How to find the probability of an outcome - PSAT Math
Card 1 of 777
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.
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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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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.
Tap to reveal answer
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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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.
Tap to reveal answer
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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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.
Tap to reveal answer
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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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.
Tap to reveal answer
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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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.
Tap to reveal answer
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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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.
Tap to reveal answer
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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There is a special contest held at a high school where the winner will receive a prize of $100. 300 seniors, 200 juniors, 200 sophomores, and 100 freshmen enter the contest. Each senior places their name in the hat 5 times, juniors 3 times, and sophmores and freshmen each only once. What is the probability that a junior's name will be chosen?
There is a special contest held at a high school where the winner will receive a prize of $100. 300 seniors, 200 juniors, 200 sophomores, and 100 freshmen enter the contest. Each senior places their name in the hat 5 times, juniors 3 times, and sophmores and freshmen each only once. What is the probability that a junior's name will be chosen?
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The first thing to do here is find the total number of students in the contest. Seniors = 300 * 5 = 1500, Juniors = 200 * 3 = 600, Sophomores = 200, and Freshmen = 100. So adding all these up you get a total of 2400 names in the hat. Out of these 2400 names, 600 of them are Juniors. So the probability of choosing a Junior's name is 600/2400 = 1/4.
The first thing to do here is find the total number of students in the contest. Seniors = 300 * 5 = 1500, Juniors = 200 * 3 = 600, Sophomores = 200, and Freshmen = 100. So adding all these up you get a total of 2400 names in the hat. Out of these 2400 names, 600 of them are Juniors. So the probability of choosing a Junior's name is 600/2400 = 1/4.
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All of Jean's brothers have red hair.
If the statement above is true, then which of the following CANNOT be true?
All of Jean's brothers have red hair.
If the statement above is true, then which of the following CANNOT be true?
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Here we have a logic statement:
If A (Jean's brother), then B (red hair).
"If Ron has red hair, then he is Jean's brother" states "If B, then A" - we do not know whether or not this is true.
"If Winston does not have red hair, then he is not Jean's brother" states "If not B, then not A" - this has to be true.
"If Paul is not Jean's brother, then he has red hair" states "If not A, then B." We do not know whether or not this is true.
"If Eddie is Jean's brother, then he does not have red hair" states "If A, then not B" - we know this cannot be true.
Here we have a logic statement:
If A (Jean's brother), then B (red hair).
"If Ron has red hair, then he is Jean's brother" states "If B, then A" - we do not know whether or not this is true.
"If Winston does not have red hair, then he is not Jean's brother" states "If not B, then not A" - this has to be true.
"If Paul is not Jean's brother, then he has red hair" states "If not A, then B." We do not know whether or not this is true.
"If Eddie is Jean's brother, then he does not have red hair" states "If A, then not B" - we know this cannot be true.
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A bag of jellybeans has 20 watermelon jellybeans, 45 sour apple jellybeans, 30 orange jellybeans and 5 cotton candy jellybeans. If you reach in and grab one jelly bean, what is the probability that it will be watermelon flavored?
A bag of jellybeans has 20 watermelon jellybeans, 45 sour apple jellybeans, 30 orange jellybeans and 5 cotton candy jellybeans. If you reach in and grab one jelly bean, what is the probability that it will be watermelon flavored?
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Add up the total number of jellybeans, 20 + 45 + 30 + 5 = 100.
Divide the number of watermelon jellybeans by the total: 20/100 and reduce the fraction to 1/5.
Add up the total number of jellybeans, 20 + 45 + 30 + 5 = 100.
Divide the number of watermelon jellybeans by the total: 20/100 and reduce the fraction to 1/5.
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Ben only goes to the park when it is sunny.
If the above statement is true, which of the following is also true?
Ben only goes to the park when it is sunny.
If the above statement is true, which of the following is also true?
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“Ben only goes to the park when it is sunny.” This means it has to be sunny out for Ben to go to the park, but it does not mean that he always is at the park when it is sunny. Looking at the first choice we can say that this is not necessarily true because it could be sunny and Ben doesn’t have to be at the park. Similar reasoning would prove the second choice wrong as well. The third choice is correct-Ben only is at the park when it is sunny, so he’s definitely not there when it’s not sunny. The fourth choice is clearly wrong because we know Ben only goes when it is sunny, not when it’s rainy.
“Ben only goes to the park when it is sunny.” This means it has to be sunny out for Ben to go to the park, but it does not mean that he always is at the park when it is sunny. Looking at the first choice we can say that this is not necessarily true because it could be sunny and Ben doesn’t have to be at the park. Similar reasoning would prove the second choice wrong as well. The third choice is correct-Ben only is at the park when it is sunny, so he’s definitely not there when it’s not sunny. The fourth choice is clearly wrong because we know Ben only goes when it is sunny, not when it’s rainy.
