Calculating Probability - Algebra II
Card 0 of 168
A biased coin is tossed. The probability of landing heads is 
, and the probability of landing tails is 
. What is the probability of tossing three tails in a row, followed by one head?
A biased coin is tossed. The probability of landing heads is
Each coin toss is an independent event, meaning the outcome of one toss does not affect the outcome of the other coin tosses. To find the overall probability, multiply the probabilities of each toss together:
Each coin toss is an independent event, meaning the outcome of one toss does not affect the outcome of the other coin tosses. To find the overall probability, multiply the probabilities of each toss together:
Compare your answer with the correct one above
What is the probability of drawing 2 aces from a standard deck of playing cards when the first card is not replaced?
What is the probability of drawing 2 aces from a standard deck of playing cards when the first card is not replaced?
There are 4 aces in a deck of 52 playing cards.
After 1 ace is removed, there are 3 aces and 51 cards remaining:
To determined the probability of both events, multiply the probability of drawing each card:
There are 4 aces in a deck of 52 playing cards.
After 1 ace is removed, there are 3 aces and 51 cards remaining:
To determined the probability of both events, multiply the probability of drawing each card:
Compare your answer with the correct one above
A bag of marbles contains 4 gold and 6 blue marbles. What is the probability that you will draw first a blue marble then a gold marble? Assume the first marble you draw is not placed back in the bag.
A bag of marbles contains 4 gold and 6 blue marbles. What is the probability that you will draw first a blue marble then a gold marble? Assume the first marble you draw is not placed back in the bag.
Selecting a specific sequence of marbles from the bag without replacement is a dependent or conditional probabiliy. If you successfully draw a blue marble, it affects the chances of subsequently drawing a gold marble.
In the notation of probability we can call the probability of drawing a blue marble 
and the probability of drawing a gold marble 
.
If we replaced the first marble we drew, we would simply multiply 
times 
, but since we aren't we want to know 
or, the probability of B, given A.
So the probability of A and B is equal to:
In words, you have a six in ten chance of drawing a blue marble. If you successfully do that, you then have a four in nine chance of drawing a gold marble. Multiply those odds together, and reduce the fraction and you get a four in fifteen probability.
Selecting a specific sequence of marbles from the bag without replacement is a dependent or conditional probabiliy. If you successfully draw a blue marble, it affects the chances of subsequently drawing a gold marble.
In the notation of probability we can call the probability of drawing a blue marble
If we replaced the first marble we drew, we would simply multiply
So the probability of A and B is equal to:
In words, you have a six in ten chance of drawing a blue marble. If you successfully do that, you then have a four in nine chance of drawing a gold marble. Multiply those odds together, and reduce the fraction and you get a four in fifteen probability.
Compare your answer with the correct one above
50 high school seniors were asked about their plans following graduation. 20 planned to attend an in state college. 12 planned to attend an out of state college. 8 did not plan to attend college and 10 were undecided.
What percent of this group of seniors is not planning to attend an in state college?
50 high school seniors were asked about their plans following graduation. 20 planned to attend an in state college. 12 planned to attend an out of state college. 8 did not plan to attend college and 10 were undecided.
What percent of this group of seniors is not planning to attend an in state college?
First, calculate the number of students not planning on attending an in state college.
Either subtract those planning on attending an in state college from the total...
50 - 20 = 30
...or add up all the other categories.
12 + 8 + 10 = 30
Then divide by the total number of students.
30/50 = 60%
First, calculate the number of students not planning on attending an in state college.
Either subtract those planning on attending an in state college from the total...
50 - 20 = 30
...or add up all the other categories.
12 + 8 + 10 = 30
Then divide by the total number of students.
30/50 = 60%
Compare your answer with the correct one above
In a sample of 100 high school students, 65 students played an organized sport. 48 students have a dog as a pet. What is the probability that a randomly selected student has a dog and plays a sport?
In a sample of 100 high school students, 65 students played an organized sport. 48 students have a dog as a pet. What is the probability that a randomly selected student has a dog and plays a sport?
Playing a sport, which we can call P(A), and owning a dog, which we can call P(B), are traits that we can assume are independent. The probability of one does not affect the probability of the other.
So, P(A and B) = P(A) * P(B)
Fill in the probabilities given into this equation.
P(A and B) = (0.65) * (0.48) = 0.312 = 31.2%
Playing a sport, which we can call P(A), and owning a dog, which we can call P(B), are traits that we can assume are independent. The probability of one does not affect the probability of the other.
So, P(A and B) = P(A) * P(B)
Fill in the probabilities given into this equation.
P(A and B) = (0.65) * (0.48) = 0.312 = 31.2%
Compare your answer with the correct one above
In a sample of cars at a parking garage, the probability of a randomly selected car being blue was found to be 0.22. The probability of a car being blue OR green was found to be 0.33. What is the probability of car from this garage being green? (Assume a car can be only one color.)
In a sample of cars at a parking garage, the probability of a randomly selected car being blue was found to be 0.22. The probability of a car being blue OR green was found to be 0.33. What is the probability of car from this garage being green? (Assume a car can be only one color.)
