Statistics - PSAT Math
Card 0 of 1435
What is the probability of choosing three hearts in three draws from a standard deck of playing cards, if replacement of cards is not allowed?
What is the probability of choosing three hearts in three draws from a standard deck of playing cards, if replacement of cards is not allowed?
The standard deck of cards has 52 cards: 13 cards in 4 suits.
Ways to choose three hearts: 13 * 12 * 11 = 1716
Ways to choose three cards: 52 * 51 * 50 = 132600
Probability is a number between 0 and 1 that is defines as the total ways of what you want ÷ by the total ways
The resulting simplified fraction is 11/850
The standard deck of cards has 52 cards: 13 cards in 4 suits.
Ways to choose three hearts: 13 * 12 * 11 = 1716
Ways to choose three cards: 52 * 51 * 50 = 132600
Probability is a number between 0 and 1 that is defines as the total ways of what you want ÷ by the total ways
The resulting simplified fraction is 11/850
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A lottery is being run at a high school to allocate parking spots. The school has 200 seniors, 300 juniors, 350 sophomores, and 450 freshmen. Each eligible senior will have their name entereted into the lottery twice, with all other eligible students' names being entered once. Only juniors and seniors will be eligible for parking spots. If there are 150 parking spots, what is the probability that any given junior will receive a spot?
A lottery is being run at a high school to allocate parking spots. The school has 200 seniors, 300 juniors, 350 sophomores, and 450 freshmen. Each eligible senior will have their name entereted into the lottery twice, with all other eligible students' names being entered once. Only juniors and seniors will be eligible for parking spots. If there are 150 parking spots, what is the probability that any given junior will receive a spot?
Find the probabilty a junior's name will be pulled for a single lottery trial. Then calculate the probability given 150 lottery trials.
(200 seniors * 2 entries each) + 300 juniors = 700 entries
For any single junior then, the odds are 1/700 for a single lottery trial.
For 150 trials, a junior will have (1/700 * 150 trials) = 150/700, which simplifies to
3/14
Find the probabilty a junior's name will be pulled for a single lottery trial. Then calculate the probability given 150 lottery trials.
(200 seniors * 2 entries each) + 300 juniors = 700 entries
For any single junior then, the odds are 1/700 for a single lottery trial.
For 150 trials, a junior will have (1/700 * 150 trials) = 150/700, which simplifies to
3/14
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A number between 1 and 15 is selected at random. What are the odds the number selected is a multiple of 6?
A number between 1 and 15 is selected at random. What are the odds the number selected is a multiple of 6?
In the set of 1 to 15, two numbers, 6 and 12, are multiples of 6. That means there are two chances out of 15 to select a multiple of 6.
2/15
In the set of 1 to 15, two numbers, 6 and 12, are multiples of 6. That means there are two chances out of 15 to select a multiple of 6.
2/15
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A big box of crayons contains a total of 120 crayons.
The box is composed of 3 colors; red, blue, and orange. 30 of the crayons are red, 40 of the crayons are blue and the rest are orange. If one picks a crayon randomly from the box, what is the probability that it will be orange?
A big box of crayons contains a total of 120 crayons.
The box is composed of 3 colors; red, blue, and orange. 30 of the crayons are red, 40 of the crayons are blue and the rest are orange. If one picks a crayon randomly from the box, what is the probability that it will be orange?
To solve the problem one must calculate that there are 50 orange crayons in the box. So 50/120 are orange. If we simplify that fraction by 10 we get 5/12.
To solve the problem one must calculate that there are 50 orange crayons in the box. So 50/120 are orange. If we simplify that fraction by 10 we get 5/12.
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A skydiver is trying to determine the probability of landing within the target of a grass field. If the field measures 1000 meters by 500 meters and the target area measures 50 meters by 50 meters, what is the probability of the skydiver landing in the target area?
A skydiver is trying to determine the probability of landing within the target of a grass field. If the field measures 1000 meters by 500 meters and the target area measures 50 meters by 50 meters, what is the probability of the skydiver landing in the target area?
