How to find the probability of an outcome - ACT Math
Card 0 of 819
A bag contains 6 yellow marbles, 5 red marbles, 4 blue marbles, and 3 green marbles. If John randomly selects a marble from the bag, what is the probability that he selects a marble that is neither red nor blue?
A bag contains 6 yellow marbles, 5 red marbles, 4 blue marbles, and 3 green marbles. If John randomly selects a marble from the bag, what is the probability that he selects a marble that is neither red nor blue?
Probability is the likelihood that an outcome will occur, and it is calculated by dividing the number of favorable events by the total number of possible events that may occur. In order to answer the question, we need to divide the number of non-red and non-blue marbles (i.e. the number of favorable events) by the total number of marbles in the bag (i.e. the total number of possible events). The total number of non-red, non-blue marbles is the same as the number of yellow and green marbles
Probability is the likelihood that an outcome will occur, and it is calculated by dividing the number of favorable events by the total number of possible events that may occur. In order to answer the question, we need to divide the number of non-red and non-blue marbles (i.e. the number of favorable events) by the total number of marbles in the bag (i.e. the total number of possible events). The total number of non-red, non-blue marbles is the same as the number of yellow and green marbles
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If two regular six-sided dice are rolled at the same time, what is the probability that the sum of the their numbers will be prime?
If two regular six-sided dice are rolled at the same time, what is the probability that the sum of the their numbers will be prime?
There number of possible outcomes is equal to six times six, or thirty-six. The sum of the two dice must be either 2, 3, 5, 7, or 11. There are 15 out of the 36 outcomes that would result in a sum that is a prime number:
\[1,1\], \[1,2\], \[1,4\], \[1,6\], \[2,1\], \[2,3\], \[2,5\], \[3,2\], \[3,4\], \[4,1\], \[4,3\], \[5,2\], \[5,6\], \[6,1\], \[6,5\]
There number of possible outcomes is equal to six times six, or thirty-six. The sum of the two dice must be either 2, 3, 5, 7, or 11. There are 15 out of the 36 outcomes that would result in a sum that is a prime number:
\[1,1\], \[1,2\], \[1,4\], \[1,6\], \[2,1\], \[2,3\], \[2,5\], \[3,2\], \[3,4\], \[4,1\], \[4,3\], \[5,2\], \[5,6\], \[6,1\], \[6,5\]
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A fair die is rolled three times. What are the odds that the die will come up a six on all three rolls?
A fair die is rolled three times. What are the odds that the die will come up a six on all three rolls?
The probability the die will turn up six on any given roll is 
. To find the probability of multiple events occurring, we multiply the individual probabilities together. 
The probability the die will turn up six on any given roll is
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There are 20 balls in a box, 10 are green, 5 are blue, and 5 are red. One red and one green ball are then removed from the box.
What is the probability that the third ball removed from the box will be green?
There are 20 balls in a box, 10 are green, 5 are blue, and 5 are red. One red and one green ball are then removed from the box.
What is the probability that the third ball removed from the box will be green?
This is a probability question, in the beginning, the odds of pulling a green ball were 10/20, which reduces down to 1/2. When 2 balls are removed, the total is now 18, and because there are only 9 green balls, we have a 9/18 chance of pulling a green ball, and this reduces down to 1/2
This is a probability question, in the beginning, the odds of pulling a green ball were 10/20, which reduces down to 1/2. When 2 balls are removed, the total is now 18, and because there are only 9 green balls, we have a 9/18 chance of pulling a green ball, and this reduces down to 1/2
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What is the probability that Jim flips a fair coin 3 times and gets 3 heads?
What is the probability that Jim flips a fair coin 3 times and gets 3 heads?
The probability of getting heads on one flip of a coin is 1/2, since the only other outcome is tails. Because the flips are independent of each other (one outcome does not depend on the others), the probability of the second coin being heads is 1/2 and the probability of the third coin being heads is 1/2. However, to calculate the probability of obtaining all of the flips ending in heads, we multiply the probabilities of the events together: (1/2)*(1/2)*(1/2) = 1/8.
The probability of getting heads on one flip of a coin is 1/2, since the only other outcome is tails. Because the flips are independent of each other (one outcome does not depend on the others), the probability of the second coin being heads is 1/2 and the probability of the third coin being heads is 1/2. However, to calculate the probability of obtaining all of the flips ending in heads, we multiply the probabilities of the events together: (1/2)*(1/2)*(1/2) = 1/8.
