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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
Answer: To find the number of ways to withdraw 5 jellybeans from the bag such that the number of green ones is less than 2, we need to consider two cases:
No green jellybeans are selected. Exactly 1 green jellybean is selected. We will calculate the number of ways for each case and then sum them to get the total number of ways.
Let’s calculate it in R:
# Given values
total_green <- 5
total_red <- 7
total_jellybeans <- total_green + total_red
jellybeans_to_withdraw <- 5
# Case 1: No green jellybeans selected
ways_case_1 <- choose(total_red, jellybeans_to_withdraw)
# Case 2: Exactly 1 green jellybean selected
ways_case_2 <- choose(total_green, 1) * choose(total_red, jellybeans_to_withdraw - 1)
# Total ways
total_ways <- ways_case_1 + ways_case_2
# Display the results
cat("Ways for Case 1 (No green jellybeans selected):", ways_case_1, "\n")## Ways for Case 1 (No green jellybeans selected): 21cat("Ways for Case 2 (Exactly 1 green jellybean selected):", ways_case_2, "\n")## Ways for Case 2 (Exactly 1 green jellybean selected): 175cat("Total Ways:", total_ways, "\n")## Total Ways: 196A 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?
Answer: To solve this problem, we can use combinations.The number of ways to choose a subcommittee of 5 members from a group of 27 members (14 senators and 13 representatives) is given by the combination formula:
C(N, n) = N!/(N - n)!
where:N is the total number of members, n is the number of members to be chosen for the subcommittee,
and ! denotes the factorial.
In this scenario, we want to find the number of ways to choose a subcommittee of 5 members where at least 4 members are representatives. This can be done by considering two cases:
Exactly 4 representatives and 1 senator.
All 5 members are representatives.
We then sum the results from these two cases to get the total number of ways.
Let’s solve this in R:
# Given values
total_senators <- 14
total_representatives <- 13
total_members <- total_senators + total_representatives
subcommittee_size <- 5
# Case 1: Exactly 4 representatives and 1 senator
ways_case_1 <- choose(total_representatives, 4) * choose(total_senators, 1)
# Case 2: All 5 members are representatives
ways_case_2 <- choose(total_representatives, 5)
# Total ways
total_ways <- ways_case_1 + ways_case_2
# Display the results
cat("Ways for Case 1:", ways_case_1, "\n")## Ways for Case 1: 10010cat("Ways for Case 2:", ways_case_2, "\n")## Ways for Case 2: 1287cat("Total Ways:", total_ways, "\n")## Total Ways: 11297If a coin is tossed 5 times, and then a standard six-sided die rolled 2 times, and finally group of three cards are drawn from a standard deck of 52 cards without replacement, how many different outcomes are possibles?
Answers: To find the total number of different outcomes, one can use the multiplication principle or product rule, which states that if there are n1n1 ways to do the first task, and n2n2 ways to do the second task, and so on, then there are n1×n2×…n1×n2×… ways to do all tasks.In this scenario:Tossing a coin 5 times has 2525 outcomes.Rolling a six-sided die 2 times has 6262 outcomes.Drawing a group of three cards from a standard deck of 52 cards without replacement has (523)(352) outcomes.The total number of different outcomes is the product of these individual outcomes:25×62×(523)25×62×(352)
Let’s calculate this in R:
# Coin toss outcomes
coin_outcomes <- 2^5
# Die roll outcomes
die_outcomes <- 6^2
# Card drawing outcomes
card_outcomes <- choose(52, 3)
# Total outcomes
total_outcomes <- coin_outcomes * die_outcomes * card_outcomes
# Display the result
cat("Total different outcomes:", total_outcomes, "\n")## Total different outcomes: 254592003 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.