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You went to order a cake but all of the cakes are in identical boxes. If there are five chocolate cakes, four carrot cakes, three vanilla cakes, and six red velvet cakes, what is the maximum number of boxes that you would have to open to ensure that you have one of each type of cake?
You went to order a cake but all of the cakes are in identical boxes. If there are five chocolate cakes, four carrot cakes, three vanilla cakes, and six red velvet cakes, what is the maximum number of boxes that you would have to open to ensure that you have one of each type of cake?
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There are four different types of cake. In this type of problem we want to guarantee we have one of each, so we need to assumbe we have very bad luck. We start with the red velvet since that is the type with the most cakes. If we open those 6 we are not guaranteed to have different ones. Then say we opened all five chocolate cakes, then all four carrot cakes. We still have only three types of cakes but opened 15 boxes. When we open the next box (16) we will be guaranteed to have one of each.
There are four different types of cake. In this type of problem we want to guarantee we have one of each, so we need to assumbe we have very bad luck. We start with the red velvet since that is the type with the most cakes. If we open those 6 we are not guaranteed to have different ones. Then say we opened all five chocolate cakes, then all four carrot cakes. We still have only three types of cakes but opened 15 boxes. When we open the next box (16) we will be guaranteed to have one of each.
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A circle is inscribed inside a square. If a point inside the square is selected at random, what is the probability that the point will also be inside the circle?
A circle is inscribed inside a square. If a point inside the square is selected at random, what is the probability that the point will also be inside the circle?
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The probability of the point being inside the circle is the ratio of the area of the circle to the area of the square. If we suppose that the circle has radius r, then the square must have side 2r. The area of the circle is πr2 and the area of the square is 〖(2r)〗2=〖4r〗2, so the proportion of the areas is (πr2)/〖4r〗2 =π/4.
The probability of the point being inside the circle is the ratio of the area of the circle to the area of the square. If we suppose that the circle has radius r, then the square must have side 2r. The area of the circle is πr2 and the area of the square is 〖(2r)〗2=〖4r〗2, so the proportion of the areas is (πr2)/〖4r〗2 =π/4.
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John has a bowl with 54 marbles in it. Half of the marbles are green and half of the marbles are blue. John takes 3 green marbles and 6 blue marbles from the jar. John then takes 13 additional marbles from the remaining marbles in the jar. What is the minimum number of these 13 marbles that must be green in order for there to be more blue marbles than green marbles left?
John has a bowl with 54 marbles in it. Half of the marbles are green and half of the marbles are blue. John takes 3 green marbles and 6 blue marbles from the jar. John then takes 13 additional marbles from the remaining marbles in the jar. What is the minimum number of these 13 marbles that must be green in order for there to be more blue marbles than green marbles left?
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The bowl has 54 marbles, half green and half blue. This gives us 27 green and 27 blue marbles:
27 G / 27 B
John then takes 3 green and 6 blue from the bowl. This leaves the bowl with:
24 G / 21 B
If there are going to be more blue than green marbles after John's 13 marbles, he has to take at least 4 more green marbles than blue marbles, because right now there are 3 less blue marbles. Therefore, we need to take at least 9 green marbles, which would mean 4 or less of the marbles would be blue (8 green and 5 blue would leave us with equal green and equal blue marbles, so it would have to be more than 8 green marbles, which gives us 9 green marbles).
We can also solve this as an inequality. You take the difference in marbles, which is 3, which means you need the difference in green and blue marbles to be greater than 3, or at least 4. You have b + g = 13 and g - b > 3, where b and g are positive integers.
b + g = 13 (Subtract g on both sides of the equation)
b = 13 - g
g - b > 3 (Substitute above equation)
g - (13 - g) > 3 (Distribute negative sign in parentheses)
g - 13 + g > 3 (Add both g variables)
2g - 13 > 3 (Add 13 to both sides of the inequality)
2g > 16 (Divide both sides of the inequality by 2)
g > 8 so g has to be 9 or greater.
The bowl has 54 marbles, half green and half blue. This gives us 27 green and 27 blue marbles:
27 G / 27 B
John then takes 3 green and 6 blue from the bowl. This leaves the bowl with:
24 G / 21 B
If there are going to be more blue than green marbles after John's 13 marbles, he has to take at least 4 more green marbles than blue marbles, because right now there are 3 less blue marbles. Therefore, we need to take at least 9 green marbles, which would mean 4 or less of the marbles would be blue (8 green and 5 blue would leave us with equal green and equal blue marbles, so it would have to be more than 8 green marbles, which gives us 9 green marbles).
