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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?

We need to find the number of ways for two possible outcomes that will fit this statement (# of green jellybeans < 2):
If the 5 jellybeans are 0 green jellybean & 5 red jellybeans, \({5}\choose{0}\)\({7}\choose{5}\)
If the 5 jellybeans are 1 green jellybean & 4 red jellybeans, \({5}\choose{1}\)\({7}\choose{4}\)

q1 = (choose(5,0) * choose(7,5)) + (choose(5,1) * choose(7,4))
cat("The # of ways 5 jellybeans can be withdrawn from the bag so that the number of green ones withdrawn will be less than 2:", q1)
## The # of ways 5 jellybeans can be withdrawn from the bag so that the number of green ones withdrawn will be less than 2: 196

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?

We need to find the number of ways for two possible outcomes that will fit this statement (# of representatives \(\ge{4}\)):
If the 5 members are 4 representatives & 1 senators, \({13}\choose{4}\)\({14}\choose{1}\)
If the 5 members are 5 representatives & 0 senators, \({13}\choose{5}\)\({14}\choose{0}\)

q2 = (choose(13,4) * choose(14,1)) + (choose(13,5) * choose(14,0))
cat("The # of ways a subcommittee of 5 can be formed if at least 4 of the members must be
representatives:", q2)
## The # of ways a subcommittee of 5 can be formed if at least 4 of the members must be
## representatives: 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?

A coin has 2 outcomes: heads or tails
So, 2 toss #1 outcomes * 2 toss #2 outcomes * 2 toss #3 outcomes * …. => \(2^5\) outcomes for 5 coin tosses
Similarly, a die has 6 outcomes So \(6^2\) outcomes for 2 die rolls
The 2 above are “with replacement” meaning we can reroll the same outcome as much as we want.
The deck is “without replacement” so, draw #1 = 52 outcomes
draw #2 = 51 outcomes
draw #3 = 50 outcomes
The outcomes combined for the card draws is 52 * 51 * 50 outcomes
Finally, we multiply all together.

q3 = (2^5)*(6^2)*(52*51*50)
cat("The # of different outcomes for 5 coin tosses, 2 six-sided die rolls, & 3 card draws:", q3)
## The # of different outcomes for 5 coin tosses, 2 six-sided die rolls, & 3 card draws: 152755200

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.

In a 52 card deck, there are 4 3-cards.
We need to find the number of ways for three possible outcomes that will fit this statement (# of 3-cards drawn \(\ge{1}\)):
If the 3 cards are 1 3-card & 2 non 3-cards, \({4}\choose{1}\)\({48}\choose{2}\)
If the 3 cards are 2 3-card & 1 non 3-cards, \({4}\choose{2}\)\({48}\choose{1}\)
If the 3 cards are 3 3-card & 0 non 3-cards, \({4}\choose{3}\)\({48}\choose{0}\)

q4 = (choose(4,1) * choose(48,2)) + (choose(4,2) * choose(48,1)) + (choose(4,3) * choose(48,0))
cat("The # of ways 3 cards can be withdrawn from the deck so that at least one of the cards drawn is a 3:", q4)
## The # of ways 3 cards can be withdrawn from the deck so that at least one of the cards drawn is a 3: 4804

To find the probability of this outcome, we divide this number by total number of outcomes \({52}\choose{3}\)

q4_prob = q4 / choose(52,3)
cat("The probability of 3 cards being withdrawn from the deck so that at least one of the cards drawn is a 3:", format(round(q4_prob, 4), nsmall = 4), "%")
## The probability of 3 cards being withdrawn from the deck so that at least one of the cards drawn is a 3: 0.2174 %

Question 5

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?

17 documentaries + 14 mysteries = 31 movies to choose from so it will be \({31}\choose{5}\)

q5_1 = choose(31,5)
cat("The # of different combinations of 5 movies he can rent:", q5_1)
## The # of different combinations of 5 movies he can rent: 169911

Step 2. How many different combinations of 5 movies can he rent if he wants at least one mystery?