Find the area of the entire field and the target area. The fraction of the field that is the target area is equal to the probability of the skydiver hitting the target area. For example, if the field were 100 m3 and the target area was 100 m3 than the probability would be 1. If the field were 100 m3 and the target area was 50 m3 than the probability would be 0.5 and so on.
(50 * 50)/(1000 * 500) = 5/1000 = 1/200
Find the area of the entire field and the target area. The fraction of the field that is the target area is equal to the probability of the skydiver hitting the target area. For example, if the field were 100 m3 and the target area was 100 m3 than the probability would be 1. If the field were 100 m3 and the target area was 50 m3 than the probability would be 0.5 and so on.
(50 * 50)/(1000 * 500) = 5/1000 = 1/200
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If a container holds 4 red balls, 3 yellow balls, and 2 blue balls, what are the odds of picking out both of the blue balls without replacement?
If a container holds 4 red balls, 3 yellow balls, and 2 blue balls, what are the odds of picking out both of the blue balls without replacement?
You take the probability of the first outcome times the probability of the second, so (2/9) * (1/8) = 2/72 = 1/36
You take the probability of the first outcome times the probability of the second, so (2/9) * (1/8) = 2/72 = 1/36
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Let R = {1, 3, 4, 12}, and let Q = {2, 6, 8, 24}. If one number is randomly selected from R, and another is randomly selected from Q, what is the probability that the product of those two randomly chosen numbers will belong to Q?
Let R = {1, 3, 4, 12}, and let Q = {2, 6, 8, 24}. If one number is randomly selected from R, and another is randomly selected from Q, what is the probability that the product of those two randomly chosen numbers will belong to Q?
First, let's consider all of the pairs that could be chosen from R and Q. There are sixteen possibilities:
1 and 2; 1 and 6; 1 and 8; 1 and 24; 3 and 2; 3 and 6; 3 and 8; 3 and 24; 4 and 2; 4 and 6; 4 and 8; 4 and 24; 12 and 2; 12 and 6; 12 and 8; 12 and 24.
Now, we need to find the product of each of these pairs, and then determine whether or not they belong to Q. The products of all of the possible pairs would be as follows:
1(2) = 2, which belongs to Q
1(6) = 6, which belongs to Q
1(8) = 8, which belongs to Q
1(24) = 24, which belongs to Q
3(2) = 6, which belongs to Q
3(6) = 18, which doesn't belong to Q
3(8) = 24, which belongs to Q
3(24) = 72, which doesn't belong
4(2) = 8, which belongs to Q
4(6) = 24, which belongs to Q
4(8) = 32, which doesn't belong
4(24) = 96, which doesn't belong
12(2) = 24, which belongs to Q
12(6) = 72, which doesn't belong
12(8) = 96, which doesn't belong
12(24) = 288, which doesn't belong
Thus, of our sixteen possible combinations, there are 9 which, when multiplied, give a number that belongs to Q. Because probability is the number of successful events out of the total number of events, the probability is 9/16.
The answer is 9/16.
First, let's consider all of the pairs that could be chosen from R and Q. There are sixteen possibilities:
1 and 2; 1 and 6; 1 and 8; 1 and 24; 3 and 2; 3 and 6; 3 and 8; 3 and 24; 4 and 2; 4 and 6; 4 and 8; 4 and 24; 12 and 2; 12 and 6; 12 and 8; 12 and 24.
Now, we need to find the product of each of these pairs, and then determine whether or not they belong to Q. The products of all of the possible pairs would be as follows:
1(2) = 2, which belongs to Q
1(6) = 6, which belongs to Q
1(8) = 8, which belongs to Q
1(24) = 24, which belongs to Q
3(2) = 6, which belongs to Q
3(6) = 18, which doesn't belong to Q
3(8) = 24, which belongs to Q
3(24) = 72, which doesn't belong
4(2) = 8, which belongs to Q
4(6) = 24, which belongs to Q
4(8) = 32, which doesn't belong
4(24) = 96, which doesn't belong
12(2) = 24, which belongs to Q
12(6) = 72, which doesn't belong
12(8) = 96, which doesn't belong
12(24) = 288, which doesn't belong
Thus, of our sixteen possible combinations, there are 9 which, when multiplied, give a number that belongs to Q. Because probability is the number of successful events out of the total number of events, the probability is 9/16.
