Outcomes - SSAT Upper Level Quantitative
Card 1 of 160
A fair six-sided die has on its faces the numbers 1, 1, 1, 2, 2, 3; a fair eight-sided die has on its faces the numbers 1, 1, 1, 1, 2, 2, 3, 4. If both dice are rolled, what is the probability that the numbers add up to 3 or less?
A fair six-sided die has on its faces the numbers 1, 1, 1, 2, 2, 3; a fair eight-sided die has on its faces the numbers 1, 1, 1, 1, 2, 2, 3, 4. If both dice are rolled, what is the probability that the numbers add up to 3 or less?
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In order for the roll to add up to 3 or less, neither die can come up 3 or 4. Therefore, we look at the probabilities that each die can come up 1 and 2.
The six-sided die comes up 1 with probability 
; it comes up 2 with probability 
.
The six-sided die comes up 1 with probability 
; it comes up 2 with probability 
.
A roll of 3 or less can happen three ways:
Case 1: Both dice come up 1. The probability of this, by the Multiplication Principle, is 
Case 2: The six-sided die comes up 1 and the eight-sided die comes up 2. The probability of this, by the Multiplication Principle, is 
Case 3: The six-sided die comes up 2 and the eight-sided die comes up 1. The probability of this, by the Multiplication Principle, is 
These are mutually exclusive events, so add these probabilities: 
In order for the roll to add up to 3 or less, neither die can come up 3 or 4. Therefore, we look at the probabilities that each die can come up 1 and 2.
The six-sided die comes up 1 with probability
The six-sided die comes up 1 with probability
A roll of 3 or less can happen three ways:
Case 1: Both dice come up 1. The probability of this, by the Multiplication Principle, is
Case 2: The six-sided die comes up 1 and the eight-sided die comes up 2. The probability of this, by the Multiplication Principle, is
Case 3: The six-sided die comes up 2 and the eight-sided die comes up 1. The probability of this, by the Multiplication Principle, is
These are mutually exclusive events, so add these probabilities:
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A fair coin is tossed nine times. Each toss comes up heads. What is the probability that the coin will come up heads a tenth time?
A fair coin is tossed nine times. Each toss comes up heads. What is the probability that the coin will come up heads a tenth time?
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Each flip of the coin is an independent event, and does not affect the other flips; all that matters is that the coin is fair. Because of that fact, the probability that the tenth flip comes up heads is 
.
Each flip of the coin is an independent event, and does not affect the other flips; all that matters is that the coin is fair. Because of that fact, the probability that the tenth flip comes up heads is
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In a jar, there are 3 blue marbles, 5 red marbles, 8 green marbles. If Bob reaches his hand in a jar, and grabs one marble, what is the likelihood he will pick up a blue marble?
In a jar, there are 3 blue marbles, 5 red marbles, 8 green marbles. If Bob reaches his hand in a jar, and grabs one marble, what is the likelihood he will pick up a blue marble?
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First, calculate the total number of marbles in the jar, which is 
. Because 3 of the marbles are blue, the chances of picking a blue marble 
.
First, calculate the total number of marbles in the jar, which is
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The threes from one standard deck of playing cards are added to another deck, and a card is drawn at random from that second, modified deck. What is the probability that that card will be an ace, a two, or a three?
The threes from one standard deck of playing cards are added to another deck, and a card is drawn at random from that second, modified deck. What is the probability that that card will be an ace, a two, or a three?
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If four more threes are added to a deck of fifty-two playing cards, then the deck will have fifty-six playing cards, including four aces, four twos, and eight threes. This makes sixteen ways to draw an ace, a two, or a three out of a possible fifty-six draws, or a probability of
.
If four more threes are added to a deck of fifty-two playing cards, then the deck will have fifty-six playing cards, including four aces, four twos, and eight threes. This makes sixteen ways to draw an ace, a two, or a three out of a possible fifty-six draws, or a probability of
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Two fair dice are rolled. What is the probability that at least one will show a 5 or a 6?
Two fair dice are rolled. What is the probability that at least one will show a 5 or a 6?
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The best way to find this probability is to find the probability that both will show a 1, 2, 3, or 4, and subtract that probability from 1.
The probability that one fair die will show a 1, 2, 3, or 4 is 
, so the probability that both will show 1, 2, 3, or 4 is
.
The probability that at least one die will show a 5 or a 6 is therefore
.
The best way to find this probability is to find the probability that both will show a 1, 2, 3, or 4, and subtract that probability from 1.