In this case, it will be easier to find the total # of combinations minus no mystery movies rented.
(than if 1 mystery is rented + if 2 mysteries are rented + … + if 5 mysteries are rented)
If the 5 movies are 5 documentaries & 0 mysteries, \({17}\choose{5}\)\({14}\choose{0}\)

q5_2 =  q5_1 - (choose(17,5)*choose(14,0))
cat("The # of different combinations of 5 movies he can rent if he wants at least
one mystery:", q5_2)
## The # of different combinations of 5 movies he can rent if he wants at least
## one mystery: 163723

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.

Since it has to be an equal # of symphonies from Brahms, Haydn, and Mendelssohn:
9 symphonies / 3 = 3 from each
# of combinations of 3 out of 4 Brahms, \({4}\choose{3}\)
# of combinations of 3 out of 104 Haydn, \({104}\choose{3}\)
# of combinations of 3 out of 17 Mendelssohn, \({17}\choose{3}\)

q6 =  choose(4,3)*choose(104,3)*choose(17,3)
cat("The # of different schedules of a 9 symphonies with 4 Brahms, 104 Haydn, and 17 Mendelssohn as options:", format(q6, scientific = TRUE, digits = 3))
## The # of different schedules of a 9 symphonies with 4 Brahms, 104 Haydn, and 17 Mendelssohn as options: 4.95e+08

\(4.95e+08\) being \(4.95*10^8\)

Question 7

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.

Since there are only 5 nonfiction books, there is only 1 option that would be outside of “no more than 4” (if 5 nonfiction books are included \({5}\choose{5}\)\({19}\choose{8}\)) so we can do total # of combinations (\({24}\choose{13}\)) minus if 5 nonfiction books

q7_1 =  choose(24,13) - (choose(5,5)*choose(19,8))
cat("The # of different schedules of 13 books if he wants to include no more than 4 nonfiction books:", format(q7_1, scientific = TRUE, digits = 3))
## The # of different schedules of 13 books if he wants to include no more than 4 nonfiction books: 2.42e+06

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.

If the 13 books are 6 plays & 7 non-plays, \({6}\choose{6}\)\({18}\choose{7}\)

q7_2 =  choose(6,6)*choose(18,7)
cat("The # of different schedules of 13 books if he wants to include all 6 plays:", format(q7_2, scientific = TRUE, digits = 3))
## The # of different schedules of 13 books if he wants to include all 6 plays: 3.18e+04

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.

There are only 2 possible outcomes that this would happen:
If we planted all 5 sycamores first and then all 5 cypress after
If we planted all 5 cypress first and then all 5 sycamores after
So we can get the probability by doing 2 / total # of combinations \({10}\choose{5}\)

q8_2 =  2 / choose(10,5)
cat("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:", format(q8_2, digits = 2), "%")
## 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: 0.0079 %

Question 9

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.)
[ 2 3 4 5 6 7 8 9 10 J Q K A ]
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 44 cards \(\le{Q}\) and 8 cards > Q.
I win - I loss => ($4 * \(\frac{44}{52}\)) - ($16 * \(\frac{8}{52}\))

q9_1 =  (4 * 44/52) - (16 * 8/52)
cat("The expected value of the proposition:", format(q9_1, digits = 2))
## The expected value of the proposition: 0.92

Step 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.

Using the expected value above, we can multiply it by the number of games played (833)

q9_2 =  q9_1 * 833
cat("If you played this game 833 times, I expect to win: $", format(q9_2, digits = 5))
## If you played this game 833 times, I expect to win: $ 768.92
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Notes

I used the choose(n, k) function for all my work. For all my \({n}\choose{k}\)s, it is the formula \(\frac{n!}{k!(n-k)!}\) (for manual).
Any number choose 0 is 1. I wrote these in for just completion of the question.