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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?

Step 1: 0 green jellybeans and 5 red jellybeans Number of ways = \(\binom{7}{5} = C(7, 5) = 21\)

Step 2: 1 green jellybean and 4 red jellybeans

Number of ways = \(\binom{5}{1} \times \binom{7}{4} = C(5, 1) \times C(7, 4) = 5 \times 35 = 175\)

Total ways = Step 1 + Step 2 = \(21 + 175 = 196\)

Answer: 196

  1. 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?

Step 1: 4 representatives and 1 senator

Number of ways = \(\binom{13}{4} \times \binom{14}{1} = C(13, 4) \times C(14, 1) = 715 \times 14 = 10,010\)

Step 2: 5 representatives

Number of ways = \(\binom{13}{5} = C(13, 5) = 1287\)

Total ways = Step 1 + Step 2 = \(10,010 + 1287 = 11,297\)

Answer: 11,297

  1. 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?

For each toss of a coin \(2^5 = 32\) outcomes.

For each roll of a die \(6^2 = 36\) outcomes.

For drawing cards \(\binom{52}{3}\) = \(C(52,3)\)

total outcome: \(32*36*2210 = 25459200\)

Answer: 25459200

  1. 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.

P(no 3 1st) = \(\frac{48}{52}\)

P(no 3 2nd) = \(\frac{47}{51}\)

P(no 3 3rd) = \(\frac{46}{50}\)

P(no 3) = \(\frac{48}{52} \times \frac{47}{51} \times \frac{46}{50}\)

P(at least one 3 ) = 1 - P(no 3) = 1 - \(\frac{48}{52} \times \frac{47}{51} \times \frac{46}{50}\) = 0.217

Answer: 0.217

  1. 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?

Total combinations of 5 movies out of 31 = \(C(31, 5)\) = 228,675

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

Combination of 5 movies with no mysteries = \(C(17, 5)\) = 5,157

Combination of 5 movies with at least one mystery = \(C(31, 5)\) - \(C(17, 5)\) = 223,518

Answer: 223,518

  1. 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.

Brahms = \(C(4,3)\) = 4

Haydn = \(C(104,3)\) = 18,136,160

Mendelssohn = \(C(17,3)\) = 680

Number of schedules =\(C(4,3)×C(104,3)×C(17,3)\) = \(4*18,136,160*680 = 3.35616×10^7\)

Answer: \(3.35616×10^7\)

  1. 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?

no_more_than_4nf <- choose(19, 13) -
                             choose(19, 9) * choose(5, 5)

# Step 2
# If he wants to include all 6 plays, how many different reading schedules are possible?

all_6_plays <- choose(18, 7)

no_more_than_4nf
## [1] -65246
all_6_plays
## [1] 31824
  1. 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.

5 sycamores next to each other = \(5!\)

5 cypress next to each other = \(5!\)

probability = \(\frac{5!*5!}{10!}\) = .0039

Answer:.0039

  1. 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.)

Step 1. Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.

Probability of winning = P(Q or Lower) = \(\frac{48}{52} = \frac{12}{13}\)

Probability of losing = P(nQ or Lower) = \(\frac{4}{52} = \frac{1}{13}\)

Expected Value = EV = (\(\frac{12}{13}\) * 4) - (\(\frac{1}{13}\) * 16) = \(\frac{32}{13}\) = 2.46

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.

\(\frac{32}{13}\) * 833 = 2049.18

Answer: $2049.18