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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?
A breakdown of the different cases:
0 green jellybeans and 5 red jellybeans. 1 green jellybean and 4 red jellybeans.
Number of ways to choose 0 green jellybeans from 5: \(\binom{5}{0} = 1\)
Number of ways to choose 5 red jellybeans from 7: \(\binom{7}{5} = 21\)
Total ways for this case: \(1 \times 21 = 21\)
Number of ways to choose 1 green jellybean from 5: \(\binom{5}{1} = 5\)
Number of ways to choose 4 red jellybeans from 7: \(\binom{7}{4} = 35\)
Total ways for this case: \(5 \times 35 = 175\)
The total number of ways to withdraw 5 jellybeans from the bag with the given condition is \(21 + 175 = 196\).
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?
Cases: 4 representatives and 1 senator. 5 members as representatives.
Number of ways to choose 4 representatives from 13: \(\binom{13}{4}\)
Number of ways to choose 1 senator from 14: \(\binom{14}{1}\)
Total ways for this case: \(\binom{13}{4} \times \binom{14}{1}\)
Number of ways to choose 5 representatives from 13: \(\binom{13}{5}\)
The total number of ways to form a subcommittee with at least 4 representatives is the sum of the ways:
\[ \text{Total Ways} = \binom{13}{4} \times \binom{14}{1} + \binom{13}{5} = 122,694 \]
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 is drawn from a standard deck of 52 cards without replacement, how many different outcomes are possible?
To find the total number of different outcomes, we calculate the product of the possible outcomes for each event:
Each coin toss has 2 possible outcomes (heads or tails). Each die roll has 6 possible outcomes. For the first card, there are 52 options, for the second card 51 options (without replacement), and for the third card 50 options.
The total number of outcomes:
\[ \text{Total Outcomes} = 2^5 \times 6^2 \times \binom{52}{3} \]
binomial_coefficient <- function(n, k) {
factorial(n) / (factorial(k) * factorial(n - k))
}
total_outcomes <- 2^5 * 6^2 * factorial(52) / (factorial(3) * factorial(52 - 3))
total_outcomes## [1] 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.
It’s easier to use the complement of the probability here.
\[ P(\text{No 3's}) = \frac{49}{52} \times \frac{48}{51} \times \frac{47}{50} \]
\[ P(\text{At Least One 3}) = 1 - P(\text{No 3's}) \]
prob_no_3 <- (49/52) * (48/51) * (47/50)
prob_at_least_one_3 <- 1 - prob_no_3
prob_at_least_one_3_rounded <- round(prob_at_least_one_3, 4)
prob_at_least_one_3_rounded## [1] 0.1663Lorenzo 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.
How many different combinations of 5 movies can he rent?
Combinations of 5 documentaries:
\[ \binom{17}{5} = \frac{17!}{5! \cdot (17 - 5)!} \]
Combinations of 5 mysteries:
\[ \binom{14}{5} = \frac{14!}{5! \cdot (14 - 5)!} \]
\[ \text{Total Combinations} = \binom{17}{5} \times \binom{14}{5} \]
binomial_coefficient <- function(n, k) {
factorial(n) / (factorial(k) * factorial(n - k))
}
combinations_documentaries <- binomial_coefficient(17, 5)
combinations_mysteries <- binomial_coefficient(14, 5)
total_combinations <- combinations_documentaries * combinations_mysteries
total_combinations## [1] 12388376How many different combinations of 5 movies can he rent if he wants at least one mystery?
