Probability - SAT Subject Test in Math I
Card 0 of 13
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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