Julia has two bags of gummy candies one at a time he really pulls eight pineapples five lines seven strawberry candies out of the first bag and put them back each time he also randomly pulled 11 mangoes three tamarind and six watermelon candies out of the second bag of put them back each time
based on this information what is the probability that Julio will select a strawberry candy from the first bag in a tamarind candy from the second bag the next time he takes a candy from each bag?
To find the probability, we need to determine the number of favorable outcomes and the total number of possible outcomes.
In the first bag, there are 7 strawberry candies out of a total of 8+5+7 = 20 candies that Julio could select. Therefore, the probability of selecting a strawberry candy from the first bag is 7/20.
In the second bag, there are 3 tamarind candies out of a total of 11+3+6 = 20 candies that Julio could select. Therefore, the probability of selecting a tamarind candy from the second bag is 3/20.
To find the probability of both events happening, we multiply the individual probabilities:
Probability = (7/20) * (3/20) = 21/400 ≈ 0.0525
Therefore, the probability that Julio will select a strawberry candy from the first bag and a tamarind candy from the second bag the next time he takes a candy from each bag is approximately 0.0525.
What's the answer in a fraction form
The probability in fraction form is 21/400.
Julia has two bags of gummy candies. one at a time he really pulls 8 pineapples, 5 limes, 7 strawberry candies out of the first bag and put them back each time. he also randomly pulled 11 mango, 3 tamarind and 6 watermelon candies out of the second bag and put them back each time.
based on this information what is the probability that Julia will select a strawberry candy from the first bag in a tamarind candy from the second bag the next time he takes a candy from each bag?
Answer in a fraction form
To find the probability, we need to determine the number of favorable outcomes and the total number of possible outcomes.
In the first bag, there are 7 strawberry candies out of a total of 8 + 5 + 7 = 20 candies that Julia could select for her first candy.
In the second bag, there are 3 tamarind candies out of a total of 11 + 3 + 6 = 20 candies that Julia could select for her second candy.
Therefore, the probability of Julia selecting a strawberry candy from the first bag and a tamarind candy from the second bag is (7/20) * (3/20) = 21/400.
So, the probability of Julia selecting a strawberry candy from the first bag and a tamarind candy from the second bag is 21/400.
A five question quiz is taken in which the first and second questions have four answer choices, the third and fourth questions have three answer choices, and the last question has five answer choices. if a student randomly marks an answer for each question what is the expected number of questions he will answer correctly.
A. 0.27
B. 1.37
C. 0.96
D. 1.00
To find the expected number of questions the student will answer correctly, we need to calculate the probability of getting each question correct and then sum them up.
For the first and second questions with four answer choices, the probability of randomly guessing the correct answer is 1/4 = 0.25. So, the expected number of correct answers for these questions is 2 * 0.25 = 0.50.
For the third and fourth questions with three answer choices, the probability of randomly guessing the correct answer is 1/3 = 0.33. So, the expected number of correct answers for these questions is 2 * 0.33 = 0.66.
For the fifth question with five answer choices, the probability of randomly guessing the correct answer is 1/5 = 0.20.
Therefore, the expected number of correct answers for the fifth question is 0.20.
Finally, we sum up the expected number of correct answers for each question:
Expected number of correct answers = 0.50 + 0.66 + 0.20 = 1.36
The closest option to the calculated expected number is B. 1.37.
on a quiz there are four multiple choice questions where three points each and two true false questions worth 1.8 each multiple choice question has five possibilities if a student randomly guesses on each question what is expected value of the student score on the test
A. 2.8
B. 1.8
C. 3.4
D. 5.4
To calculate the expected value of the student's score on the test, we need to calculate the probability of obtaining each possible score and then multiply it by the corresponding score.
For the multiple-choice questions, there are 4 questions worth 3 points each. Since there are 5 possibilities for each question, the probability of randomly guessing the correct answer is 1/5 = 0.20. Therefore, the expected value for the multiple-choice questions is 4 * (0.20 * 3) = 2.4.
For the true/false questions, there are 2 questions worth 1.8 points each. Since there are 2 possibilities for each question, the probability of randomly guessing the correct answer is 1/2 = 0.50. Therefore, the expected value for the true/false questions is 2 * (0.50 * 1.8) = 1.8.
Finally, we sum up the expected values for each section:
Expected value of the student's score = 2.4 + 1.8 = 4.2
The closest option to the calculated expected value is D. 5.4.
Compare the rules of two games.
Game A: 2 6-sided dice are thrown the numbers on the dice determine which player gets a point player one gets a point if the product of the numbers on the dice is even player two gets a point if the product is odd.
Game B: two six-sided dice are thrown the numbers on the dice determine which player gets a point player one gets a point if the sum of the numbers on the dice is even player 2 gets a point at the sum is odd.
Which statement accurately describes whether these games are fair?
