problem asks for the probability of one event or another, so we will be using the addition rule. Because the event drawing a heart is not mutually exclusive from the event drawing a Jack, we must subtract the probability of getting both. Otherwise, we will double-count the Jack of Hearts.
First, find the probability of each event occuring
Now, find the probability of drawing both a Jack and a Heart. There is one card that is both.
Now, solve the equation.
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You have a deck of cards and you draw one card. What is the probability of drawing an ace or a spade?
You have a deck of cards and you draw one card. What is the probability of drawing an ace or a spade?
You must use the addition rule of probability which is the probability of either of two dependent events happening is the sum of the probabilities of each dependent event minus the probabilty of both happening (this eliminates double counting possible out comes). The probabilty of getting an ace is 
and the probability of getting a spade is 
. The probability of getting an ace of spades is 
. The final answer would then be 
= 
.
You must use the addition rule of probability which is the probability of either of two dependent events happening is the sum of the probabilities of each dependent event minus the probabilty of both happening (this eliminates double counting possible out comes). The probabilty of getting an ace is
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Billy likes to play sports. He plays baseball 30% of his afternoons, and soccer 40%. But he gets tired so he only plays both sports in the same day 15% of the time.
How many days a week does Billy play sports? (with rounding)
Billy likes to play sports. He plays baseball 30% of his afternoons, and soccer 40%. But he gets tired so he only plays both sports in the same day 15% of the time.
How many days a week does Billy play sports? (with rounding)
Billy plays baseball 40%+football 30% gets 70%.
But he plays both 15% so:
%
Billy plays baseball 40%+football 30% gets 70%.
But he plays both 15% so:
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If 
and 
, what is 
?
If
2 ways: 1st, sum of 
, since 
anything above 1 represents the overlap, or 
2nd: since 
, the portion of the data in B and not A (
) is 
(think venn diagram), and thus the portion of 
included in A is 
2 ways: 1st, sum of
2nd: since
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