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Robert has a lot of loose change in his pocket. He pulled it all out and found that he had 5 nickels, 8 dimes, 3 quarters, and 13 pennies in his pocket. Robert then put all the change back in his pocket except for the quarters. What is the probability of Robert pulling out a dime at random? (assume that the physical size of the coins has no effect on their probability of being picked)
Robert has a lot of loose change in his pocket. He pulled it all out and found that he had 5 nickels, 8 dimes, 3 quarters, and 13 pennies in his pocket. Robert then put all the change back in his pocket except for the quarters. What is the probability of Robert pulling out a dime at random? (assume that the physical size of the coins has no effect on their probability of being picked)
The total number of coins in Robert's pocket is 26. because there are 8 dimes, he has an 8/26 chance of choosing a dime. This reduces down to 4/13 when you divide the numerator and denominator by 2 (GCF)
The total number of coins in Robert's pocket is 26. because there are 8 dimes, he has an 8/26 chance of choosing a dime. This reduces down to 4/13 when you divide the numerator and denominator by 2 (GCF)
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Maria is planning the seating for the head table at a college gala. There are eight speakers that will be seated along one side of the table. Richard wants to sit beside Hang, and Maria knows that Thomas and Lily should not be seated together. In how many ways can Maria make up the seating plan?
Maria is planning the seating for the head table at a college gala. There are eight speakers that will be seated along one side of the table. Richard wants to sit beside Hang, and Maria knows that Thomas and Lily should not be seated together. In how many ways can Maria make up the seating plan?
The simplest way to solve this is to find the number of seating arrangements in which Richard and Hang are seated together and then subtract those in which Thomas and Lily are also seated together. Consider Richard and Hang as a unit. This pair can be arranged with the other six speakers in 7_P_7 ways. For each of these ways, Hang could be either on Richard’s left or his right. Thus, there are 7_P_7 × 2 = 10 080 arrangements in which Richard and Hang are seated together. Now also consider Lily and Thomas as a unit. The two pairs can be arranged with the remaining four speakers in 6_P_6 ways, and the total number of arrangements with each of the pairs together is 6_P_6 × 2 × 2 = 2880.
Therefore, the number of seating arrangements in which Richard and Hang are adjacent but Thomas and Lily are not is 10 080 − 2880 = 7 200.
The simplest way to solve this is to find the number of seating arrangements in which Richard and Hang are seated together and then subtract those in which Thomas and Lily are also seated together. Consider Richard and Hang as a unit. This pair can be arranged with the other six speakers in 7_P_7 ways. For each of these ways, Hang could be either on Richard’s left or his right. Thus, there are 7_P_7 × 2 = 10 080 arrangements in which Richard and Hang are seated together. Now also consider Lily and Thomas as a unit. The two pairs can be arranged with the remaining four speakers in 6_P_6 ways, and the total number of arrangements with each of the pairs together is 6_P_6 × 2 × 2 = 2880.
Therefore, the number of seating arrangements in which Richard and Hang are adjacent but Thomas and Lily are not is 10 080 − 2880 = 7 200.
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A coin is tossed five times. What is the probability of tails occuring at least once?
A coin is tossed five times. What is the probability of tails occuring at least once?
Probability of getting a head: 0.5
Probability of getting a tail: 0.5
The probability of not getting ANY tails = 0.5 * 0.5 * 0.5 * 0.5 * 0.5= 0.03125
The probability of getting AT LEAST 1 tail = 1 - 0.03125 = 0.969
Probability of getting a head: 0.5
Probability of getting a tail: 0.5
The probability of not getting ANY tails = 0.5 * 0.5 * 0.5 * 0.5 * 0.5= 0.03125
The probability of getting AT LEAST 1 tail = 1 - 0.03125 = 0.969
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Elena took a math test with 15 questions, each question having four choices. Elena is sure that she got exactly 9 of the first 12 questions correct, but she guessed randomly on the last 3 questions. What is the probability that she will get at least 80% on the test?
Elena took a math test with 15 questions, each question having four choices. Elena is sure that she got exactly 9 of the first 12 questions correct, but she guessed randomly on the last 3 questions. What is the probability that she will get at least 80% on the test?
To get 80%, Elena needs to get 12 out of the 15 questions right. If she answered 9 out of the first 12 questions right, she can score 80% only if she guessed all 3 of the remaining questions correctly.
Therefore Elena has a (1/4)^3 chance of getting all 3 questions correct.
To get 80%, Elena needs to get 12 out of the 15 questions right. If she answered 9 out of the first 12 questions right, she can score 80% only if she guessed all 3 of the remaining questions correctly.
Therefore Elena has a (1/4)^3 chance of getting all 3 questions correct.