The probability that one fair die will show a 1, 2, 3, or 4 is
The probability that at least one die will show a 5 or a 6 is therefore
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The tens from one standard deck of playing cards are added to another deck, and a card is drawn at random from that second, modified deck. What is the probability that that card will be an ace, a two, or a three?
The tens from one standard deck of playing cards are added to another deck, and a card is drawn at random from that second, modified deck. What is the probability that that card will be an ace, a two, or a three?
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If four more tens are added to a deck of fifty-two playing cards, then the deck will have fifty-six playing cards, including four aces, four twos, and four threes. This makes twelve ways to draw an ace, a two, or a three out of a possible fifty-six draws, or a probability of
.
If four more tens are added to a deck of fifty-two playing cards, then the deck will have fifty-six playing cards, including four aces, four twos, and four threes. This makes twelve ways to draw an ace, a two, or a three out of a possible fifty-six draws, or a probability of
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Ten balls - five green, three blue, two yellow - are placed in a hat. A ball is drawn, then replaced; then, a ball is drawn again. What is the probability that a blue ball will be drawn at least once?
Ten balls - five green, three blue, two yellow - are placed in a hat. A ball is drawn, then replaced; then, a ball is drawn again. What is the probability that a blue ball will be drawn at least once?
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This can more easily be solved by finding the probability that neither draw will result in a blue ball, then subtracting it from 1.
The probability that the first ball will not be blue will be seven (balls that are not blue) out of ten (balls total), or 
. Since the ball is replaced, this is also the probability that the second will not be blue. Therefore, the probability that neither will be blue is 
.
The probability that at least one draw will result in a blue ball is therefore 
.
This can more easily be solved by finding the probability that neither draw will result in a blue ball, then subtracting it from 1.
The probability that the first ball will not be blue will be seven (balls that are not blue) out of ten (balls total), or
The probability that at least one draw will result in a blue ball is therefore
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Ten balls - five green, three blue, two yellow - are placed in a hat. Two balls are drawn, one after the other, without replacement. What is the probability that at least one will be green?
Ten balls - five green, three blue, two yellow - are placed in a hat. Two balls are drawn, one after the other, without replacement. What is the probability that at least one will be green?
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This can more easily be solved by finding the probability that neither will be green, then subtracting it from 1.
The probability that the first ball will be not be green is five out of ten, or 
.
This will leave nine balls, four not green, making the probability that the second ball will then not be green 
.
Multiply these probabilities:
This is the probability that neither ball is green; subtract this from 1: 
This can more easily be solved by finding the probability that neither will be green, then subtracting it from 1.
The probability that the first ball will be not be green is five out of ten, or
This will leave nine balls, four not green, making the probability that the second ball will then not be green
Multiply these probabilities:
This is the probability that neither ball is green; subtract this from 1:
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Three fair dice are rolled. What is the probability that none of them will show a 5 or a 6?
Three fair dice are rolled. What is the probability that none of them will show a 5 or a 6?
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The equivalent question is this: What is the probability that all three dice will show 1, 2, 3, or 4?
The probability that one fair die will show a 1, 2, 3, or 4 is 
, so the probability that all three will show 1, 2, 3, or 4 is
.
The equivalent question is this: What is the probability that all three dice will show 1, 2, 3, or 4?
The probability that one fair die will show a 1, 2, 3, or 4 is
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Six boxes are marked with numbers 1 to 6. Each box contains ten marbles, with the number of red marbles in each box corresponding to its number, and the remainder being white.
A fair die is rolled. The box corresponding to the roll is chosen and a marble is drawn at random from that box. What is the probability of drawing a red marble?
Six boxes are marked with numbers 1 to 6. Each box contains ten marbles, with the number of red marbles in each box corresponding to its number, and the remainder being white.
A fair die is rolled. The box corresponding to the roll is chosen and a marble is drawn at random from that box. What is the probability of drawing a red marble?
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The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are
red marbles out of
marbles total, the probability of choosing a red marble is
The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are
red marbles out of
marbles total, the probability of choosing a red marble is
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Six boxes are marked with numbers 1 to 6. Each box contains ten marbles, with the number of red marbles in each box corresponding to its number, and the remainder being white.
A fair die is rolled. The box corresponding to the roll is chosen and a marble is drawn at random from that box. What is the probability of drawing a white marble?
Six boxes are marked with numbers 1 to 6. Each box contains ten marbles, with the number of red marbles in each box corresponding to its number, and the remainder being white.
A fair die is rolled. The box corresponding to the roll is chosen and a marble is drawn at random from that box. What is the probability of drawing a white marble?
