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