A survey taken by 150 people revealed that 65 like apple juice while 85 dislike it. One person is randomly chosen from this group. What is the chance that the chosen person dislikes apple juice? Write your answer as a ratio in simplest form.
The total number of people surveyed is 150. Out of those, 85 dislike apple juice. Therefore, the probability that a randomly chosen person dislikes apple juice is:
85/150
We can simplify this ratio by dividing both the numerator and denominator by 5:
17/30
Therefore, the probability that a randomly chosen person dislikes apple juice is 17:30 or 17/30.
A letter is to be randomly picked from the word MISSISSIPPI. Which set of letters have equal chances to be selected?
M, I, and P
M and P
S and I
M and I
S and I have equal chances to be selected. There are four instances of each letter in MISSISSIPPI, so the probability of selecting an S or an I is 4/11. The probability of selecting M is 1/11 and the probability of selecting P is 2/11. Therefore, S and I have equal chances to be selected from the word.
A six-sided number cube is rolled 30 times and lands on 3 ten times and on 5 eight times. Calculate the experimental probability of landing on a 3. Write your answer in the simplest form of a fraction.
The number cube is rolled a total of 30 times, and lands on 3 ten times. Therefore, the experimental probability of landing on a 3 is:
10/30
This fraction can be simplified by dividing both the numerator and denominator by 10:
1/3
Therefore, the experimental probability of landing on a 3 is 1/3 or one-third.
The experiment involved tossing three coins simultaneously. The experiment was carried out 100 times, and it was noted that three heads occurred 40 times. What is the difference between the experimental probability of getting three heads and its theoretical probability? Write the answer in the simplest form of fraction.
The theoretical probability of getting three heads in three coin tosses is (1/2) x (1/2) x (1/2) = 1/8, or one-eighth. This is because the probability of getting heads on a single coin toss is 1/2, and the probability of getting heads on three independent tosses is the product of the probabilities of getting heads on each toss.
The experimental probability of getting three heads in 100 trials is 40/100 = 2/5, or two-fifths.
The difference between the experimental probability and the theoretical probability is:
2/5 - 1/8 = 16/40 - 5/40
= 11/40
Therefore, the difference between the experimental probability of getting three heads and its theoretical probability is 11/40.
An experiment involves picking a card from the number cards 2, 4, 6, 10. In equation form. What is the probability model for this experiment?(1 point)
f(x)=
where x=2, 4, 6, 10
The probability model for this experiment can be represented in equation form as:
f(x) =
- 1/4 for x = 2
- 1/4 for x = 4
- 1/4 for x = 6
- 1/4 for x = 10
This probability model shows that each of the four number cards has an equal probability of 1/4 of being picked in a single draw.
Suppose the probability of selecting a supermarket shopper who prefers plastic bags instead of paper bags is 50%. Out of 150 shoppers, how many can you expect will prefer plastic bags?
If the probability of selecting a supermarket shopper who prefers plastic bags is 50%, then the expected proportion of shoppers who will prefer plastic bags is also 50%. To find how many out of 150 shoppers can be expected to prefer plastic bags, we can simply multiply the total number of shoppers by the expected proportion:
150 x 0.5 = 75
Therefore, out of 150 shoppers, we can expect 75 shoppers to prefer plastic bags.
A spinner has 8 equally sized sections labelled as A, B, C, D, E, F, G, H. In 160 spins, how many times can you expect to spin on a consonant?(1 point)
? times
Consonants are all the letters in the spinner except for A, E, and I, which are vowels. Therefore, there are 5 consonants out of a total of 8 letters on the spinner. The probability of spinning on a consonant in one spin is therefore 5/8.
To find the expected number of spins on a consonant in 160 spins, we can multiply the total number of spins by the probability of spinning on a consonant:
160 x 5/8 = 100
Therefore, we can expect to spin on a consonant 100 times out of 160 spins.
A single coin is tossed 300 times. Heads were observed 180 times. What is the long-run relative frequency of tails? Express the answer in decimal form.
The long-run relative frequency of tails is equal to the number of tails observed divided by the total number of coin tosses. We can find the number of tails observed by subtracting the number of heads from the total number of coin tosses:
300 - 180 = 120
Therefore, tails were observed 120 times out of 300 coin tosses. The long-run relative frequency of tails can be found by dividing the number of tails by the total number of coin tosses:
120/300 = 0.4
Expressed as a decimal, the long-run relative frequency of tails is 0.4.
An experiment involves picking a card from a selection of cards numbered 5, 10, and 15. In equation form, what is the probability model for this experiment? List the values of x in ascending order.
The probability model for this experiment can be represented in equation form as:
f(x) =
- 1/3 for x = 5
- 1/3 for x = 10
- 1/3 for x = 15
This probability model shows that each of the three cards has an equal probability of 1/3 of being picked in a single draw.
The values of x in ascending order are: 5, 10, 15.
An experiment involves picking a card from a selection of cards numbered 5, 10, and 15. In equation form, what is the probability model for this experiment? List the values of x in ascending order.
f(x)= ? where x= ?
The probability model for this experiment can be represented in equation form as:
f(x) =
- 1/3 for x = 5
- 1/3 for x = 10
- 1/3 for x = 15
This probability model shows that each of the three cards has an equal probability of 1/3 of being picked in a single draw.
The values of x in ascending order are: 5, 10, 15.
If the table below represents a probability model based on observed frequencies, what is the value of x?
Y C A R E S
P(Y) 0.25 0.18 0.30 0.17 x(1 point)
1 0.01 0.10 0.20
The sum of the probabilities must equal 1. Therefore:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Simplifying the equation, we have:
0.9 + x = 1
Subtracting 0.9 from both sides, we get:
x = 0.1
Therefore, the value of x is 0.1.
Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly selected a ball from both bowls and recorded the results in a table. If there are 500 balls in Bowl B, what is the estimated difference in the expected number of white and red balls in Bowl B?
100
300
200
50
The probability of selecting a white ball from Bowl A is 45/(45+55) = 0.45, and the probability of selecting a white ball from Bowl B is 60/(60+40) = 0.6.
If there are 500 balls in Bowl B, we would expect that approximately:
- 0.6 x 500 = 300 of the balls are white, and
- 0.4 x 500 = 200 of the balls are red.
Therefore, we would expect the difference between the expected number of white and red balls in Bowl B to be:
300 - 200 = 100
The estimated difference in the expected number of white and red balls in Bowl B is 100.
What is the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll?
2
1
14
12
The probability of rolling an odd number on the first roll of a six-sided cube is 3/6, or 1/2. This is because there are three odd numbers (1, 3, 5) out of six possible outcomes.
The probability of rolling an even number on the second roll is also 1/2, since the first roll does not affect the probability of rolling an even number on the second roll.
To find the probability of both events happening together, we now multiply the individual probabilities:
(1/2) x (1/2) = 1/4
Therefore, the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll is 1/4 or 0.25. The answer is 1.
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.
The probability of selecting one carnation out of the 18 flowers in the vase is 6/18, or 1/3. After one carnation is selected, there will be 5 begonias, 5 carnations, and 7 tulips remaining.
Therefore, the probability of selecting a second carnation out of the remaining 17 flowers is 5/17.
To find the probability of selecting two carnations, we multiply the probability of selecting one carnation by the probability of selecting another carnation after one carnation has already been selected:
(1/3) x (5/17) = 5/51
Therefore, the probability of selecting two carnations without replacement is 5/51.