A coin is tossed, and a standard number cube is rolled. What is the probability that the coin shows heads and the number cube shows an even number?
A. start fraction 1 over 6 end fraction
B. 1
C. one-fourth
D. one-half
There are 2 possible outcomes for the coin toss (heads or tails) and 3 possible outcomes for the number cube (even numbers are 2, 4, or 6). So there are 2 x 3 = 6 possible outcomes in total. Only one of those outcomes is heads and an even number: heads and 2. Therefore, the probability is 1/6.
Answer: A. start fraction 1 over 6 end fraction
A bag contains 3 blue marbles, 9 green marbles, and 11 yellow marbles. Twice you draw a marble and replace it. Find P(blue, then green).
A. start fraction 27 over 529 end fraction
B. start fraction 27 over 23 end fraction
C. start fraction 15 over 529 end fraction
D. Start Fraction 12 over 23 End Fraction
The probability of drawing a blue marble on the first draw is 3/23 (since there are 3 blue marbles out of 23 total). After replacing the first marble, the probabilities are reset, so the probability of drawing a green marble on the second draw is also 9/23.
To find the probability of both events happening (drawing blue then green), we multiply the probabilities:
P(blue, then green) = (3/23) x (9/23) = 27/529
So the answer is A. start fraction 27 over 529 end fraction.
The diagram shows the contents of a jar of marbles. You select two marbles at random. One marble is drawn and not replaced. Then a second marble is drawn. What is the probability of selecting a red marble and then another red marble?
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
A. start fraction 3 over 64 end fraction
B. one-fourth
C. Fraction is 1 over 20
D. start fraction 1 over 30 end fraction
There are 4 red marbles out of a total of 16 marbles. If one marble is drawn and not replaced, there will be a total of 15 marbles left in the jar, 3 of which are red. Therefore, the probability of drawing a red marble first is 4/16, or 1/4, and the probability of drawing another red marble second is 3/15.
To find the probability of both events happening (drawing a red marble and then another red marble), we multiply the probabilities:
P(red, then red) = (1/4) x (3/15) = 1/20
So the answer is C. Fraction is 1 over 20.
From a committee of 6 girls and 4 boys, a name is selected. Then another name is selected. What is the probability that both people drawn will be girls?
A. nine over twenty-five
B. two-fifteenths
C. three-fifths
D. one-third
There are 6 girls and 4 boys on the committee, so the probability of selecting a girl on the first draw is 6/10 and the probability of selecting another girl on the second draw (assuming the first selection is not replaced) is 5/9, since there will be one less person overall and one less girl.
To find the probability of both events happening (drawing two girls), we multiply the probabilities:
P(girl, then girl) = (6/10) x (5/9) = 1/3
So the answer is D. one-third.
Write the number of permutations in factorial form. Then simplify.
J K L M N O P
A. 7!; 720
B. 8!: 40,320
C. 7!; 49
D. 7!; 5,040
There are 7 letters in the given set, so the number of permutations is 7!.
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040
So the answer is D. 7!; 5,040.
How many 2-digit numbers can be formed using only the digits 2, 3, 5, and 6, if the digits are not be repeated within a number?
A. 11
B. 12
C. 10
D. 2
There are four choices for the first digit, and three choices for the second digit (because the digit cannot be repeated). Therefore, the total number of possible 2-digit numbers is:
4 x 3 = 12
So the answer is B. 12.
A juice company decides to test five different brands of juice. The different brands have been labeled A¸ B¸ C¸ D¸ and E. The company decides to compare each brand with the other brands by pairing together different brands. How many different pairs will result by selecting two different brands at a time?
A. 11
B. 15
C. 120
D. 10
There are 5 choices for the first brand and 4 choices for the second brand (since a brand cannot be paired with itself). However, the order in which the pairs are listed does not matter (i.e. the pair AB is the same as the pair BA), so we need to divide by 2 to adjust for overcounting.
Therefore, the number of different pairs that can be formed is:
(5 x 4)/2 = 10
So the answer is D. 10.
There are 9 candidates running for 3 seats on a committee. How many different election results are possible?
A. 36
B. 84
C. 504
D. 56
There are 9 choices for the first seat, 8 choices for the second seat (since one candidate has already been elected), and 7 choices for the third seat. However, the order in which the candidates are elected does not matter (i.e. electing candidates A, B, and C is the same as electing candidates C, B, and A), so we need to divide by 3! to adjust for overcounting.
Therefore, the number of different possible election results is:
(9 x 8 x 7)/(3 x 2 x 1) = 84
So the answer is B. 84.