Answer:
To find the probability that at least one of the cards drawn is a 3, we can use the complement rule. The complement of “at least one 3” is “none of the cards are 3s.” We calculate this probability and subtract it from 1.The probability of drawing a card that is not a 3 on the first draw is 48525248 (since there are 4 cards with a value of 3 in a standard deck of 52 cards). On the second draw, the probability is 47515147, and on the third draw, it is 46505046.The overall probability of not getting a 3 on any of the draws is the product of these probabilities:P(No 3s)=4852×4751×4650P(No 3s)=5248×5147×5046The probability of getting at least one 3 is then given by the complement:P(At least one 3)=1−P(No 3s)P(At least one 3)=1−P(No 3s)
# Probability of not getting a 3 on any draw
prob_no_3s <- (48/52) * (47/51) * (46/50)
# Probability of at least one 3
prob_at_least_one_3 <- 1 - prob_no_3s
# Display the result
cat("Probability of at least one 3:", round(prob_at_least_one_3, 4), "\n")## Probability of at least one 3: 0.2174Lorenzo 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.
# Given values
documentaries <- 17
mysteries <- 14
total_movies <- documentaries + mysteries
movies_to_rent <- 5
# Number of different combinations of 5 movies
combinations_step_1 <- choose(total_movies, movies_to_rent)
# Display the result
cat("Step 1: Different combinations of 5 movies:", combinations_step_1, "\n")## Step 1: Different combinations of 5 movies: 169911Answer:
# Combinations with no mysteries
combinations_no_mystery <- choose(documentaries, movies_to_rent)
# Number of different combinations of 5 movies with at least one mystery
combinations_step_2 <- combinations_step_1 - combinations_no_mystery
# Display the result
cat("Step 2: Different combinations of 5 movies with at least one mystery:", combinations_step_2, "\n")## Step 2: Different combinations of 5 movies with at least one mystery: 163723In 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.
**Answers: **
To find the number of different schedules Cory can create, we can use
the multinomial coefficient. The multinomial coefficient generalizes the
concept of combinations and is calculated
as:C(n;n1,n2,…,nk)=n!n1!⋅n2!⋅…⋅nk!C(n;n1,n2,…,nk)=n1!⋅n2!⋅…⋅nk!n!where
nn is the total number of items, and n1,n2,…,nkn1,n2,…,nk are the counts
of each distinct type of item.In this case, Cory needs to choose 9
symphonies out of a total of 4+104+17=1254+104+17=125 symphonies, with
the constraint of having an equal number of symphonies from Brahms,
Haydn, and Mendelssohn.Let’s calculate this in R:
are the counts of each distinct type of item.
In this case, Cory needs to choose 9 symphonies out of a total of 4 + 104 + 17 = 125 4+104+17=125 symphonies, with the constraint of having an equal number of symphonies from Brahms, Haydn, and Mendelssohn.
Let’s calculate this in R:
library(gtools)
# Given values
brahms_symphonies <- 4
haydn_symphonies <- 104
mendelssohn_symphonies <- 17
total_symphonies <- brahms_symphonies + haydn_symphonies + mendelssohn_symphonies
symphonies_to_choose <- 9
# Multinomial coefficient for choosing 9 symphonies with equal counts from each composer
multinomial_coefficient <- choose(total_symphonies, symphonies_to_choose) * choose(symphonies_to_choose, c(brahms_symphonies, haydn_symphonies, mendelssohn_symphonies))
# Display the result in scientific notation rounded to the hundredths place
cat("Number of different schedules in scientific notation:", format(multinomial_coefficient, scientific = TRUE, digits = 2), "\n")## Number of different schedules in scientific notation: 1.9e+15 0.0e+00 0.0e+00An 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. Answer:
# Given values
novels <- 6
plays <- 6
poetry <- 7
nonfiction <- 5
total_books <- novels + plays + poetry + nonfiction
books_to_choose <- 13
# Multinomial coefficient for choosing 13 books with no more than 4 nonfiction books
multinomial_coefficient_step_1 <- choose(total_books, books_to_choose) * choose(books_to_choose, c(novels, plays, poetry, min(4, nonfiction)))
# Display the result in scientific notation rounded to the hundredths place
result_step_1 <- format(multinomial_coefficient_step_1, scientific = TRUE, digits = 2)
cat("Step 1: Different reading schedules with no more than 4 nonfiction books:", result_step_1, "\n")## Step 1: Different reading schedules with no more than 4 nonfiction books: 4.3e+09 4.3e+09 4.3e+09 1.8e+09Step 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. Answer:
# Multinomial coefficient for choosing 13 books with all 6 plays
multinomial_coefficient_step_2 <- choose(total_books, books_to_choose) * choose(books_to_choose - plays, c(novels, plays, poetry, nonfiction))
# Display the result in scientific notation rounded to the hundredths place
result_step_2 <- format(multinomial_coefficient_step_2, scientific = TRUE, digits = 2)
cat("Step 2: Different reading schedules with all 6 plays:", result_step_2, "\n")## Step 2: Different reading schedules with all 6 plays: 1.7e+07 1.7e+07 2.5e+06 5.2e+07Zane 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.