The answer is 9/16.
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If John does not have freckles, he is not Tim's sibling.
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If Jake does not have freckles, he is Tim's sibling.
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If Suzy is not Tim's sibling, she will not have freckles.
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All of the other answers are false.
All of Tim's siblings have freckles. Which of the above statements must be true?
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If John does not have freckles, he is not Tim's sibling.
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If Jake does not have freckles, he is Tim's sibling.
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If Suzy is not Tim's sibling, she will not have freckles.
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All of the other answers are false.
All of Tim's siblings have freckles. Which of the above statements must be true?
Only John's statement must be true. Suzy could have freckles and not be Tim's sibling and if Jake doesn't have freckles then he is not Tim's sibling.
Only John's statement must be true. Suzy could have freckles and not be Tim's sibling and if Jake doesn't have freckles then he is not Tim's sibling.
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A bag contains red marbles and blue marbles only. Let r equal the number of red marbles and b equal the number of blue marbles in the bag. Mark draws one marble from the bag and then a second one without replacing the first. Which of the following expressions is equivalent to the probability that Mark will draw a red marble and then a blue marble?
A bag contains red marbles and blue marbles only. Let r equal the number of red marbles and b equal the number of blue marbles in the bag. Mark draws one marble from the bag and then a second one without replacing the first. Which of the following expressions is equivalent to the probability that Mark will draw a red marble and then a blue marble?
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In a bag, there are 5 blue marbles, 3 red marbles, and 2 green marbles. If three marbles are chosen consecutively at random without replacement, what is the probability that the color of the chosen marbles would be blue-red-green in that order?
In a bag, there are 5 blue marbles, 3 red marbles, and 2 green marbles. If three marbles are chosen consecutively at random without replacement, what is the probability that the color of the chosen marbles would be blue-red-green in that order?
The probability that a blue marble is chosen first is 5/10 because at the outset there are 5 blue marbles in a total of 10 marbles. After one marble has been chosen, there are 9 total marbles remaining since there is no replacement. The probability that a red marble is chosen in the second pick is 3/9 (because there are 3 red marbles out of a total of 9). There are now 8 total marbles remaining and 2 green marbles. Thus, the probability of picking a green marble on the third pick is 2/8.
Therefore, the probability of picking a blue-red-green marble outcome (in that order) is:
5/10 * 3/9 * 2/8 = 30/720 = 1/24
The probability that a blue marble is chosen first is 5/10 because at the outset there are 5 blue marbles in a total of 10 marbles. After one marble has been chosen, there are 9 total marbles remaining since there is no replacement. The probability that a red marble is chosen in the second pick is 3/9 (because there are 3 red marbles out of a total of 9). There are now 8 total marbles remaining and 2 green marbles. Thus, the probability of picking a green marble on the third pick is 2/8.
Therefore, the probability of picking a blue-red-green marble outcome (in that order) is:
5/10 * 3/9 * 2/8 = 30/720 = 1/24
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A book shelf has 15 theology books, 20 philosophy books, and 5 history books. If someone were to draw two texts at random, what is the chance that they would draw at least one theology book?
A book shelf has 15 theology books, 20 philosophy books, and 5 history books. If someone were to draw two texts at random, what is the chance that they would draw at least one theology book?
Thinking through our data, we know that we are looking for the following combination of events, where T is a theology book, and < > represents a draw from the shelf:
1. <T><T>
2. <T><Non-T>
3. <Non-T><T>
To save us quite a bit of trouble, let us note that only one event is excluded:
<Non-T><Non-T>
The easiest way to solve this would be to solve for the probability of this one case and subtract that from 1. This will give us the "remaining probability" that applies to the three cases that we want.
