Probability - SAT Math
Card 0 of 72
A red die is altered so that it comes up a "1" 
of the time; the other five numbers come up with equal probability to each other. A blue die is altered similarly.
Both dice are rolled. What are the odds against rolling doubles (both numbers are the same)?
A red die is altered so that it comes up a "1"
Both dice are rolled. What are the odds against rolling doubles (both numbers are the same)?
Each die comes up "1" with probability 
, and the probabilities of the other five rolls are equal, so the probability of a "2" on either die (or any other roll) is 
.
The probability of rolling a double 1 is 
.
The probability of rolling any other given double is 
.
Therefore, the probability of rolling any ofthe six doubles is
,
which is 
odds against doubles - that is, 40 to 9.
Each die comes up "1" with probability
The probability of rolling a double 1 is
The probability of rolling any other given double is
Therefore, the probability of rolling any ofthe six doubles is
which is
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A penny is altered so that the odds are 3 to 2 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of both coins coming up heads?
A penny is altered so that the odds are 3 to 2 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of both coins coming up heads?
3 to 2 odds in favor of heads is equal to a probability of 
, which is the probability that the penny will come up heads. Similarly, 4 to 3 odds in favor of heads is equal to a probability of 
, which is the probability that the nickel will come up heads.
Since the tosses of the two coins are independent, multiply the probabilities. The probability that there will be two heads is
This is 
to 12 against both coins coming up heads.
3 to 2 odds in favor of heads is equal to a probability of
Since the tosses of the two coins are independent, multiply the probabilities. The probability that there will be two heads is
This is
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A penny is altered so that the odds are 5 to 4 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of there being one head and one tail?
A penny is altered so that the odds are 5 to 4 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of there being one head and one tail?
5 to 4 odds in favor of heads is equal to a probability of 
, which is the probability that the penny will come up heads. The probability that the penny will come up tails is 
.
Similarly, 4 to 3 odds in favor of heads is equal to a probability of 
, which is the probability that the nickel will come up heads. The probablity that the nickel will come up tails is 
.
The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.
The probability of the penny being heads and the nickel being tails is 
; that of the reverse is 
.
These are disjoint, so add these probabilities; the probability that one head and one tail will come up will be 
. This translates to 
to 31 odds against this event.
5 to 4 odds in favor of heads is equal to a probability of
Similarly, 4 to 3 odds in favor of heads is equal to a probability of
The outcomes of the tosses of the penny and the nickel are independent, so the probabilities can be multiplied.
The probability of the penny being heads and the nickel being tails is
These are disjoint, so add these probabilities; the probability that one head and one tail will come up will be
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What is the probability of getting an ace in any standard deck of cards? 
What is the probability of getting an ace in any standard deck of cards?
There are four aces in a standard fifty-two card deck. So we set up a fraction.
The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.
Answer is 
.
There are four aces in a standard fifty-two card deck. So we set up a fraction.
Answer is
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I have five white balls, three blue balls, and two yellow balls. What's the probability I get either white or blue?
I have five white balls, three blue balls, and two yellow balls. What's the probability I get either white or blue?
When we see the word or, we are adding probabilities. Since we are looking for white or blue, our chance is:
The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.
Answer is 
.
When we see the word or, we are adding probabilities. Since we are looking for white or blue, our chance is:
Answer is
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I have five white balls, three blue balls, and two yellow balls. What's the probability I DON'T get a blue ball?
I have five white balls, three blue balls, and two yellow balls. What's the probability I DON'T get a blue ball?
Since I don't want a blue ball, I just subtract the total number of balls and the number of blue balls in the set. The probability is:
The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.
Answer is 
.
Since I don't want a blue ball, I just subtract the total number of balls and the number of blue balls in the set. The probability is:
Answer is
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Joe has five apples, four oranges, six bananas, and five limes. What's the probability he gets two apples without replacement?
Joe has five apples, four oranges, six bananas, and five limes. What's the probability he gets two apples without replacement?
Since we are trying to get two apples without replacement, this is a condition we need to meet. This means we need to multiply probabilities. For his first pick he has five apples out of twenty fruits in which the probability is 
or 
. With one apple gone, there are only nineteen fruits left. Since he wants to get another apple, there areonly four apples left out of nineteen fruits or 
. Now, we multiply our probabilities:
.
Since we are trying to get two apples without replacement, this is a condition we need to meet. This means we need to multiply probabilities. For his first pick he has five apples out of twenty fruits in which the probability is
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If there are 
blue balls, 
black balls, 
green balls, and 
red balls and I want to pick two balls without replacement. what’s the probability I get a black ball on the first pull and a red ball on the second?