The probability of finding either a blue car (A) or a green car (B) is written
P(A or B).
This type of probability can be found using the equation:
P(A or B) = P(A) + P(B) - P(A and B)
We need to solve this equation for P(B).
Two of the terms are given in the prompt:
P(A or B) = 0.33
P(A) = 0.22
The third, P(A and B) can be inferred. Since a car can be only one color,
P(A and B) = 0
Thus,
0.33 = 0.22 + P(B) - 0
P(B) = 0.11
The probability of finding either a blue car (A) or a green car (B) is written
P(A or B).
This type of probability can be found using the equation:
P(A or B) = P(A) + P(B) - P(A and B)
We need to solve this equation for P(B).
Two of the terms are given in the prompt:
P(A or B) = 0.33
P(A) = 0.22
The third, P(A and B) can be inferred. Since a car can be only one color,
P(A and B) = 0
Thus,
0.33 = 0.22 + P(B) - 0
P(B) = 0.11
Compare your answer with the correct one above
In a standard deck of card, what is the probality of drawing either a face card or a red card?
In a standard deck of card, what is the probality of drawing either a face card or a red card?
The probability of drawing a face card P(A) or a red card P(B) can be written as P(A or B), and be calculated using:
P(A or B) = P(A) + P(B) - P(A and B)
The first two terms on the right side of the equation are fairly straightforward. Since we'll be adding and subtracting fractions, I will keep the denominator the same and won't reduce the fractions until the end.
P(A) = 12/52
P(B) = 26/52 = 1/2
The third, P(A and B), can be calculated using:
P(A and B) = P(A) * P(B)
P(A and B) = 12/52 * 1/2 = 12/104 = 6/52
(Conversely, think there are 12 face cards in a deck of 52, half of which \[6\] are red.)
Back to our original equation!
P(A or B) = 12/52 + 26/52 - 6/52
= 32/52 = 8/13
The probability of drawing a face card P(A) or a red card P(B) can be written as P(A or B), and be calculated using:
P(A or B) = P(A) + P(B) - P(A and B)
The first two terms on the right side of the equation are fairly straightforward. Since we'll be adding and subtracting fractions, I will keep the denominator the same and won't reduce the fractions until the end.
P(A) = 12/52
P(B) = 26/52 = 1/2
The third, P(A and B), can be calculated using:
P(A and B) = P(A) * P(B)
P(A and B) = 12/52 * 1/2 = 12/104 = 6/52
(Conversely, think there are 12 face cards in a deck of 52, half of which \[6\] are red.)
Back to our original equation!
P(A or B) = 12/52 + 26/52 - 6/52
= 32/52 = 8/13
Compare your answer with the correct one above
If a coin is tossed in the air and a 6-sided die is rolled, what is the probability of getting a tail on the coin and a 3 on the die?
If a coin is tossed in the air and a 6-sided die is rolled, what is the probability of getting a tail on the coin and a 3 on the die?
Multiply these two probabilities together:
Therefore the probability of flipping a tail on a coin and rolling a 3 is 
.
Multiply these two probabilities together:
Therefore the probability of flipping a tail on a coin and rolling a 3 is
Compare your answer with the correct one above
A restaurant serves apple pie and cherry pie. On average, 60% of the customers choose apple pie. If the next 4 customers order pie, what is the probability 3 of them order the cherry pie?
A restaurant serves apple pie and cherry pie. On average, 60% of the customers choose apple pie. If the next 4 customers order pie, what is the probability 3 of them order the cherry pie?
Binomial probability
Binomial probability
Compare your answer with the correct one above
If I flip a coin twice, what's the probability it lands on tails twice?
If I flip a coin twice, what's the probability it lands on tails twice?
Let's list all possible outcomes.
We have H T, H, H, T T, T H.
There are four possible outcomes so that represents the denominator.
Since we are looking for two tails, we only have one possible outcome which is the numerator.
Our final answer is
.
Let's list all possible outcomes.
We have H T, H, H, T T, T H.
There are four possible outcomes so that represents the denominator.
Since we are looking for two tails, we only have one possible outcome which is the numerator.
Our final answer is
Compare your answer with the correct one above
Suppose a loaded coin lands on heads 
times.
What is the probability that the coin will land HTTH?
Suppose a loaded coin lands on heads
What is the probability that the coin will land HTTH?
The probability of flipping the coin in this particular order is calculated by multiplying the probabilities of each turn by each other. Because we have a "loaded" coin the probabilities are as follows:
.
Then simplify:
.
This fraction is in its simplest form.
The probability of flipping the coin in this particular order is calculated by multiplying the probabilities of each turn by each other. Because we have a "loaded" coin the probabilities are as follows:
Then simplify:
This fraction is in its simplest form.
Compare your answer with the correct one above
Dylan rolls a six-sided die. What is the probability that he will roll a five on the first roll and a six on the second roll?
Dylan rolls a six-sided die. What is the probability that he will roll a five on the first roll and a six on the second roll?
To calculate the probability of two independent events occuring, you multiply the probabilities of each event occuring independently.
