There are two spinners. Each spinner is numbered 1-4. Each spinner is spun once. Find the probability of spinning each of the following:2 on the first spinner and a 4 on the second spinner, both numbers the same, 3 on exactly one spinner, one number less than the other.
prob(2 on 1st, 4 on 2nd) = (1/4)(1/4) = 1/16
prob(both a 3) = (1/4)(1/4) = 1/16
third question: 3 on one, one less on the other ???
do you mean (3,2) and (2,3)
if so , then prob = 1/16 + 1/16 = 1/8
multiplie 3x4 = 12 (1/12)
answer: 1/12
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
7 / 13
First, we need to identify the consonants: L, Z, and O. So, the probability of landing on a consonant on the first spin is 3/6 or 1/2. After the first spin, there are 5 sections left on the spinner. Since we need to land on a consonant again, there are only 2 consonants left out of the 5 sections. So, the probability of landing on a consonant on the second spin, given that the first spin was a consonant, is 2/5.
To find the probability of both events happening, we multiply the probabilities:
1/2 × 2/5 = 1/5
Therefore, the probability of the spinner stopping on a consonant and then again on a consonant is 1/5, or option A: two-ninths.
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
The probability of drawing a purple tile on the first draw is 10/20. However, since we don't replace the tile on the first draw, the probability of drawing a purple tile on the second draw is 9/19.
To find the probability of both events happening, we multiply the probabilities:
10/20 × 9/19 = 9/38
Therefore, the probability of drawing a purple tile, then drawing another purple tile without replacing the first one, is 9/38, or option 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
To choose the president, there are 18 options. Once the president is chosen, there are 17 remaining options for the vice president since the same person cannot hold two positions. Finally, there are 16 options left for the treasurer.
Using the multiplication principle, we multiply these numbers together:
18 × 17 × 16 = 4,896
Therefore, there are 4,896 different ways to choose the positions of president, vice president, and treasurer from the 18-member math team, or option D.
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 her first topic from the 10 options available. After selecting her first topic, there are only 9 options remaining for her second topic, since she cannot choose the same topic twice.
Using the multiplication principle, we multiply these numbers together:
10 × 9 = 90
Therefore, Juliana can choose from 90 different pairs of topics, or option A.
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, the number 4 came up three times.
The experimental probability of tossing 4 is calculated by dividing the number of times 4 came up by the total number of tosses:
3/10
Therefore, the experimental probability of tossing 4 is 3/10, or option A.
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 guessing one question correctly is 1/5, since there are 5 possible answers and only one is correct.
To find the probability of guessing all 5 questions correctly, we multiply the probabilities together:
(1/5) × (1/5) × (1/5) × (1/5) × (1/5) = 1/31,250
Therefore, the probability of answering all 5 questions correctly by guessing randomly is 1/31,250, or option A.
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
To determine the probability of tossing "heads" at least three times out of five, we need to count the number of outcomes from the table that have at least three odd digits. We can see that there are 10 such outcomes:
32766, 53855, 34591, 27732, 47406, 35521, 26658, 81704, 56212, 65311
Out of the 16 total outcomes, the probability of getting at least three heads is:
10/16 = 5/8
Therefore, the probability of tossing "heads" at least 3 times out of 5 is 5/8, or option A: three-eighths.
4 were wrong
I apologize for my mistake. You are correct that if 4 were wrong, then only 1 was correct, and the probability of answering all 5 questions correctly by guessing randomly would be:
(1/5) × (1/5) × (1/5) × (1/5) × (1/5) = 1/3125
Therefore, the correct answer is not option A, but rather:
A. one over three thousand one hundred twenty-five.