How many different ways are there to rearrange the letters of the word EXAM, if you don't care if the result is a recognizable word?
A. 4
B. 12
C. 24
D. 256
The word EXAM has 4 letters, so there are 4 choices for the first letter, 3 choices for the second letter (since one letter has already been used), 2 choices for the third letter, and only 1 choice for the final letter. Therefore, the total number of possible arrangements is:
4 x 3 x 2 x 1 = 24
So the answer is C. 24.
Suppose you spin the spinner once. Find the probability.
A circle is divided equally into eight sections.· Two of the sections are labeled with an upper B.
· One of the sections is labeled with an upper G.
· One of the sections is labeled with an upper Y.
· Four of the sections are labeled with an upper R.
· An arrow originating from the center of the circle is pointing at one of the sections with an upper R.
P(yellow)
A. one-eighth
B. start fraction 1 over 6 end fraction
C. one-fourth
D. one-half
There is only one section labeled with a Y out of a total of 8 sections, so the probability of landing on yellow is:
P(yellow) = 1/8
So the answer is A. one-eighth.
Drink Survey
Drink
Number of Shoppers Who Preferred It
A
10
B
15
C
7
D
3
E
6
What is the probability that 1 shopper, selected at random, preferred neither Drink E nor Drink C?
A. thirty-five over forty-one
B. thirteen over forty-one
C. twenty-eight over forty-one
D. thirteen over twenty-eight
To find the probability that a shopper preferred neither E nor C, we need to add up the number of shoppers who preferred A, B, and D, since those are the only remaining choices.
10 + 15 + 3 = 28
Therefore, out of the total number of shoppers surveyed (10 + 15 + 7 + 3 + 6 = 41), 28 preferred a drink other than E or C.
So the probability that a shopper, selected at random, preferred neither drink E nor drink C is:
28/41
So the answer is C. twenty-eight over forty-one.
A farmer examines a sample of 25 cartons of eggs and finds that 3 contain cracked eggs. What is the best prediction of the number of cartons with cracked eggs in a delivery with 500 cartons?
A. 6
B. 12
C. 60
D. 120
We can use proportion to make a prediction. The proportion of cartons containing cracked eggs in the sample is:
3/25
To predict the number of cartons containing cracked eggs in a delivery with 500 cartons, we can set up a proportion:
3/25 = x/500
where x is the number of cartons containing cracked eggs in the delivery.
Cross-multiplying gives:
25x = 3 x 500
Simplifying gives:
25x = 1500
Dividing both sides by 25 gives:
x = 60
Therefore, the best prediction of the number of cartons containing cracked eggs in a delivery with 500 cartons is 60.
So the answer is C. 60.
A coin is tossed. If heads appears, a spinner that can land on any number from 1 to 4 is spun. If tails appears, a second coin is tossed instead of spinning the spinner. What are the possible outcomes?
A. H1 H2 H3 H4
B. H1 H2 H3
C. H1 H2 H3 H4 TH TT
D. HH HT
The possible outcomes can be listed as combinations of a coin toss and a spinner spin/coin toss:
- Head, spinner lands on 1
- Head, spinner lands on 2
- Head, spinner lands on 3
- Head, spinner lands on 4
- Tail, first coin lands on heads and second coin lands on heads
- Tail, first coin lands on heads and second coin lands on tails
- Tail, first coin lands on tails and second coin lands on heads
- Tail, first coin lands on tails and second coin lands on tails
So the answer is:
A. H1, H2, H3, H4
TH, HH, HT, TT
A lunch menu has 4 different sandwiches, 2 different soups, and 5 different drinks. How many different lunches consisting of a sandwich, a soup, and a drink can you choose?
A. 10
B. 11
C. 40
D. 13
There are 4 choices for the sandwich, 2 choices for the soup, and 5 choices for the drink. Therefore, the total number of possible lunches is:
4 x 2 x 5 = 40
So the answer is C. 40.
If the spinner is spun twice, what is the probability that the spinner will stop on a consonant and then again on a consonant?
The spinner is a circle divided into 6 equal sections. The sections are labeled L U Z O E and I.
A. two-ninths
B. start fraction 1 over 3 end fraction
C. start fraction 1 over 6 end fraction
D. one-ninth
There are 4 consonants (L, Z, O, and I) and 2 vowels (U and E) on the spinner.
The probability of spinning a consonant on the first spin is 4/6 = 2/3. If a consonant is spun on the first spin, there will be 3 consonants and 2 vowels remaining for the second spin. Therefore, the probability of spinning a consonant on the second spin (given that a consonant was spun on the first spin) is 3/5.
