1. A bag contains 5 green and 7 red jellybeans. How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
# Total jellybeans of each color
green_jellybeans <- 5
red_jellybeans <- 7
# Scenario 1: No green jellybeans
ways_no_green <- choose(red_jellybeans, 5)
# Scenario 2: 1 green jellybean
ways_one_green <- choose(green_jellybeans, 1) * choose(red_jellybeans, 4)
# Total ways
total_ways <- ways_no_green + ways_one_green
cat("Total number of ways to withdraw 5 jellybeans with less than 2 green ones:", total_ways)## Total number of ways to withdraw 5 jellybeans with less than 2 green ones: 1962. A certain congressional committee consists of 14 senators and 13 representatives. How many ways can a subcommittee of 5 be formed if at least 4 of the members must be representatives?
# Total senators and representatives
senators <- 14
representatives <- 13
# Scenario 1: 4 representatives and 1 senator
ways_4_rep_1_sen <- choose(representatives, 4) * choose(senators, 1)
# Scenario 2: All 5 members are representatives
ways_5_rep <- choose(representatives, 5)
# Total ways to form the subcommittee
total_ways <- ways_4_rep_1_sen + ways_5_rep
cat("Total number of ways to form a subcommittee with at least 4 representatives:", total_ways)## Total number of ways to form a subcommittee with at least 4 representatives: 112973. If a coin is tossed 5 times, and then a standard six-sided die is rolled 2 times, and finally a group of three cards are drawn from a standard deck of 52 cards without replacement, how many different outcomes are possible?
# Coin tosses
coin_outcomes <- 2^5
# Die rolls
die_outcomes <- 6^2
# Card draws
card_outcomes <- choose(52, 3)
# Total outcomes
total_outcomes <- coin_outcomes * die_outcomes * card_outcomes
cat("Total number of different outcomes:", total_outcomes)## Total number of different outcomes: 254592004.3 cards are drawn from a standard deck without replacement. What is the probability that at least one of the cards drawn is a 3? Express your answer as a fraction or a decimal number rounded to four decimal places.
# Probability of not drawing a '3' in each draw
p_not_3_first_draw <- 48 / 52
p_not_3_second_draw <- 47 / 51
p_not_3_third_draw <- 46 / 50
# Probability of not drawing a '3' in any of the 3 draws
p_no_3s <- p_not_3_first_draw * p_not_3_second_draw * p_not_3_third_draw
# Rounding the result to four decimal places
p_at_least_one_3_rounded <- round(1 - p_no_3s, 4)
# Output the result
p_at_least_one_3_rounded## [1] 0.21745.Lorenzo is picking out some movies to rent, and he is
primarily interested in documentaries and mysteries. He has narrowed
down his selections to 17 documentaries and 14 mysteries.
Step 1. How many different combinations of 5 movies can he
rent?
# Total movies
total_movies <- 17 + 14
# Combinations of 5 movies from the total
total_combinations <- choose(total_movies, 5)
cat("Total combinations of 5 movies:", total_combinations)## Total combinations of 5 movies: 169911Step 2.How many different combinations of 5 movies can he rent if he wants at least one mystery?
# Combinations of 5 documentaries (no mysteries)
doc_combinations <- choose(17, 5)
# Combinations with at least one mystery
combinations_with_mystery <- total_combinations - doc_combinations
cat("Combinations of 5 movies with at least one mystery:", combinations_with_mystery)## Combinations of 5 movies with at least one mystery: 1637236. In choosing what music to play at a charity fund raising event, Cory needs to have an equal number of symphonies from Brahms, Haydn, and Mendelssohn. If he is setting up a schedule of the 9 symphonies to be played, and he has 4 Brahms, 104 Haydn, and 17 Mendelssohn symphonies from which to choose, how many different schedules are possible? Express your answer in scientific notation rounding to the hundredths place
# Combinations of choosing symphonies
brahms_combinations <- choose(4, 3)
haydn_combinations <- choose(104, 3)
mendelssohn_combinations <- choose(17, 3)
# Total possible schedules
total_schedules <- brahms_combinations * haydn_combinations * mendelssohn_combinations * factorial(9)
# Expressing the result in scientific notation
total_schedules_sci <- formatC(total_schedules, format = "e", digits = 2)
cat("Total possible schedules: ", total_schedules_sci)## Total possible schedules: 1.80e+147.An English teacher needs to pick 13 books to put on his
reading list for the next school year, and he needs to plan the order in
which they should be read. He has narrowed down his choices to 6 novels,
6 plays, 7 poetry books, and 5 nonfiction books.
Step 1. If he wants to include no more than 4 nonfiction books, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
total_schedules_step_1_sci <- (choose(24,13)-choose(19,8)*choose(5,5))*factorial(13)
cat("Total reading schedules for Step 1: ", total_schedules_step_1_sci)## Total reading schedules for Step 1: 1.507289e+16
Step 2. If he wants to include all 6 plays, how many different reading schedules are possible? Express your answer in scientific notation rounding to the hundredths place.
# Choose 7 books from the remaining 18 (excluding the 6 mandatory plays)
# Total reading schedules including permutations
total_schedules_step_2 <- choose(18, 7) * factorial(13)
# Express the result in scientific notation
total_schedules_step_2_sci <- formatC(total_schedules_step_2, format = "e", digits = 2)
cat("Total reading schedules for Step 2: ", total_schedules_step_2_sci)## Total reading schedules for Step 2: 1.98e+148.Zane is planting trees along his driveway, and he has 5 sycamores and 5 cypress trees to plant in one row. What is the probability that he randomly plants the trees so that all 5 sycamores are next to each other and all 5 cypress trees are next to each other?Express your answer as a fraction or a decimal number rounded to four decimal places.
#When we consider planting all 5 sycamores together and all 5 cypress trees together, we can think of each group of 5 trees as a single unit. Thus, we have 2 units to arrange, which can be done in 2! ways (either sycamores first or cypress first).
# Total ways to plant the trees
total_ways <- factorial(10) / (factorial(5) * factorial(5))
# Ways with sycamores and cypress together
favorable_ways <- factorial(2)
# Probability calculation
rounded_probability <- round(favorable_ways / total_ways,4)
rounded_probability## [1] 0.00799.If you draw a queen or lower from a standard deck of cards, I will pay you $4. If not, you pay me $16. (Aces are considered the highest card in the deck.) Step 1. Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.
# Step 1: Calculate the expected value per game
probability_win <- 44 / 52
probability_lose <- 8 / 52
expected_value <- probability_win * 4 + probability_lose * (-16)
# Step 2: Calculate the expected total from 833 games
expected_total <- expected_value * 833
# Rounding the results
expected_value_rounded <- round(expected_value, 2)
expected_total_rounded <- round(expected_total, 2)
# Displaying the results
cat("Expected value per game: $", expected_value_rounded)## Expected value per game: $ 0.92Step 2. If you played this game 833 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be expressed as negative values.
cat("Expected total from 833 games: $", expected_total_rounded)## Expected total from 833 games: $ 768.92