Answers:
To calculate the probability that all 5 sycamores are next to each other and all 5 cypress trees are next to each other, we can treat each set of sycamores and cypress trees as single entities.There are two possible arrangements: sycamores followed by cypress trees or cypress trees followed by sycamores.Let’s calculate the probability using the total number of ways to arrange the 10 trees:Total arrangements=10!Total arrangements=10!Now, considering that each set of 5 trees is treated as a single entity, there are 5!5! ways to arrange the sycamores within the sycamore set and 5!5! ways to arrange the cypress trees within the cypress set.So, the probability is given by:P(Desired arrangement)=2×5!×5!10!P(Desired arrangement)=10!2×5!×5!
# Calculate the probability
probability <- (2 * factorial(5) * factorial(5)) / factorial(10)
# Display the result as a decimal rounded to four decimal places
cat("Probability:", round(probability, 4), "\n")## Probability: 0.0079If 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.
To calculate the expected value of the proposition, we need to determine the probability of drawing a queen or lower and the corresponding payoffs.Step 1:Let’s calculate the probability of drawing a queen or lower. There are 4 suits in a standard deck, and each suit has one queen. Additionally, each suit has three cards lower than the queen (10, 9, 8). So, the total number of favorable outcomes is 4×4=164×4=16 (four queens and three cards lower than the queen in each suit).The total number of possible outcomes in a standard deck is 52 cards.The probability of drawing a queen or lower is given by:P(Queen or lower)=Numbe of favorabl outcomes Total number of possible outcomesP(Queen or lower)=Total number of possible outcomesNumber of favorable outcomesStep 2:The expected value (EVEV) is then given by:EV=P(Queen or lower)×Payoff if successful+P(Not Queen or lower)×Payoff if unsuccessfulEV=P(Queen or lower)×Payoff if successful+P(Not Queen or lower)×Payoff if unsuccessful
# Given values
payoff_successful <- 4
payoff_unsuccessful <- -16
# Probability of drawing a queen or lower
probability_queen_or_lower <- 16 / 52
# Expected value calculation
expected_value <- probability_queen_or_lower * payoff_successful + (1 - probability_queen_or_lower) * payoff_unsuccessful
# Display the result rounded to two decimal places
cat("Step 1: Expected value of the proposition: $", round(expected_value, 2), "\n")## Step 1: Expected value of the proposition: $ -9.85Step 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.
Answer:
For Step 2, if you played the game 833 times, the expected total winnings or losses would be:Total Expected Value=Number of times played×Expected ValueTotal Expected Value=Number of times played×Expected Value
# Given value for Step 2
number_of_times_played <- 833
# Total expected value calculation
total_expected_value <- number_of_times_played * expected_value
# Display the result rounded to two decimal places
cat("Step 2: Total expected winnings or losses after playing 833 times: $", round(total_expected_value, 2), "\n")## Step 2: Total expected winnings or losses after playing 833 times: $ -8201.85