We can also solve this as an inequality. You take the difference in marbles, which is 3, which means you need the difference in green and blue marbles to be greater than 3, or at least 4. You have b + g = 13 and g - b > 3, where b and g are positive integers.
b + g = 13 (Subtract g on both sides of the equation)
b = 13 - g
g - b > 3 (Substitute above equation)
g - (13 - g) > 3 (Distribute negative sign in parentheses)
g - 13 + g > 3 (Add both g variables)
2g - 13 > 3 (Add 13 to both sides of the inequality)
2g > 16 (Divide both sides of the inequality by 2)
g > 8 so g has to be 9 or greater.
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Michelle is randomly drawing cards from a deck of of 52 cards. What is the chance she will draw a black queen followed by a 5 of any color, without replacing the cards?
Michelle is randomly drawing cards from a deck of of 52 cards. What is the chance she will draw a black queen followed by a 5 of any color, without replacing the cards?
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There are 2 black queens in the deck, one of spades and one of clubs, so there is a 2/52 chance a black Queen will be drawn and 4/51 chance of drawing a 5 of any color, since the queen has already been removed from the deck. Thus: 2/52 * 4/51 = 8/2652 → 2/663.
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There are 2 black queens in the deck, one of spades and one of clubs, so there is a 2/52 chance a black Queen will be drawn and 4/51 chance of drawing a 5 of any color, since the queen has already been removed from the deck. Thus: 2/52 * 4/51 = 8/2652 → 2/663.
1
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If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?
If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?
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If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?
Here we have 5 possible choices for x and 4 possible choices for y, giving us 5 * 4 = 20 possible outcomes.
We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where x and y are odd. We have 3 possibilities of odd numbers for x, and 3 possibilities of odd numbers for y, so we will have 9 outcomes of our total 20 outcomes where xy is odd, giving us a probability of 9/20.
If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?
Here we have 5 possible choices for x and 4 possible choices for y, giving us 5 * 4 = 20 possible outcomes.
We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where x and y are odd. We have 3 possibilities of odd numbers for x, and 3 possibilities of odd numbers for y, so we will have 9 outcomes of our total 20 outcomes where xy is odd, giving us a probability of 9/20.
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Mike has a bag of marbles, 4 white, 8 blue, and 6 red. He pulls out one marble from the bag and it is red. What is the probability that the second marble he pulls out of the bag is white?
Mike has a bag of marbles, 4 white, 8 blue, and 6 red. He pulls out one marble from the bag and it is red. What is the probability that the second marble he pulls out of the bag is white?
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There are 18 marbles in total. One of them is removed so now there are 17 marbles. This is our denominator. All of the original white marbles are still in the bag so there is a 4 out of 17 or 4/17 chance that the next marble taken out of the bag will be white.
There are 18 marbles in total. One of them is removed so now there are 17 marbles. This is our denominator. All of the original white marbles are still in the bag so there is a 4 out of 17 or 4/17 chance that the next marble taken out of the bag will be white.
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Michael tosses three fair coins. What is the probability that at least one of these coins will land on heads?
Michael tosses three fair coins. What is the probability that at least one of these coins will land on heads?
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Michael can toss either one head, two heads, or three heads.
If Michael tosses one head, then it could be on either the first, second, or third toss. We could model it like this, where H represents heads and T represents tails.
HTT, THT, or TTH
If Michael tosses two heads, then there are three possible combinations:
HHT, HTH, or THH
If Michael tosses three heads, then there is only one possible combination:
HHH
Thus, there are seven ways that Michael can toss at least one head. We must find the probability of each of these ways and then add them together.
The probability of rolling a head is ½ and the probability of rolling a tail is ½. Because each coin toss is independent, we can multiply the probabilities together.
For example, the probability of the combination HTT is (1/2)(1/2)(1/2) = 1/8
Probability of HTT = 1/8
Probability of THT = (1/2)(1/2)(1/2) = 1/8
Probability of TTH = (1/2)(1/2)(1/2) = 1/8
Probability of HHT = 1/8
Probability of HTH = 1/8
Probability of THH = 1/8
Probability of HHH = 1/8
So, there are seven possible ways that Michael can toss at least one head. The probability of each of these seven ways is equal to 1/8. Thus, the total probability of all seven events is 7/8.
ALTERNATE SOLUTION:
Michael can toss at least one head, or he can toss zero heads. The sum of these two probabilities must equal one, because they represent all of the ways that Michael could toss the coins. He could either toss at least on head, or he could toss no heads at all.