Tap to reveal answer
The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are
white marbles out of
marbles total, the probability of choosing a white marble is
The die is fair, making each box equally likely to be chosen; also, each marble is equally likely to be chosen after the box is chosen. This makes each marble overall equally likely to be chosen. Since there are
white marbles out of
marbles total, the probability of choosing a white marble is
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A card is drawn at random from a standard 52-card deck. The person drawing wins if the card he draws is a club, a king, or a queen. What are the odds against drawing a winning card?
A card is drawn at random from a standard 52-card deck. The person drawing wins if the card he draws is a club, a king, or a queen. What are the odds against drawing a winning card?
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There are 19 winning cards: the 13 clubs, the other 3 queens, and the other 3 kings (the queen and king of clubs are already counted). There are subsequently 33 losing cards, so the odds against a winning card are the number of losing cards, 33, to the number of winning cards, 19. This cannot be reduced, so the correct choice is 33 to 19.
There are 19 winning cards: the 13 clubs, the other 3 queens, and the other 3 kings (the queen and king of clubs are already counted). There are subsequently 33 losing cards, so the odds against a winning card are the number of losing cards, 33, to the number of winning cards, 19. This cannot be reduced, so the correct choice is 33 to 19.
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A card is drawn at random from a standard 52-card deck. The person drawing wins if the card he draws is a two, a three, or a black four. What are the odds against drawing a winning card?
A card is drawn at random from a standard 52-card deck. The person drawing wins if the card he draws is a two, a three, or a black four. What are the odds against drawing a winning card?
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There are 10 winning cards: the 4 twos, the 4 threes, and the 2 black fours. There are subsequently 42 losing cards, so the odds against a winning card are the number of losing cards, 42, to the number of winning cards, 10. This can be reduced to
or 21 to 5.
There are 10 winning cards: the 4 twos, the 4 threes, and the 2 black fours. There are subsequently 42 losing cards, so the odds against a winning card are the number of losing cards, 42, to the number of winning cards, 10. This can be reduced to
or 21 to 5.
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A card is drawn at random from a standard 52-card deck. The person drawing wins if the card she draws is a red four or a black nine. What are the odds against drawing a winning card?
A card is drawn at random from a standard 52-card deck. The person drawing wins if the card she draws is a red four or a black nine. What are the odds against drawing a winning card?
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There are 4 winning cards: the 2 red fours and the 2 black nines. There are subsequently 48 losing cards, so the odds against a winning card are the number of losing cards, 48, to the number of winning cards, 4. This can be reduced to
or 12 to 1.
There are 4 winning cards: the 2 red fours and the 2 black nines. There are subsequently 48 losing cards, so the odds against a winning card are the number of losing cards, 48, to the number of winning cards, 4. This can be reduced to
or 12 to 1.
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John participates in a charity fundraiser in which he pays $1 to draw a card from a standard 52-card deck. If he draws the ace of spades, he wins $25; if he draws any other ace, he wins $5; if he draws any other card, he does not win. To the nearest cent, what is the expected value of the game to John?
John participates in a charity fundraiser in which he pays $1 to draw a card from a standard 52-card deck. If he draws the ace of spades, he wins $25; if he draws any other ace, he wins $5; if he draws any other card, he does not win. To the nearest cent, what is the expected value of the game to John?
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Since there is only one card out of 52 (ace of spades) that wins John the $25 prize, the probability of this happening is 
. The value of this outcome to John is $24 - the $25 prize minus the $1 he paid to play.
Since there are three cards out of 52 (ace of clubs, ace of diamonds, ace of hearts) that win John a $5 prize, the probability of this happening is 
. The value of this outcome to John is $4 - the $5 prize minus the $1 he paid to play.
Since there are 48 out of 52 cards that do not win John a prize, the probability of this happening is 
. The value of this outcome to John is 
, since he had to pay $1 to play.
To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to John is therefore
dollars, or
The expected value of the game to Johnny rounds to 
.
Since there is only one card out of 52 (ace of spades) that wins John the $25 prize, the probability of this happening is
Since there are three cards out of 52 (ace of clubs, ace of diamonds, ace of hearts) that win John a $5 prize, the probability of this happening is
Since there are 48 out of 52 cards that do not win John a prize, the probability of this happening is
To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to John is therefore
The expected value of the game to Johnny rounds to
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Sharon participates in a charity fundraiser in which she pays $2 to draw a card from a deck of cards; the deck is a standard 52-card deck plus the joker. If she draws the joker, she wins $25; if she draws the ace of spades, she wins $10; if she draws any other ace, she wins $7; if she draws any other card, she does not win. To the nearest cent, what is the expected value of the game to Sharon?