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You roll two dice. The first die shows a 3 and the other die rolls under the table and you cannot see it. What is the probability that both die show 3 ?
You roll two dice. The first die shows a 3 and the other die rolls under the table and you cannot see it. What is the probability that both die show 3 ?
A die has six sides and you already know that the first die shows the number 3. There is is a 1/6 probability that the second die under the table shows a 3.
A die has six sides and you already know that the first die shows the number 3. There is is a 1/6 probability that the second die under the table shows a 3.
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{10, 12, 24, 50, 60, 100, 260, 480, 606, 1000}
What is the probability that a number selected randomly from the set will be divisible by both 4 and 6?
{10, 12, 24, 50, 60, 100, 260, 480, 606, 1000}
What is the probability that a number selected randomly from the set will be divisible by both 4 and 6?
First, find the numbers that in the set that are divisible by 4. 12, 24, 60, 100, 260, 480, and 1000 are all divisible by 4. Now find the numbers that are divisible by 6. 12, 24, 60, 480, 606 are all divisible by 6. The numbers that are divisible by both 4 and 6 are 12, 24, 60, and 480, or 4 total numbers from the set. So 4 out of the 10 numbers are divisible by 4 and 6. The probablility is 4/10, which reduces to 2/5. The correct answer is 2/5.
First, find the numbers that in the set that are divisible by 4. 12, 24, 60, 100, 260, 480, and 1000 are all divisible by 4. Now find the numbers that are divisible by 6. 12, 24, 60, 480, 606 are all divisible by 6. The numbers that are divisible by both 4 and 6 are 12, 24, 60, and 480, or 4 total numbers from the set. So 4 out of the 10 numbers are divisible by 4 and 6. The probablility is 4/10, which reduces to 2/5. The correct answer is 2/5.
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If you flip a coin 3 times, what are the odds that the coin will be heads all three times?
If you flip a coin 3 times, what are the odds that the coin will be heads all three times?
If you flip a coin, the chances of you getting heads is 1/2. This is true every time you flip the coin so if you flip it 3 times, the chances of you getting heads every time is 1/2 * 1/2 * 1/2, or 1/8.
If you flip a coin, the chances of you getting heads is 1/2. This is true every time you flip the coin so if you flip it 3 times, the chances of you getting heads every time is 1/2 * 1/2 * 1/2, or 1/8.
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If there are 490 marbles in a bag, divided evenly into the seven colors of the rainbow, what is the probability of picking out a marble that is one of the three primary colors?
If there are 490 marbles in a bag, divided evenly into the seven colors of the rainbow, what is the probability of picking out a marble that is one of the three primary colors?
Since the rainbow consists of seven colors, and there are 490 marble, there are 70 of each color (490/7 = 70).
The probability of choosing a marble that is a primary color would be (70 * 3)/490 = 210/490 = 3/7.
Since the rainbow consists of seven colors, and there are 490 marble, there are 70 of each color (490/7 = 70).
The probability of choosing a marble that is a primary color would be (70 * 3)/490 = 210/490 = 3/7.
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There is a box with 24 handkerchiefs, some are blue, some are red, and some are green.
If there are 4 red handkerchiefs in the box, and the probability of pulling a blue handkerchief is 3 times more likely than pulling a red one, how many green handkerchiefs are in the box?
There is a box with 24 handkerchiefs, some are blue, some are red, and some are green.
If there are 4 red handkerchiefs in the box, and the probability of pulling a blue handkerchief is 3 times more likely than pulling a red one, how many green handkerchiefs are in the box?
If the probability of pulling out a blue handkerchief is 3 times that of pulling a red one, then it means there are 3 times as many blue handkerchiefs as there are red.
4 x 3 =12 blue handkerchiefs.
24 total – 12 blue – 4 red = 8 green handkerchiefs
If the probability of pulling out a blue handkerchief is 3 times that of pulling a red one, then it means there are 3 times as many blue handkerchiefs as there are red.
4 x 3 =12 blue handkerchiefs.
24 total – 12 blue – 4 red = 8 green handkerchiefs
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There is a box with 24 handkerchiefs, some are blue, some are red, and some are green.
If there are 4 red handkerchiefs in the box, and the probability of pulling a blue handkerchief is 3 times more likely than pulling a red one, what is the probability of pulling a green handkerchief?
There is a box with 24 handkerchiefs, some are blue, some are red, and some are green.
If there are 4 red handkerchiefs in the box, and the probability of pulling a blue handkerchief is 3 times more likely than pulling a red one, what is the probability of pulling a green handkerchief?
