data_605_hw6

Question 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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?

Solution: The answer is 6. “Less than 2” means that it can be either 1 or 0 green jellybean chosen from the bag. If we are going to pick 5 out in random from the bag and there’s only 1 green jellybean, that green jellybean can be chosen in any order (or position), i.e. 1st, 2nd, 3rd, 4th or 5th. Therefore, we would have 5 combinations if there’s only 1 green jellybean chosen. In addition, we could have chosen 0 green jellybean. Together, we end up having 6 ways of satisfying the requirement.

Question 2

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?

sen <- 14
rep <- 13
subcom <- 5
scenarios <- c(4, 5)
tempList <- vector(mode = "list", length = length(scenarios))

for(i in 1:length(scenarios)){
    
    tempList[[i]] = choose(rep, scenarios[i]) * choose(sen, subcom - scenarios[i])    
    
}

solution = reduce(tempList, `+`)

print(paste0(glue("There are ", {solution}, " ways of chooseing at least 4 out of ", {rep}, " representatives and 1 or 0 out of ", {sen}, " senators to form a 5-member subcommittee")))

## [1] "There are 11297 ways of chooseing at least 4 out of 13 representatives and 1 or 0 out of 14 senators to form a 5-member subcommittee"

Solution: To form a 5-member subcommittee and have at least 4 representatives, that means we would have chosen either 4 or 5 representatives out of 13. There are only 2 scenarios, i.e. either 4 representatives and 1 senator or 5 representatives and 0 senators. We need to sum up the number of combinations for each of these two scenarios. The solution is 11297.

Question 3

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?

solution <- nrow(gtools::combinations(2, 5, repeats.allowed = TRUE)) *
nrow(gtools::combinations(6, 2, repeats.allowed = TRUE)) *
nrow(gtools::combinations(52, 3, repeats.allowed = TRUE))

print(glue("Assuming that order does not matter, e.g. when we flip a coin 3 times, Tail-Head-Tail and Tail-Tail-Head would be considered the same outcome, then there are ", {solution}, " different outcomes possible in this situation"))

## Assuming that order does not matter, e.g. when we flip a coin 3 times, Tail-Head-Tail and Tail-Tail-Head would be considered the same outcome, then there are 3125304 different outcomes possible in this situation

Question 4

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.

first_draw_not_three = 48 / 52
second_draw_not_three = 47 / 51
third_draw_not_three = 46 / 50

at_least_one_three = 1 - first_draw_not_three * second_draw_not_three * third_draw_not_three
print(glue("The probability that at least one of the cards drawn from a standard deck without replacement is a 3 is approximately ", round({at_least_one_three}, 4)))

## The probability that at least one of the cards drawn from a standard deck without replacement is a 3 is approximately 0.2174

Question 6

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.

b <- choose(4, 3)
h <- choose(104, 3)
m <- choose(17, 3)
solution = b * h * m
print(glue("There are ", {formatC(solution, format = "e", digits = 2)}, " different schedules possible"))

## There are 4.95e+08 different schedules possible

Question 8

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.

total_choices = choose(10, 5)
the_outcome_we_care_about = 1
types_of_trees = 2

print(glue("The probability of planting the trees next to each other with the same kind is approximately ", {round(types_of_trees * the_outcome_we_care_about / total_choices, 4)}, ". Consider you have to pick 5 of the same kind out of 10 trees in random. That's a simple combination formula, i.e. choose(10, 5) but with only one desireable outcome, i.e. SSSSS or CCCCC. Now, after you finish picking 5 of the same kind out of 10, i.e. 1 / choose(10, 5), the remaining 5 would also be the same kind. Therefore, we just need to care about finding the probability of picking 5 of the same kind out of 10 in random. Since there are two types of trees, we would add that probability twice. In this case, the probability would be 1 / choose(10, 5) + 1 / choose(10, 5)."))

## The probability of planting the trees next to each other with the same kind is approximately 0.0079. Consider you have to pick 5 of the same kind out of 10 trees in random. That's a simple combination formula, i.e. choose(10, 5) but with only one desireable outcome, i.e. SSSSS or CCCCC. Now, after you finish picking 5 of the same kind out of 10, i.e. 1 / choose(10, 5), the remaining 5 would also be the same kind. Therefore, we just need to care about finding the probability of picking 5 of the same kind out of 10 in random. Since there are two types of trees, we would add that probability twice. In this case, the probability would be 1 / choose(10, 5) + 1 / choose(10, 5).