To find the probability of both events happening (spinning a consonant and then spinning another consonant), we multiply the probabilities:
P(consonant, then consonant) = (2/3) x (3/5) = 2/5
So the answer is not one of the choices provided. The closest answer is A. two-ninths, but that is not correct for this problem.
A box contains 4 yellow tiles, 6 green tiles, and 10 purple tiles. Without looking, you draw out a tile and then draw out a second tile without returning the first tile.
Find P(purple, then purple).
A. nine over thirty-eight
B. one-fourth
C. three-hundredths
D. three over nineteen
There are 20 tiles in the box before the first draw, so the probability of drawing a purple tile first is 10/20 = 1/2. After the first tile is drawn, there are 19 tiles left, of which 9 are purple (since one purple tile has already been drawn). Therefore, the probability of drawing another purple tile on the second draw is 9/19.
To find the probability of both events happening (drawing a purple tile and then another purple tile), we multiply the probabilities:
P(purple, then purple) = (1/2) x (9/19) = 9/38
So the answer is A. nine over thirty-eight.
In how many different ways can the following positions be chosen from the 18-member math team?
president, vice president, and treasurer
A. 4,080
B. 306
C. 5,814
D. 4,896
There are 18 choices for the first position (president), since any of the 18 team members could be chosen. However, once the first position is filled, there are only 17 team members left to choose from for the second position (vice president), since one team member has already been chosen. Similarly, there are only 16 team members left to choose from for the third position (treasurer), since two team members have already been chosen.
Therefore, the total number of ways of filling the three positions is:
18 x 17 x 16 = 4,896
So the answer is D. 4,896.
In Mr. Tarkiainen’s history class, each student chooses two of the topics listed in the chart above for their papers. How many different pairs can Juliana choose?
A chart with 2 columns is shown.
The first column is titled Year. The second column is titled Invention.
Row 1: 1807 Steamboat
Row 2: 1825 Photography
Row 3: 1837 Telegraph
Row 4: 1846 Sewing Machine
Row 5: 1869 Hydroelectric power
Row 6: 1876 Telephone
Row 7: 1879 light bulb
Row 8: 1888 Rabies Vaccine
Row 9: 1890 Diesel Engine
Row 10: 1897 Airplane
A. 90
B. 10
C. 45
D. 30
Juliana can choose any of the 10 topics for the first part of her paper and any of the remaining 9 topics for the second part (since she can't choose the same topic twice). However, the order in which she chooses the topics does not matter (i.e. choosing steamboat and telegraph is the same as choosing telegraph and steamboat), so we need to divide by 2 to adjust for overcounting.
Therefore, the total number of different pairs Juliana can choose is:
(10 x 9)/2 = 45
So the answer is C. 45.
Below are the results of tossing a number cube 10 times. Find the experimental probability of tossing 4.
2 6 3 5 4 4 1 2 4 3
A. start fraction 3 over 10 end fraction
B. start fraction 1 over 6 end fraction
C. start fraction 1 over 5 end fraction
D. two-thirds
Out of the 10 tosses, there are 3 instances where 4 is rolled.
Therefore, the experimental probability of rolling a 4 is:
3/10
So the answer is A. start fraction 3 over 10 end fraction.
A multiple choice test has 5 questions each with 5 possible answers. Find the probability of answering all the questions correctly by guessing randomly.
A. one over three-thousand-one-hundred-twenty-five
B. one over twenty-five
C. one over-six-hundred-twenty-five
D. one over one-hundred-twenty-five
The probability of getting any one question right by random guessing is 1/5. Since each question is independent (i.e. getting one question right doesn't affect the probability of getting another question right), we can multiply the probabilities of getting each question right:
P(all 5 correct) = (1/5)^5
Simplifying, we get:
P(all 5 correct) = 1/3125
So the answer is A. one over three-thousand-one-hundred-twenty-five.
Eva wants to know the probability of tossing “heads” at least three times out of five. She used the random number table to simulate flipping a coin. She assigned odd digits to represent “heads” and even digits, including zero, to represent “tails.”
32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 65311
Use the table to find P(“heads” at least 3 out of 5 times).
A. three-eighths
B. seven-eighths
C. five-sixteenths
D. thirteen-sixteenths
bot answer this question
Eva wants to know the probability of tossing “heads” at least three times out of five. She used the random number table to simulate flipping a coin. She assigned odd digits to represent “heads” and even digits, including zero, to represent “tails.”
32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 65311
Use the table to find P(“heads” at least 3 out of 5 times).
A. three-eighths
B. seven-eighths
C. five-sixteenths
D. thirteen-sixteenths