The total number of combinations of 5 movies from all available movies:
\[ \text{Total Combinations} = \binom{17}{5} \times \binom{14}{0} + \binom{17}{4} \times \binom{14}{1} + \binom{17}{3} \times \binom{14}{2} + \binom{17}{2} \times \binom{14}{3} + \binom{17}{1} \times \binom{14}{4} + \binom{17}{0} \times \binom{14}{5} \]
Combinations of 5 movies with 0 mysteries:
\[ \binom{17}{5} \times \binom{14}{0} \]
Calculate the difference: \[ \text{Combinations with at Least One Mystery} = \text{Total Combinations} - \text{Combinations with No Mysteries} \]
combinations_no_mysteries <- binomial_coefficient(17, 5) * binomial_coefficient(14, 0)
total_combinations <- binomial_coefficient(17, 5) * binomial_coefficient(14, 0) +
binomial_coefficient(17, 4) * binomial_coefficient(14, 1) +
binomial_coefficient(17, 3) * binomial_coefficient(14, 2) +
binomial_coefficient(17, 2) * binomial_coefficient(14, 3) +
binomial_coefficient(17, 1) * binomial_coefficient(14, 4) +
binomial_coefficient(17, 0) * binomial_coefficient(14, 5)
combinations_at_least_one_mystery <- total_combinations - combinations_no_mysteries
combinations_at_least_one_mystery## [1] 163723In choosing what music to play at a charity fundraising 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.
\[ \text{Multinomial Coefficient} = \frac{(4 + 104 + 17)!}{4! \cdot 104! \cdot 17!} \]
schedules <- factorial(4 + 104 + 17) / (factorial(4) * factorial(104) * factorial(17))
schedules_rounded <- round(schedules, 2)
schedules_rounded## [1] 2.141413e+27\[ 4.71×10^{176} \]
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.
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.
\[ \text{Multinomial Coefficient} = \frac{(6 + 6 + 7 + 4)!}{6! \cdot 6! \cdot 7! \cdot 4!} \]
multinomial_coefficient <- function(a, b, c, d) {
factorial(a + b + c + d) / (factorial(a) * factorial(b) * factorial(c) * factorial(d))
}
schedules <- multinomial_coefficient(6, 6, 7, 4)
schedules## [1] 412275623760\[ 4.12×10^{12} \]
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.
Since the English teacher wants to include all 6 plays in the reading schedule, we only need to consider the arrangements of the remaining 7 books (6 novels, 7 poetry books, and 5 nonfiction books).
To find the number of different reading schedules for these 7 books, we calculate the multinomial coefficient:
\[ \text{Multinomial Coefficient} = \frac{(6 + 7 + 5)!}{6! \cdot 7! \cdot 5!} \]
schedules_all_plays <- factorial(6 + 7 + 5) / (factorial(6) * factorial(7) * factorial(5))
schedules_all_plays## [1] 14702688\[ 1.47×10^{7} \]
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.
The total number of ways to arrange the 10 trees is \(10!\)
There are \(5!\) ways to arrange 5 sycamore trees, and \(5!\) to arrange 5 cypress trees. Then you have the choice of putting either the sycamore trees first, or the cypress trees first, so multiply by 2.
The probability is:
\[ P = \frac{2 \times 5! \times 5!}{10!} \]
probability <- (2 * factorial(5) * factorial(5)) / factorial(10)
probability_rounded <- round(probability, 4)
probability_rounded## [1] 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.) Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.
There are four cards lower than the queen in each suit (10, 9, 8, 7). There are a total of \(4 + 4 \times 4 = 20\) cards that can be drawn to win.
The probability of winning (\(P(\text{win})\)) is the number of winning outcomes divided by the total number of outcomes, which is 52 (the total number of cards in the deck).
The probability of losing (\(P(\text{lose})\)) is \(1 - P(\text{win})\).
The expected value (\(E\)) is calculated as follows:
\[ E = P(\text{win}) \times \text{4} + P(\text{lose}) \times \text{-16} \]
prob_win <- 20/52
prob_lose <- 1 - prob_win
expected_value <- prob_win * 4 + prob_lose * (-16)
expected_value_rounded <- round(expected_value, 2)
expected_value_rounded## [1] -8.31If 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.
Multiply the expected value for one play (\(E\)) by the number of plays (833):
\[ E_{\text{total}} = E \times \text{Number of Plays} \]
num_plays <- 833
total_expected_value <- expected_value * num_plays
total_expected_value_rounded <- round(total_expected_value, 2)
total_expected_value_rounded## [1] -6920.31