Sharon participates in a charity fundraiser in which she pays $2 to draw a card from a deck of cards; the deck is a standard 52-card deck plus the joker. If she draws the joker, she wins $25; if she draws the ace of spades, she wins $10; if she draws any other ace, she wins $7; if she draws any other card, she does not win. To the nearest cent, what is the expected value of the game to Sharon?
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Since there is only one card out of 53 (joker) that wins Sharon the $25 prize, the probability of this happening is 
. The value of this outcome to Sharon is $23 - the $25 prize minus the $2 she paid to play.
Since there is only one card out of 53 (ace of spades) that wins Sharon the $10 prize, the probability of this happening is 
. The value of this outcome to Sharon is $8 - the $10 prize minus the $2 she paid to play.
Since there are three cards out of 53 (ace of clubs, ace of diamonds, ace of hearts) that win Sharon a $7 prize, the probability of this happening is 
. The value of this outcome to Sharon is $5 - the $7 prize minus the $2 she paid to play.
Since there are 48 out of 53 cards that do not win Sharon a prize, the probability of this happening is 
. The value of this outcome to Sharon is 
, since he had to pay $2 to play.
To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to Sharon is therefore
dollars, or
The expected value of the game to Sharon rounds to 
.
Since there is only one card out of 53 (joker) that wins Sharon the $25 prize, the probability of this happening is
Since there is only one card out of 53 (ace of spades) that wins Sharon the $10 prize, the probability of this happening is
Since there are three cards out of 53 (ace of clubs, ace of diamonds, ace of hearts) that win Sharon a $7 prize, the probability of this happening is
Since there are 48 out of 53 cards that do not win Sharon a prize, the probability of this happening is
To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to Sharon is therefore
The expected value of the game to Sharon rounds to
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Jeff collects basketball cards of players on his three favorite teams. He decides to put 5 cards from each team in a paper bag and then to draw out 3 cards at random. What are the odds of him getting one player from each team?
Jeff collects basketball cards of players on his three favorite teams. He decides to put 5 cards from each team in a paper bag and then to draw out 3 cards at random. What are the odds of him getting one player from each team?
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For this problem, we will multiply together the odds of each draw (assuming he draws one card at a time...the odds won't change if he draws three at once, but it's easier to visualize this way) resulting in a card that works for Jeff's goal of having one player from each team. The first draw cannot fail, as he needs one player from each team and the first card he draws must be from one of the teams. After this draw, he has 14 cards remaining, and 10 of these are players on the two teams that can still offer a player.
So the odds of a successful second draw are 
.
The last draw is the trickiest, as there would now be 13 cards remaining, with only 5 being players from the team that he still needs represented. When we multiply all of these odds together, we get
which is 27.5%.
For this problem, we will multiply together the odds of each draw (assuming he draws one card at a time...the odds won't change if he draws three at once, but it's easier to visualize this way) resulting in a card that works for Jeff's goal of having one player from each team. The first draw cannot fail, as he needs one player from each team and the first card he draws must be from one of the teams. After this draw, he has 14 cards remaining, and 10 of these are players on the two teams that can still offer a player.
So the odds of a successful second draw are
The last draw is the trickiest, as there would now be 13 cards remaining, with only 5 being players from the team that he still needs represented. When we multiply all of these odds together, we get
which is 27.5%.
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Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?
Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?
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A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is 
.
There are thirteen spades in the deck, so the probability of drawing a spade is 
.
Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is 
Thus the probability of drawing a face card or a spade is:
A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is
There are thirteen spades in the deck, so the probability of drawing a spade is
Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is
Thus the probability of drawing a face card or a spade is:
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Set A: 
Set B: 
One letter is picked from Set A and Set B. What is the probability of picking two consonants?
Set A:
Set B:
One letter is picked from Set A and Set B. What is the probability of picking two consonants?
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Set A: 
Set B: 
In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is 
.
In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is 
.
The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:
Set A:
Set B:
In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is
In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is
The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:
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Set A: 
Set B: 
One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?
Set A:
Set B:
One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?
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Set A: 
Set B: 
In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is 
.
In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is 
.
The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:
The interesection is:
So, we can find the probability of drawing at least one consonant:
Set A:
Set B:
In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is
In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is
The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:
The interesection is:
So, we can find the probability of drawing at least one consonant:
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