The <Non-T><Non-T> case would be calculated:
First draw: 25/40 = 5/8
Second draw: 24/39 = 8/13
Total probability: (5/8) * (8/13) = 5/13
The probability of our case is 1 – (5/13) = (13 – 5)/13 = 8/13.
Thinking through our data, we know that we are looking for the following combination of events, where T is a theology book, and < > represents a draw from the shelf:
1. <T><T>
2. <T><Non-T>
3. <Non-T><T>
To save us quite a bit of trouble, let us note that only one event is excluded:
<Non-T><Non-T>
The easiest way to solve this would be to solve for the probability of this one case and subtract that from 1. This will give us the "remaining probability" that applies to the three cases that we want.
The <Non-T><Non-T> case would be calculated:
First draw: 25/40 = 5/8
Second draw: 24/39 = 8/13
Total probability: (5/8) * (8/13) = 5/13
The probability of our case is 1 – (5/13) = (13 – 5)/13 = 8/13.
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A drawing is being held for concert tickets at a high school. There are 200 seniors, 150 juniors, 200 sophomores and 150 freshmen. Each senior's name is placed in the drawing 4 times, each junior's 3, sophomores 2, and freshmen one time. What is the probability that a senior's name will be chosen?
A drawing is being held for concert tickets at a high school. There are 200 seniors, 150 juniors, 200 sophomores and 150 freshmen. Each senior's name is placed in the drawing 4 times, each junior's 3, sophomores 2, and freshmen one time. What is the probability that a senior's name will be chosen?
You multiply each number of students in each class by how many times that class is entered into the drawing, then add up the totals. There will be a total of 1800 entrants and 800 will be seniors.
You multiply each number of students in each class by how many times that class is entered into the drawing, then add up the totals. There will be a total of 1800 entrants and 800 will be seniors.
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Set S = {0, 1, 5, 9}, and set T = {2, 3, 4, 7}. If one number is chosen randomly from S and another is chosen randomly from T, what is the probability that the sum of these two numbers will be prime?
Set S = {0, 1, 5, 9}, and set T = {2, 3, 4, 7}. If one number is chosen randomly from S and another is chosen randomly from T, what is the probability that the sum of these two numbers will be prime?
There are four possible numbers that can be chosen from S, and there are four that can be chosen from T. This means that there are 4 * 4, or 16, pairs of numbers that can be drawn from S and T. We need to find the sum of these pairs and determine if each sum is prime. Let's find the sum of the 16 pairs. Remember that a number is prime if it divisible only by itself and 1.
0 + 2 = 2, which is prime
0 + 3 = 3, prime
0 + 4 = 4, not prime
0 + 7 = 7, prime
1 + 2 = 3, prime
1 + 3 = 4, not
1 + 4 = 5, prime
1 + 7 = 8, not
5 + 2 = 7, prime
5 + 3 = 8, not
5 + 4 = 9, not
5 + 7 = 12, not
9 + 2=11, prime
9 + 3 = 12, not
9 + 4 = 13, prime
9 + 7 = 16, not
Of the sixteen pairs, 8 have a sum that equals a prime number. Thus, because each of these pairs has an equal chance of being drawn randomly, the probability that the sum will be prime is 8 out of 16, or 8/16 = 1/2.
The answer is 1/2.
There are four possible numbers that can be chosen from S, and there are four that can be chosen from T. This means that there are 4 * 4, or 16, pairs of numbers that can be drawn from S and T. We need to find the sum of these pairs and determine if each sum is prime. Let's find the sum of the 16 pairs. Remember that a number is prime if it divisible only by itself and 1.