In this case, the probability of rolling any number is 
, so the probability of rolling this particular combination is:
.
To calculate the probability of two independent events occuring, you multiply the probabilities of each event occuring independently.
In this case, the probability of rolling any number is
Compare your answer with the correct one above
What is the probability of rolling two dice and have a sum that is no greater than three?
What is the probability of rolling two dice and have a sum that is no greater than three?
A die has 6 faces with a number on each face from one to six.
Calculate the total number of combinations of the 2 dice.
Write out the combination sets of the condition that satisfy the scenario.
There are only three sets that will achieve the sum of 2 or 3, but no greater than 3. Divide the total number of possibilities by the total number of combinations.
The probability of rolling a sum that is no greater than 3 is: 
A die has 6 faces with a number on each face from one to six.
Calculate the total number of combinations of the 2 dice.
Write out the combination sets of the condition that satisfy the scenario.
There are only three sets that will achieve the sum of 2 or 3, but no greater than 3. Divide the total number of possibilities by the total number of combinations.
The probability of rolling a sum that is no greater than 3 is:
Compare your answer with the correct one above
If there are ten marbles with only two colors and six are red, what's the probability that you pick one marble and is red?
If there are ten marbles with only two colors and six are red, what's the probability that you pick one marble and is red?
There are ten marbles in the bag and there are only two colors. One of the colors is red and there are six of them.
Because it's asking for red, we write a fraction.
The top of the fraction represents what the question is asking and the bottom represents the total possible outcomes.
So our answer is
.
There are ten marbles in the bag and there are only two colors. One of the colors is red and there are six of them.
Because it's asking for red, we write a fraction.
The top of the fraction represents what the question is asking and the bottom represents the total possible outcomes.
So our answer is
Compare your answer with the correct one above
What's the probability I pick a face card in a standard 
card deck?
What's the probability I pick a face card in a standard
There are three face cards which are J Q K.
Next, there are four suits for every face card which gives us a total of 
cards we are looking for.
Finally, since there are 
cards in a deck, we do a fraction of 
.
There is no answer given however if we divide top and bottom by 
we get an answer of
.
There are three face cards which are J Q K.
Next, there are four suits for every face card which gives us a total of
Finally, since there are
There is no answer given however if we divide top and bottom by
Compare your answer with the correct one above
In a bag has slips of paper each numbered from 
. What's the probability you pick a perfect square?
In a bag has slips of paper each numbered from
Our perect squares are
.
There are ten of them. Since there are 
slips in the bag, our fraction becomes 
.
If we cut a zero from both top and bottom, which is the same as dividing by ten, our final answer is
.
Our perect squares are
There are ten of them. Since there are
If we cut a zero from both top and bottom, which is the same as dividing by ten, our final answer is
Compare your answer with the correct one above
What's the probability of a die rolled once and shows a prime number?
What's the probability of a die rolled once and shows a prime number?
In a die, there are six sides numbered
.
Prime numbers are numbers with a factor of one and itself.
The prime numbers are 
.
We have three outcomes and it's divided over six possible outcomes.
So our fraction is 
or 
.
In a die, there are six sides numbered
Prime numbers are numbers with a factor of one and itself.
The prime numbers are
We have three outcomes and it's divided over six possible outcomes.
So our fraction is
Compare your answer with the correct one above
Bob picks a random card in a standard 
card deck. After keeping the card, he takes another card and keeps it again. What's the probability that he has a pair in his hand?
Bob picks a random card in a standard
In a standard deck, there are thirteen kind of cards from 
. There are four suits. To have a pair is to have the same kind of card but with different suit. Since we are looking for any pair, the first card Bob picks is anything so the probability is 
.
Because he keeps the card, we have 
cards remaining. To get that same card, there are only 
of them left in the deck. So the chance of getting a pair becomes 
.
We don't have this choice but if we divide both sides by 
we get 
.
In a standard deck, there are thirteen kind of cards from
Because he keeps the card, we have
We don't have this choice but if we divide both sides by
Compare your answer with the correct one above
If for each question on an exam, there is only one right answer and four wrong answers, what's the probability of getting the wrong answer for two questions?
If for each question on an exam, there is only one right answer and four wrong answers, what's the probability of getting the wrong answer for two questions?
The probability of getting a question wrong is 
.
Since we want to get it wrong twice, we need to multiply as these are events that must occur.
So we do
.
The probability of getting a question wrong is
Since we want to get it wrong twice, we need to multiply as these are events that must occur.
So we do
Compare your answer with the correct one above
There are four red socks, five blue socks, eight white socks, and two yellow socks, What's the probability of picking a blue or yellow sock?
There are four red socks, five blue socks, eight white socks, and two yellow socks, What's the probability of picking a blue or yellow sock?
When we see or, we must add probabilities. Since there are 
socks and we are interested in blue or yellow, we add those totals up.
There are 
blue or yellow socks to get.
Our fraction now becomes
.
When we see or, we must add probabilities. Since there are
There are
Our fraction now becomes
Compare your answer with the correct one above