If the probability of pulling out a blue handkerchief is 3 times that of pulling a red one, then it means there are 3 times as many blue handkerchiefs as there are red.
4 x 3 =12 blue handkerchiefs.
24 total – 12 blue – 4 red = 8 green handkerchiefs
8/24 simplifies into 1/3 (divide both numbers by 8)
If the probability of pulling out a blue handkerchief is 3 times that of pulling a red one, then it means there are 3 times as many blue handkerchiefs as there are red.
4 x 3 =12 blue handkerchiefs.
24 total – 12 blue – 4 red = 8 green handkerchiefs
8/24 simplifies into 1/3 (divide both numbers by 8)
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Caroline is given instructions to plant five different flowers in a specific order in a garden. When she goes to plant the flowers, Caroline can only remember which flowers are to be planted in the first two positions in the garden, and she cannot remember which flowers to plant in the last three positions in the garden. If Caroline plants the first two flowers correctly, and then randomly plants the last three flowers, what is the probability that Caroline has planted all of the flowers in the correct order?
Caroline is given instructions to plant five different flowers in a specific order in a garden. When she goes to plant the flowers, Caroline can only remember which flowers are to be planted in the first two positions in the garden, and she cannot remember which flowers to plant in the last three positions in the garden. If Caroline plants the first two flowers correctly, and then randomly plants the last three flowers, what is the probability that Caroline has planted all of the flowers in the correct order?
If the flowers that go in the first two locations are already determined, there are only three positions that run any risk of having the wrong flower planted in them. First, determine the number of possible ways in which Caroline can finish planting the garden. There are 3 possible flowers that she could plant in the third position. After she plants a flower there, there will be 2 flowers she can plant in the fourth position. After planting a flower in the fourth position, there will be only one flower left to plant in the fifth position; therefore, the number of possible ways she could finish the garden is:
There are 6 possible outcomes. Only one of these outcomes will be successful; thus, the probability that Caroline finishes planting the garden in the correct order is one out of six:
If the flowers that go in the first two locations are already determined, there are only three positions that run any risk of having the wrong flower planted in them. First, determine the number of possible ways in which Caroline can finish planting the garden. There are 3 possible flowers that she could plant in the third position. After she plants a flower there, there will be 2 flowers she can plant in the fourth position. After planting a flower in the fourth position, there will be only one flower left to plant in the fifth position; therefore, the number of possible ways she could finish the garden is:
There are 6 possible outcomes. Only one of these outcomes will be successful; thus, the probability that Caroline finishes planting the garden in the correct order is one out of six:
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Jacob was 27 years old when his son Mike was born. Mike was 23 years old when his son Sam was born. Sam will celebrate his seventh birthday in 2014. What year was Jacob born?
Jacob was 27 years old when his son Mike was born. Mike was 23 years old when his son Sam was born. Sam will celebrate his seventh birthday in 2014. What year was Jacob born?
If Sam will celebrate his seventh birthday in 2014, then he was born in 2007. So Mike was born 23 years before in 1984 and Jacob was born 27 years before that in 1957.
If you answered 1964, then you did not factor that Sam was 7 years old at the time of calculation.
If you answered 1984, then you found the year that Mike was born, not Jacob.
If you answered 2007, then you found the year that Sam was born, not Jacob.
If you answered 1987, then you just subtracted 27 years old from the day of the party in 2014.
If Sam will celebrate his seventh birthday in 2014, then he was born in 2007. So Mike was born 23 years before in 1984 and Jacob was born 27 years before that in 1957.
If you answered 1964, then you did not factor that Sam was 7 years old at the time of calculation.
If you answered 1984, then you found the year that Mike was born, not Jacob.
If you answered 2007, then you found the year that Sam was born, not Jacob.
If you answered 1987, then you just subtracted 27 years old from the day of the party in 2014.
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Ilene has a bag of gum drops and a bag of gummy bears. The bag of gum drops contains 23 gum drops in the following colors: 3 red, 3 orange, 4 brown, 4 green, 6 blue and 3 yellow. The bag of gummy bears contains 26 gummy bears in the following colors: 4 red, 6 orange, 5 purple, 3 green, 5 yellow and 3 pink. If she dumps all of the gum drops and gummy bears into one bag, what is the probability that she will pick up either a red or green gum drop or gummy bear?
Ilene has a bag of gum drops and a bag of gummy bears. The bag of gum drops contains 23 gum drops in the following colors: 3 red, 3 orange, 4 brown, 4 green, 6 blue and 3 yellow. The bag of gummy bears contains 26 gummy bears in the following colors: 4 red, 6 orange, 5 purple, 3 green, 5 yellow and 3 pink. If she dumps all of the gum drops and gummy bears into one bag, what is the probability that she will pick up either a red or green gum drop or gummy bear?