A. Game a is not fair and game B is fair
B. Neither game a nor game B is fair
C. Both game a and game B are fair
D. Game a is fair and game B is not fair
To determine if the games are fair, we need to assess if the probabilities of each outcome are equal.
Game A:
In this game, player one gets a point if the product of the numbers on the dice is even, and player two gets a point if the product is odd. For the product to be even, we need at least one even number on the dice. Out of the 36 possible outcomes (6 choices for the first die * 6 choices for the second die), there are 18 outcomes where the product is even and 18 outcomes where the product is odd. Therefore, the probabilities of each outcome are equal, and Game A is fair.
Game B:
In this game, player one gets a point if the sum of the numbers on the dice is even, and player two gets a point if the sum is odd. For the sum to be even, we need both numbers on the dice to be either even or odd. Out of the 36 possible outcomes, there are 18 outcomes where the sum is even and 18 outcomes where the sum is odd. Therefore, the probabilities of each outcome are equal, and Game B is also fair.
Thus, the statement that accurately describes whether these games are fair is C. Both Game A and Game B are fair.
A school fundraiser sells $1,200 raffle tickets each ticket cost $2 there is one grand prize worth $100 and 5 smaller prizes worth $20 each what is the expected gain or loss for buying one ticket
A. Loss of $1.91
B. loss of $1.99
C. gain of $2.16
D. loss of $1.83
To calculate the expected gain or loss for buying one ticket, we need to compare the potential outcomes with their respective probabilities.
The total cost of one raffle ticket is $2.
The probability of winning the grand prize worth $100 is 1/1200 since there is only one grand prize among the 1200 tickets.
The probability of winning one of the five smaller prizes worth $20 each is 5/1200 since there are five smaller prizes among the 1200 tickets.
Therefore, the expected gain from winning the grand prize is (1/1200) * $100 = $0.0833.
The expected gain or loss from winning a smaller prize is (5/1200) * $20 = $0.0833.
The expected loss from not winning any prize is 1 - (1/1200) - (5/1200) = 1184/1200.
To calculate the expected gain or loss, we subtract the expected loss from the expected gains:
Expected gain/loss = $0.0833 + $0.0833 - ($2 * (1184/1200)) = -$1.91
Therefore, the expected gain or loss for buying one ticket is a loss of $1.91, which corresponds to option A. Loss of $1.91.
Shasta is going to toss two number cubes each labeled 1 through 6 what is the probability that Shasta will toss a sum of six
A. 1/6
B. 1/9
C. 1/12
D. 5/36
To find the probability of Shasta tossing a sum of six with two number cubes, we need to determine the number of favorable outcomes and the total number of possible outcomes.
When tossing two number cubes, there are 36 possible outcomes, as there are 6 possible outcomes for each cube (6 sides on each cube).
To calculate the favorable outcomes, we list all the possible pairs of numbers on the cubes that sum up to six: (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1). Therefore, there are 5 favorable outcomes.
The probability is then determined by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = 5/36
Hence, the probability that Shasta will toss a sum of six is 5/36, which corresponds to option D. 5/36.
Sean wants to invest a gift of $4,250 in the stock market but broker suggests two companies solar solutions and Frontline medical he provides the possibilities for an annual return on $4,250 based on his their historical gains solar solutions 50% chance of a thousand gain 35% chance of $200 gain and 15% chance of $100 loss Frontline medical 80% chance of a $750 gain 20% chance of $100 loss which is the best analysis of these investments?
A solar solutions is the better investment because it's expected annual return is $950 greater than Frontline medical
B. Frontline medical is a better investment because it's expected annual return is $580 greater than solar solutions
C. Frontline medical has the higher risk of loss but is likely the better investment over time because the annual expected return is $100 greater than solar solutions
D. solar solutions is a better investment because it has a lower risk of loss and its Annual return is $80 dollars greater than Frontline medical.
To determine which investment is better, we need to calculate the expected annual return for each company and assess the risk involved.
For Solar Solutions:
- 50% chance of a $1000 gain
- 35% chance of a $200 gain
- 15% chance of a $100 loss
Expected annual return for Solar Solutions = (0.50 * $1000) + (0.35 * $200) + (0.15 * -$100) = $500 + $70 - $15 = $555
For Frontline Medical:
- 80% chance of a $750 gain
- 20% chance of a $100 loss
Expected annual return for Frontline Medical = (0.80 * $750) + (0.20 * -$100) = $600 + (-$20) = $580
Based on the expected annual return, Solar Solutions has a higher expected return of $555 compared to Frontline Medical's expected return of $580.
Additionally, Solar Solutions has a lower risk of loss since it only has a 15% chance of a $100 loss compared to Frontline Medical's 20% chance of a $100 loss.
Therefore, the best analysis of these investments is D. Solar Solutions is a better investment because it has a lower risk of loss and its annual return is $80 greater than Frontline Medical.