If there are
Since we are trying to get first black then red balls without replacement, this is a condition we need to meet. This means we need to multiply probabilities. For my first pick I have five balls out of eighteen in which the probability is 
. With one ball gone, there are only seventeen balls left. Since I want a red ball next, there are three left out of seventeen balls or 
. Now, we multiply our probabilities:
.
Since we are trying to get first black then red balls without replacement, this is a condition we need to meet. This means we need to multiply probabilities. For my first pick I have five balls out of eighteen in which the probability is
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I pick a number from 
to 
. What's the probability I get a prime number
I pick a number from
Prime numbers are numbers with factors of 
and itself. Those numbers are 
. Our probability:
The numerator represents how many possibilites we are looking for. The denominator represents the total in the given data.
Answer is .
Prime numbers are numbers with factors of
Answer is
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Sheila flips a coin 
times. What’s the probability she doesn’t get heads on all the flips?
Sheila flips a coin
To not get any heads, this would mean Sheila would need to get tails on all the flips. Since there are two choices in a flip and Sheila only want tails, our chances is 
. Since she needs this done twice more, this becomes a condition and we need to multiply out our probabiities. The odds are the same so our answer is just 
.
To not get any heads, this would mean Sheila would need to get tails on all the flips. Since there are two choices in a flip and Sheila only want tails, our chances is
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If you draw an ace from and deck then place it back in the deck, what is the probability of drawing another ace?
If you draw an ace from and deck then place it back in the deck, what is the probability of drawing another ace?
Since the first card was placed back into the deck, it will have no effect on the second draw.
Since there are 
aces in a deck of 
cards, the probability would be
or 
.
Since the first card was placed back into the deck, it will have no effect on the second draw.
Since there are
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A fair 6 sided die is rolled twice. What is the probability of observing a 6 both times?
A fair 6 sided die is rolled twice. What is the probability of observing a 6 both times?
The probability of rolling a 6 once is 
because 6 is one of 6 different possible outcomes.
The probability of two independent events happening can be multiplied together. The probability of getting 2 6's is:
The probability of rolling a 6 once is
The probability of two independent events happening can be multiplied together. The probability of getting 2 6's is:
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If an integer 
is randomly selected such that 
, what is the probability that 
is a prime number?
If an integer
Probability is defined as specified outcomes divided by total possible outcomes.
Recall that a prime numbered is defined as a number that is divisible only by one and itself, and 
is not prime. There are 
prime numbers in the range of possible 
values:
Notice that the range of values, includes zero , so there are 
total numbers to select from.
Probability is defined as specified outcomes divided by total possible outcomes.
Recall that a prime numbered is defined as a number that is divisible only by one and itself, and
Notice that the range of values,
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Two dice, one red and one blue, are altered. The red die comes up a "1" 
of the time and a "2" 
of the time, and the blue die comes up a "1" 
of the time and a "2" 
of the time. What are the odds against rolling a "2", a "3", or a "4" with the two dice?
Two dice, one red and one blue, are altered. The red die comes up a "1"
A "2" can only be rolled on the two dice with a double "1";, a "3", with a "1-2" or "2-1", and a "4", with a "1-3", a "2-2", or a "3-1". This makes six favoriable rolls. To answer the question, we must know the probabilities that the red die will come up "1", "2", and "3"; we must also know the same for the blue die. We know the probabilities for "1" and "2" for each die, but not "3". Therefore, insufficent information is given in the problem.
A "2" can only be rolled on the two dice with a double "1";, a "3", with a "1-2" or "2-1", and a "4", with a "1-3", a "2-2", or a "3-1". This makes six favoriable rolls. To answer the question, we must know the probabilities that the red die will come up "1", "2", and "3"; we must also know the same for the blue die. We know the probabilities for "1" and "2" for each die, but not "3". Therefore, insufficent information is given in the problem.
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What is the probability of choosing 2 67's in a row out of the following data set? Assume replacement.
What is the probability of choosing 2 67's in a row out of the following data set? Assume replacement.
What is the probability of choosing 2 67's in a row out of the following data set? Assume replacement.
Let's begin by recalling that probability can be thought of as follows:
Meaning that the probability of an event is the number of times the desired outcome exists, over the total number of outcomes.
In this case, each pick has two possible desired outcomes, and 9 total outcomes. (there are two 67's, and 9 total terms).
So the probability of each choice being a 67 is:
However, we are not done yet. We need the probability of choosing two 67's in a row. Assuming replacement means that each probability will be the same, so to find the probability of events, we will multiply each probability together. This gets:
What is the probability of choosing 2 67's in a row out of the following data set? Assume replacement.
Let's begin by recalling that probability can be thought of as follows:
Meaning that the probability of an event is the number of times the desired outcome exists, over the total number of outcomes.