0 + 2 = 2, which is prime
0 + 3 = 3, prime
0 + 4 = 4, not prime
0 + 7 = 7, prime
1 + 2 = 3, prime
1 + 3 = 4, not
1 + 4 = 5, prime
1 + 7 = 8, not
5 + 2 = 7, prime
5 + 3 = 8, not
5 + 4 = 9, not
5 + 7 = 12, not
9 + 2=11, prime
9 + 3 = 12, not
9 + 4 = 13, prime
9 + 7 = 16, not
Of the sixteen pairs, 8 have a sum that equals a prime number. Thus, because each of these pairs has an equal chance of being drawn randomly, the probability that the sum will be prime is 8 out of 16, or 8/16 = 1/2.
The answer is 1/2.
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Judy is practicing to be a magician and has an ordinary deck of 52 playing cards, a regular 6-sided die, and a fair coin. What is the probability that Judy rolls a 6 on the die, then flips the coin head face up, and then draws a spade from the deck of cards?
Judy is practicing to be a magician and has an ordinary deck of 52 playing cards, a regular 6-sided die, and a fair coin. What is the probability that Judy rolls a 6 on the die, then flips the coin head face up, and then draws a spade from the deck of cards?
We can find the individual probabilities of these three events occuring first.
P(rolling a 6) = 1/6
P(head) = 1/2
P(spade) = 13/52 = 1/4
Now, to find the probability of rolling a 6 AND flipping a head AND drawing a spade, we must multiply the individual probabilities. So the answer is 1/6 * 1/2 * 1/4 = 1/48.
We can find the individual probabilities of these three events occuring first.
P(rolling a 6) = 1/6
P(head) = 1/2
P(spade) = 13/52 = 1/4
Now, to find the probability of rolling a 6 AND flipping a head AND drawing a spade, we must multiply the individual probabilities. So the answer is 1/6 * 1/2 * 1/4 = 1/48.
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A bag of marbles has 7 yellow marbles, 5 red marbles, 3 blue marbles, and 6 white marbles. What is the probability of choosing a yellow marble, putting it back and choosing a blue marble, and then NOT putting the blue marble back and picking a white marble?
A bag of marbles has 7 yellow marbles, 5 red marbles, 3 blue marbles, and 6 white marbles. What is the probability of choosing a yellow marble, putting it back and choosing a blue marble, and then NOT putting the blue marble back and picking a white marble?
There are a total of 7 + 5 + 3 + 6 = 21 marbles. The probability of picking a yellow marble is 7/21 = 1/3. Then we put it back and choose a blue marble with probability 3/21 = 1/7. We do NOT put this blue marble back, but then we grab for a white. The probability of picking a white is now 6/20 = 3/10, because now we are choosing from 20 marbles instead of 21. So putting it together, the probability of choosing a yellow marble, replacing it and then choosing a blue and a white, is 1/3 * 1/7 * 3/10 = 1/70.
There are a total of 7 + 5 + 3 + 6 = 21 marbles. The probability of picking a yellow marble is 7/21 = 1/3. Then we put it back and choose a blue marble with probability 3/21 = 1/7. We do NOT put this blue marble back, but then we grab for a white. The probability of picking a white is now 6/20 = 3/10, because now we are choosing from 20 marbles instead of 21. So putting it together, the probability of choosing a yellow marble, replacing it and then choosing a blue and a white, is 1/3 * 1/7 * 3/10 = 1/70.
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In a regular 52-card deck, what is the probability of drawing three aces in a row, with replacement?
In a regular 52-card deck, what is the probability of drawing three aces in a row, with replacement?
There are 4 aces in the 52 card deck, so the probability of drawing an ace is 4/52. Then we put this ace back in the deck and draw again. The probability of drawing an ace is again 4/52. Similarly, on the third draw, the probability of getting an ace is 4/52. So the probability of drawing an ace on the 1st draw AND the 2nd draw AND the 3rd draw is 4/52 * 4/52 * 4/52.
There are 4 aces in the 52 card deck, so the probability of drawing an ace is 4/52. Then we put this ace back in the deck and draw again. The probability of drawing an ace is again 4/52. Similarly, on the third draw, the probability of getting an ace is 4/52. So the probability of drawing an ace on the 1st draw AND the 2nd draw AND the 3rd draw is 4/52 * 4/52 * 4/52.
