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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
1 Withdraw 1 green out of 5 green jellybeans and 4 red out of 7 red jellybeans as follows: \[ \left(\begin{array}{l} 5 \\ 1 \end{array}\right) *\left(\begin{array}{l} 7 \\ 4 \end{array}\right)=\frac{5 !}{1 !(5-1) !} * \frac{7 !}{4 !(7-4) !}=175 \] 2 Withdraw 0 green jellybeans and 5 reds out of 7 red jellybeans as follows: \[ \left(\begin{array}{l} 7 \\ 5 \end{array}\right)=\frac{7 !}{5 !(7-5) !}=21 \] Therefore, the total number of ways is \(175+21=196\)
# Choosing 1 green out of 5 green and 4 reds out of 7 reds
# Choosing 0 green out of 5 and 5 reds out of 7 reds
print ("Total Number of ways of 5 jellybeans taken out from the bag are: ") ## [1] "Total Number of ways of 5 jellybeans taken out from the bag are: "choose(5,1) * choose(7,4) + choose(5,0) * choose(7,5)## [1] 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?
1 Choosing 4 representatives out of 13 and 1 senator out of 14 as follows: \[ \left(\begin{array}{l} 13 \\ 4 \end{array}\right) *\left(\begin{array}{l} 14 \\ 1 \end{array}\right)=\frac{13 !}{4 !(13-4) !} * \frac{14 !}{1 !(14-1) !}=10,010 \] 2 Choosing all 5 representatives out of 13 as follows: \[ \left(\begin{array}{l} 13 \\ 5 \end{array}\right)=\frac{13 !}{5 !(13-5) !}=1287 \]
print("Total Number of ways a subcommittee of 5 can be formed if at least 4 of the members must be representatives are:")## [1] "Total Number of ways a subcommittee of 5 can be formed if at least 4 of the members must be representatives are:"choose(13,4)*choose(14,1) + choose(13,5)## [1] 11297If 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<- 2^5 # 2 possibilities (H or T) for each of 5 tosses
die<- 6^2 # 6 possibilities (1 to 6) for each of 2 rolls
card <- choose(52, 3) #Number of ways to choose 3 cards out of 52 without replacement
# The total number of different outcomes
total <- coin * die * card
cat("Total outcomes for all events combined is :", total)## Total outcomes for all events combined is : 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.
The probability that at least one is a 3 can be calculated by taking the complement of the probability of no card being a 3.
Probability of not drawing a 3 in the first draw: 48/52.
Probability of not drawing a 3 in the second draw, after not drawing a 3 in the first draw: 47/51.
Probability of not drawing a 3 in the third draw, after not drawing a 3 in the first two draws: 46/50.
#Probability of drawing at least one 3 from a standard deck when drawing 3 cards without replacement
three <- (48/52) * (47/51) * (46/50)
#Taking the complement of the probability of no card being a 3
round(1-three, digits = 4) ## [1] 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.
The combination formula \(\mathrm{C}\) is \[ C(n, k)=n ! /(k !(n-k) !) \] \[ \begin{aligned} & \mathrm{n}=31 \text { movies } \\ & \mathrm{k}=5 \text { movies } \end{aligned} \]
total_movies <- 31
movies_to_select <- 5
documentaries <- 17
# Calculate the total number of combinations
C <- choose(total_movies, movies_to_select)
cat("Number of different combinations of 5 movies Lorenzo can rent:", C, "\n")## Number of different combinations of 5 movies Lorenzo can rent: 169911# Combinations without any mysteries
No_mysteries <- choose(documentaries, movies_to_select)
# The number of combinations with at least one mystery
One_mystery <- C - No_mysteries
cat("Number of different combinations of 5 movies with at least one mystery Lorenzo can rent:", One_mystery, "\n")## Number of different combinations of 5 movies with at least one mystery Lorenzo can rent: 163723Step 1. How many different combinations of 5 movies can he rent? Answer: 169911
Step 2. How many different combinations of 5 movies can he rent if he wants at least one mystery? Answer:163723
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.
If he is setting up a schedule of the 9 symphonies to be played, there will be 3 symphonies of each composer
total_symphonies <- 9
symphonies_to_play <- 3
brahms <- 4
haydn <- 104
mendelssohn <- 17
# Scientific notation rounded to the hundredths place
result <- format((choose(4,3)*choose(104,3)*choose(17,3)), scientific = TRUE, digits=3)
# Print the result
cat("Number of different possible schedules:", result, "\n")## Number of different possible schedules: 4.95e+08Answer: Number of different possible schedules: 4.95e+08
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_books <- 13
novels <- 6
plays <- 6
poetry <- 7
nonfiction <- 5
# Choose 4 nonfiction books out of 5 and 9 other books out of 19 (6 novels, 6 plays, 7 poetry books)
C4 <- choose(5,4)*choose(19,9)
# Choose 3 nonfiction books out of 5 and 10 other books out of 19 (6 novels, 6 plays, 7 poetry books)
C3 <- choose(5,3)*choose(19,10)
# Choose 2 nonfiction books out of 5 and 11 other books out of 19 (6 novels, 6 plays, 7 poetry books)
C2 <- choose(5,2)*choose(19,11)
# Choose 1 nonfiction books out of 5 and 12 other books out of 19 (6 novels, 6 plays, 7 poetry books)
C1 <- choose(5,1)*choose(19,12)
# Choose 0 nonfiction books out of 5 and 13 other books out of 19 (6 novels, 6 plays, 7 poetry books)
C0 <- choose(5,0)*choose(19,13)
# Print the result
result <- format(sum(C4,C3, C2, C1, C0), scientific = TRUE, digits = 3 )
cat("Number of different possible reading schedules:", result, "\n")## Number of different possible reading schedules: 2.42e+06Step 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 6 plays out of 6 and 7 other books out of 18 (6 novels, 7 poetry, 5 notification books)
Cp <- choose(6,6)*choose(18,7)
result <- format(Cp, scientific = TRUE, digits = 2)
# Print the result
cat("Number of different possible reading schedules with all 6 plays included:", result, "\n")## Number of different possible reading schedules with all 6 plays included: 3.2e+04Zane 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.
#There are total 10 trees (5 sycamores and 5 cypress trees) and 2 ways to arrange them (sycamores first or cypress first)
# Calculate the probability of total amount of ways to plant 10 trees
probability <- 2 / choose(10,5)
# Round the probability to four decimal places
probability_rounded <- round(probability, digits = 4)
# Print the probability
cat("Probability:", probability_rounded, "\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.
# There are 52 cards, 44 of them are a queen or lower, 8 of them are above queen.
probability_winning <- 44 / 52
probability_losing <- 8 / 52
pay_winning <- 4
pay_losing <- -16
expected_value <- (4 * (44/52) - (16 * (8/52)) )
expected_value_rounded <- round(expected_value, digits = 4)
cat("Expected Value:", expected_value_rounded, "\n")## Expected Value: 0.9231Step 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.
played_times <- 833
# Calculate the expected total winnings/loss
Expectation <- round(expected_value_rounded * played_times, digits = 2)
cat("Expected Amount of Total Winnings/Losses:", Expectation, "\n")## Expected Amount of Total Winnings/Losses: 768.94