The total number of gum drops and gummy bears is 49. There are 7 red gum drops and gummy bears and 7 green gum drops and gummy bears. There is a 2 in 7 probability that you will pick up a green or red gum drop or gummy bear.
If you answered 1/7, then you only accounted for choosing either red gum drop or gummy bear or green ones.
If you answered 2/47 then you did not add the number of gum drops and gummy bears correctly.
If you answered 7 in 14 then you found the number of red or green gummy bears/gum drops out of the total red and green gummy bears and gum drops.
And if you answered 3 out of 49, you found only the probability of choosing either red gum drops or green gummy bears.
The total number of gum drops and gummy bears is 49. There are 7 red gum drops and gummy bears and 7 green gum drops and gummy bears. There is a 2 in 7 probability that you will pick up a green or red gum drop or gummy bear.
If you answered 1/7, then you only accounted for choosing either red gum drop or gummy bear or green ones.
If you answered 2/47 then you did not add the number of gum drops and gummy bears correctly.
If you answered 7 in 14 then you found the number of red or green gummy bears/gum drops out of the total red and green gummy bears and gum drops.
And if you answered 3 out of 49, you found only the probability of choosing either red gum drops or green gummy bears.
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There are 42 candies in a jar. One-third of the candies are chocolate, one-sixth of the candies are caramel and one-sixth of candies are coconut. The rest of the candies are mint. If you select three pieces of candy from the jar, what is the approximate probability that two are chocolate and one is coconut?
There are 42 candies in a jar. One-third of the candies are chocolate, one-sixth of the candies are caramel and one-sixth of candies are coconut. The rest of the candies are mint. If you select three pieces of candy from the jar, what is the approximate probability that two are chocolate and one is coconut?
Total Chocolates = 1/3 * 42 = 14
Total Coconuts = 1/6 * 42 = 7
There are 14 chocolates and 7 coconuts to choose from 42 total candies, and note that each time you select one candy, the total number of candies decreases by one.
So assume that to arrive at the desired outcome of two chocolate and one coconut, you chose two chocolates and then a coconut. After picking out the first chocolate, there are now 13 chocolates to choose from 41 total candies. And after picking out the second chocolate, that leaves 7 coconuts to choose from 40 total candies.
The probability of that outcome = (14/42)(13/41)(7/40) = 0.01849, or just under 1.85%
However, there are two other sequences that could arrive at that desired outcome: chocolate, coconut, chocolate (which would be 14/42 * 7/41 * 13/40 = 1.85%) and coconut, chocolate, chocolate (which would be 7/42 * 14/41 * 13/40 = 1.85%). So summing all three probabilities of favorable outcomes gives you approximately 5.5%
Total Chocolates = 1/3 * 42 = 14
Total Coconuts = 1/6 * 42 = 7
There are 14 chocolates and 7 coconuts to choose from 42 total candies, and note that each time you select one candy, the total number of candies decreases by one.
So assume that to arrive at the desired outcome of two chocolate and one coconut, you chose two chocolates and then a coconut. After picking out the first chocolate, there are now 13 chocolates to choose from 41 total candies. And after picking out the second chocolate, that leaves 7 coconuts to choose from 40 total candies.
The probability of that outcome = (14/42)(13/41)(7/40) = 0.01849, or just under 1.85%
However, there are two other sequences that could arrive at that desired outcome: chocolate, coconut, chocolate (which would be 14/42 * 7/41 * 13/40 = 1.85%) and coconut, chocolate, chocolate (which would be 7/42 * 14/41 * 13/40 = 1.85%). So summing all three probabilities of favorable outcomes gives you approximately 5.5%
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What is the probability of rolling a sum of four or less on a pair of regular six-sided dice?
What is the probability of rolling a sum of four or less on a pair of regular six-sided dice?
Probability = what you want ÷ total number
There are 6 ways to get a sum of four or less: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)
The total would be 6 * 6 = 36
So the probability of getting four or less (4, 3, 2, 1) is 6/36 = 1/6
The probability of getting less than four (3, 2, 1) is 3/36 = 1/12
Probability = what you want ÷ total number
There are 6 ways to get a sum of four or less: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)
The total would be 6 * 6 = 36
So the probability of getting four or less (4, 3, 2, 1) is 6/36 = 1/6
The probability of getting less than four (3, 2, 1) is 3/36 = 1/12
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