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A bag contains four blue marbles, eight red marbles, and six orange marbles. If a marble is randomly selected from the bag, what is the probability that the marble will be either red or orange?
A bag contains four blue marbles, eight red marbles, and six orange marbles. If a marble is randomly selected from the bag, what is the probability that the marble will be either red or orange?
The probability of an event is the ratio of the number of desired outcomes to the total number of possible outcomes. In this problem, the total number of outcomes is equal to the total number of marbles in the bag. There are four blue, eight red, and six orange marbles, so the total number of marbles is the sum of four, eight, and six, or eighteen.
We are asked to find the probability of choosing a marble that is either red or orange. This means we have to consider the number of marbles that are either red or orange. Because there are eight red and six orange, there are fourteen marbles that are either red or orange.
The probability is thus fourteen out of eighteen, because there are fourteen red or orange marbles, out of a total of eighteen marbles. We will need to simplify the fraction 14/18.
Probability = 14/18 = 7/9
The answer is 7/9.
The probability of an event is the ratio of the number of desired outcomes to the total number of possible outcomes. In this problem, the total number of outcomes is equal to the total number of marbles in the bag. There are four blue, eight red, and six orange marbles, so the total number of marbles is the sum of four, eight, and six, or eighteen.
We are asked to find the probability of choosing a marble that is either red or orange. This means we have to consider the number of marbles that are either red or orange. Because there are eight red and six orange, there are fourteen marbles that are either red or orange.
The probability is thus fourteen out of eighteen, because there are fourteen red or orange marbles, out of a total of eighteen marbles. We will need to simplify the fraction 14/18.
Probability = 14/18 = 7/9
The answer is 7/9.
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Two dice are rolled. Find the probability that the numbers sum to 4.
Two dice are rolled. Find the probability that the numbers sum to 4.
The possible dice combinations that sum to 4 are 
.
The number of all possible dice combinations is 
. (6 numbers on each of the two dice.)
So the probability that the numbers sum to 4 =
The possible dice combinations that sum to 4 are
The number of all possible dice combinations is
So the probability that the numbers sum to 4 =
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Jackie is a contestant on a gameshow. She has to pick marbles out of a big bag to win various amounts of money. The bag contains a total of 200 marbles.
There are 100 red marbles worth $10 each, 50 blue marbles worth $20 each, 30 green marbles worth $50 each, 15 white marbles worth $100 each and 5 black marbles worth $1000 each. If she picks once, what percent chance does she have of picking a $1000 marble?
Jackie is a contestant on a gameshow. She has to pick marbles out of a big bag to win various amounts of money. The bag contains a total of 200 marbles.
There are 100 red marbles worth $10 each, 50 blue marbles worth $20 each, 30 green marbles worth $50 each, 15 white marbles worth $100 each and 5 black marbles worth $1000 each. If she picks once, what percent chance does she have of picking a $1000 marble?
There are 5 black marbles worth $1000 dollars out of the total of 200 marbles. 
or 2.5%
There are 5 black marbles worth $1000 dollars out of the total of 200 marbles.
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What is the probability that you will pull out 3 diamonds in a standard deck without replacement.
What is the probability that you will pull out 3 diamonds in a standard deck without replacement.
There are 13 diamonds in a standard deck of 52 cards. So, you have a 13/52 (1/4) chance of getting a diamond; then a 12/51 (4/17) chance of pulling the next diamond; last, there is a 11/50 chance of getting the third diamond. When you combine probabilities, you multiply the individual probabilities together
There are 13 diamonds in a standard deck of 52 cards. So, you have a 13/52 (1/4) chance of getting a diamond; then a 12/51 (4/17) chance of pulling the next diamond; last, there is a 11/50 chance of getting the third diamond. When you combine probabilities, you multiply the individual probabilities together
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