In this case, each pick has two possible desired outcomes, and 9 total outcomes. (there are two 67's, and 9 total terms).
So the probability of each choice being a 67 is:
However, we are not done yet. We need the probability of choosing two 67's in a row. Assuming replacement means that each probability will be the same, so to find the probability of events, we will multiply each probability together. This gets:
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A fair coin is tossed 4 times. What is the total number ways the results can be observed (ie HHHH, HHHT, HHTH.....).
A fair coin is tossed 4 times. What is the total number ways the results can be observed (ie HHHH, HHHT, HHTH.....).
In general, the number of outcomes that can be observed when flipping a coin is found using the formula 
where n is the number of tosses. In this case there are 
ways to observe the outcome.
Alternatively, if n is small enough you can list out the possible outcomes:
1. HHHH
2. HHHT
3. HHTH
4. HTHH
5. THHH
6. HHTT
7. HTHT
8. HTTH
9. THTH
10. TTHH
11. THHT
12. TTTH
13. TTHT
14. THTT
15. HTTT
16. TTTT
In general, the number of outcomes that can be observed when flipping a coin is found using the formula
Alternatively, if n is small enough you can list out the possible outcomes:
1. HHHH
2. HHHT
3. HHTH
4. HTHH
5. THHH
6. HHTT
7. HTHT
8. HTTH
9. THTH
10. TTHH
11. THHT
12. TTTH
13. TTHT
14. THTT
15. HTTT
16. TTTT
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In a carnival game, a player must spin two independent spinners. If the first spinner has a probability of 
landing on red, and the second spinner has a probability of 
landing on red, what is the probability of the first spinner NOT landing on red, and the second spinner landing on red?
In a carnival game, a player must spin two independent spinners. If the first spinner has a probability of
Since the spinners are independent of each other, we will need to multiply the probability of not landing on red on the first spinner and the probability of landing on red on the second spinner together.
The probability of not landing on red for the first spinner is found by subtracting the probability of landing on red from 
.
Thus, 
Since the spinners are independent of each other, we will need to multiply the probability of not landing on red on the first spinner and the probability of landing on red on the second spinner together.
The probability of not landing on red for the first spinner is found by subtracting the probability of landing on red from
Thus,
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Suppose there is a bag of marbles. There are two reds, two greens, and one orange marbles. Without replacement, what is the probability of choosing a red, and another red?
Suppose there is a bag of marbles. There are two reds, two greens, and one orange marbles. Without replacement, what is the probability of choosing a red, and another red?
There are five marbles altogether.
The first marble is a red, which is two of the five marbles in the bag.
The probability of choosing a red for the first trial is: 
Without replacement, there are four marbles remaining, and only one red marble remains.
The probability of choosing a red for the second trial is: 
Multiply the two probabilities.
The answer is: 
There are five marbles altogether.
The first marble is a red, which is two of the five marbles in the bag.
The probability of choosing a red for the first trial is:
Without replacement, there are four marbles remaining, and only one red marble remains.
The probability of choosing a red for the second trial is:
Multiply the two probabilities.
The answer is:
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A red die is altered so that it comes up a "1" of the time; the other five numbers come up with equal probability to each other. A blue die is altered similarly.
Both dice are rolled. What are the odds against rolling doubles (both numbers are the same)?
A red die is altered so that it comes up a "1"
Both dice are rolled. What are the odds against rolling doubles (both numbers are the same)?
Each die comes up "1" with probability , and the probabilities of the other five rolls are equal, so the probability of a "2" on either die (or any other roll) is .
The probability of rolling a double 1 is .
The probability of rolling any other given double is .
Therefore, the probability of rolling any ofthe six doubles is
,
which is odds against doubles - that is, 40 to 9.
Each die comes up "1" with probability
The probability of rolling a double 1 is
The probability of rolling any other given double is
Therefore, the probability of rolling any ofthe six doubles is
which is
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A penny is altered so that the odds are 3 to 2 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of both coins coming up heads?
A penny is altered so that the odds are 3 to 2 against it coming up tails when tossed; a nickel is altered so that the odds are 4 to 3 against it coming up tails when tossed. If both coins are tossed; what are the odds of both coins coming up heads?
3 to 2 odds in favor of heads is equal to a probability of , which is the probability that the penny will come up heads. Similarly, 4 to 3 odds in favor of heads is equal to a probability of , which is the probability that the nickel will come up heads.
Since the tosses of the two coins are independent, multiply the probabilities. The probability that there will be two heads is
This is to 12 against both coins coming up heads.
3 to 2 odds in favor of heads is equal to a probability of
Since the tosses of the two coins are independent, multiply the probabilities. The probability that there will be two heads is
This is
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