Probability of tossing at least one head + probability of tossing no heads = 1
The probability of tossing no heads is only possible with the combination TTT. The probability of tossing three tails is equal to (1/2)(1/2)(1/2) = 1/8
Probability of tossing at least one head + 1/8 = 1
Probability of tossing at least one head = 1 – 1/8 = 7/8 .
Michael can toss either one head, two heads, or three heads.
If Michael tosses one head, then it could be on either the first, second, or third toss. We could model it like this, where H represents heads and T represents tails.
HTT, THT, or TTH
If Michael tosses two heads, then there are three possible combinations:
HHT, HTH, or THH
If Michael tosses three heads, then there is only one possible combination:
HHH
Thus, there are seven ways that Michael can toss at least one head. We must find the probability of each of these ways and then add them together.
The probability of rolling a head is ½ and the probability of rolling a tail is ½. Because each coin toss is independent, we can multiply the probabilities together.
For example, the probability of the combination HTT is (1/2)(1/2)(1/2) = 1/8
Probability of HTT = 1/8
Probability of THT = (1/2)(1/2)(1/2) = 1/8
Probability of TTH = (1/2)(1/2)(1/2) = 1/8
Probability of HHT = 1/8
Probability of HTH = 1/8
Probability of THH = 1/8
Probability of HHH = 1/8
So, there are seven possible ways that Michael can toss at least one head. The probability of each of these seven ways is equal to 1/8. Thus, the total probability of all seven events is 7/8.
ALTERNATE SOLUTION:
Michael can toss at least one head, or he can toss zero heads. The sum of these two probabilities must equal one, because they represent all of the ways that Michael could toss the coins. He could either toss at least on head, or he could toss no heads at all.
Probability of tossing at least one head + probability of tossing no heads = 1
The probability of tossing no heads is only possible with the combination TTT. The probability of tossing three tails is equal to (1/2)(1/2)(1/2) = 1/8
Probability of tossing at least one head + 1/8 = 1
Probability of tossing at least one head = 1 – 1/8 = 7/8 .
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A jar of marbles has 125 marbles in it. 25 are blue marbles, 65 are red marble, 15 are green marbles and 20 are yellow marbles. What is the probability that the first three marbles picked are green or blue?
A jar of marbles has 125 marbles in it. 25 are blue marbles, 65 are red marble, 15 are green marbles and 20 are yellow marbles. What is the probability that the first three marbles picked are green or blue?
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Probability of each event = (# green marbles + # blue marbles)/ Total # of Marbles
P1 = (15 + 25) / 125 = 40 / 125
Second event assumes a blue or green was chosen for first event so there is one fewer marble on top and also one fewer marble in the total number of marbles.
P2 = (14 + 25) / 124 = 39 / 124
Third event assumes a blue or green was chosen for first and second events so there are two fewer marbles on top and also two fewer marbles in the total number of marbles.
P3 = (13 + 25) / 124 = 38 / 123
Probability for multiple events = P1 x P2 x P3
(40 / 125) * (39 / 124) * (38 / 123)
( 40 * 39 * 38) / (125 * 124 * 123 ) = 59280 / 1906500 = 0.031
Probability of each event = (# green marbles + # blue marbles)/ Total # of Marbles
P1 = (15 + 25) / 125 = 40 / 125
Second event assumes a blue or green was chosen for first event so there is one fewer marble on top and also one fewer marble in the total number of marbles.
P2 = (14 + 25) / 124 = 39 / 124
Third event assumes a blue or green was chosen for first and second events so there are two fewer marbles on top and also two fewer marbles in the total number of marbles.
P3 = (13 + 25) / 124 = 38 / 123
Probability for multiple events = P1 x P2 x P3
(40 / 125) * (39 / 124) * (38 / 123)
( 40 * 39 * 38) / (125 * 124 * 123 ) = 59280 / 1906500 = 0.031
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You have a full deck of 52 cards. If there are four queens in the deck, what is the probability that out of two cards dealt to you both are queens?
You have a full deck of 52 cards. If there are four queens in the deck, what is the probability that out of two cards dealt to you both are queens?
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Probability of each event = (# queens)/ Total # of cards
P1 = 4 / 52
Second event assumes a queen was chosen for first event so there is one less queen and also one less card:
P2 = 3 / 51
Probability for multiple events = P1 x P2
(4 / 52) * (3 / 51)
( 4 * 3) / (52 * 51 ) = 12 / 2652 = 0.0045
Probability of each event = (# queens)/ Total # of cards
P1 = 4 / 52
Second event assumes a queen was chosen for first event so there is one less queen and also one less card:
P2 = 3 / 51
Probability for multiple events = P1 x P2
(4 / 52) * (3 / 51)
( 4 * 3) / (52 * 51 ) = 12 / 2652 = 0.0045
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