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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?

Combination - order doesn’t matter here:

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

Number of green jellybeans less than 2 mean that we need to find out number of ways to withdraw 0 green jellybeans and 1 green jellybeans

0 green jellybean - all red ball are withdrawn:

\[\binom{7}{5} = \frac{7!}{5!(7-5)!} =\]

C_7_5 = factorial(7) / (factorial(5)*(factorial(7-5)))
C_7_5
## [1] 21

1 green jellybean are withdrawn;

\[\binom{7}{4}*\binom{5}{1} = \frac{7!}{4!(7-4)!}*\frac{5!}{1!(5-1)!} =\]

C_7_4 = factorial(7) / (factorial(4)*(factorial(7-4)))
C_5_1 = factorial(5) / (factorial(1)*(factorial(5-1)))

C_7_4*C_5_1
## [1] 175

Total number of ways to withdraw 5 jellybeans: 21 + 175 = 196 ways

  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?

Combination - order doesn’t matter here:

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

At least 4 of the members must be representatives mean that we need to find out number of ways of:

4 members are representatives and 1 senator

5 members are representatives and 0 senator

4 members are representatives and 1 senator:

\[\binom{13}{4}*\binom{14}{1} = \frac{13!}{4!(13-4)!}*\frac{14!}{1!(14-1)!} = \]

C_13_4 = factorial(13) / (factorial(4)*(factorial(13-4)))
C_14_1 = factorial(14) / (factorial(1)*(factorial(14-1)))

C_13_4*C_14_1
## [1] 10010

5 members are representatives and 0 senator:

\[\binom{13}{5} = \frac{13!}{5!(13-5)!} =\]

C_13_5 = factorial(13) / (factorial(5)*(factorial(13-5)))

C_13_5
## [1] 1287

Total number of ways to to form a subcommittee of 5 members with at least 4 representatives: 10,010 + 1287 = 11,297 ways

  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?

Coin is tossed 5 times - Each time there are 2 possible outcomes (Head/Tail):

22222 = 2^5 =

Six-sided die is rolled 2 times - Each time there are 6 possible outcomes:

6*6 = 6^2

A group of three cards are drawn from a standard deck of 52 cards without replacement -

\[\binom{52}{3} = \frac{52!}{3!(52-3)!} =\]

C_52_3 = factorial(52) / (factorial(3)*(factorial(52-3)))

C_52_3
## [1] 22100

Total possible different outcomes:

(2^5)*(6^2)*C_52_3
## [1] 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(at least one 3)= 1 − P(none of the cards is a 3)

P(none of the cards is a 3):

There are 4 cards that are “3”, the probability of getting non-3 card on first three draw:

1st Draw: 48/52

2nd Draw: 47/51 (51 card left after 1st draw, there are still 4 cards that are “3” in the deck, 51-4 = 47 non-3 cards left)

3rd Draw: 46/50 (50 card left after 2nd draw, there are still 4 cards that are “3” in the deck, 50-4 = 46 non-3 cards left)

Therefore, the probability that at least one of the cards drawn is a 3:

round(1-((48/52)*(47/51)*(46/50)), 4)
## [1] 0.2174
  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 movies = 14 + 17 = 31

C_31_5 = factorial(31) / (factorial(5)*(factorial(31-5)))
C_31_5
## [1] 169911

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

At least 1 mystery means we can eliminate the combinations of selecting all 5 documentaries:

Combinations of 5 documentaries:

\[\binom{17}{5} = \frac{17!}{5!(17-5)!} =\]

C_17_5 = factorial(17) / (factorial(5)*(factorial(17-5)))
C_17_5
## [1] 6188

Total combination - Combinations of 5 documentaries:

C_31_5-C_17_5 
## [1] 163723
  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.

Given that Cory needs to have an equal number of symphonies, each composer will be assigned 3 symphonies.

Brahms: \[\binom{4}{3}\] Haydn: \[\binom{104}{3}\] Mendelssohn: \[\binom{17}{3}\]

Total number of schedules:

C_4_3 = choose(4, 3)
C_104_3 = choose(104, 3)
C_17_3 = choose(17, 3)

formatC(C_4_3*C_104_3*C_17_3, format = "e")
## [1] "4.9532e+08"
  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? Express your answer in scientific notation rounding to the hundredths place.

Total books = 6 + 6 + 7 + 5 = 24

No more than 4 nonfiction books means we can use total possible outcome minus total outcomes with 5 nonfiction books

Total possible outcomes:

C_24_13 = choose(24, 13)
C_24_13
## [1] 2496144

Outcomes with 5 nonfiction books:

C_5 = choose(5, 5)*choose(19,8)
C_5
## [1] 75582

Possible reading schedules to include no more than 4 nonfiction books:

formatC(C_24_13-C_5, format = "e")
## [1] "2.4206e+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.

Outcomes with 6 play books:

# 6 choose 6 represent 6 play books, 18 choose 7 represent the outcomes for choosing (13-6) from the rest (6 + 7 + 5)
C_6 = choose(6, 6)*choose(18,7)
formatC(C_6, format = "e")
## [1] "3.1824e+04"
  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.

Ways for arranging sycamores themselves: 5! = 120 ways

Ways for cypress trees themselves: 5! = 120 ways

Total number of favorable arrangements for the sycamores and cypress trees: 5! X 5!

Ways to arrange 2 entities (sycamores and cypress trees): 2!

Total number of favorable arrangements: 2! X 5! X 5!

Total number of possible arrangements of all 10 trees: 10!

The probability is:

round((factorial(2)*factorial(5)*factorial(5))/factorial(10), 4)
## [1] 0.0079
  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.

Total number of cards: 52

Number of cards that are queen or lower: 12 (4 queens, 4 jacks, 4 kings)

Probability of drawing a queen or lower = 12 / 52

If you draw a queen or lower, you win. Othewise, you loose.

Probability of drawing a queen or lower = P(Win)

P(Loose) = 1 - P(Win)

Expected Value = P(Win)$4 + P(Loose)(-$16)

Expected Value is $ -11.38

P_Win = 12/52
P_Loose = 1-P_Win

EV = P_Win*4 + P_Loose*(-16)
EV
## [1] -11.38462

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.

Expected outcome = Expected value * Number of times played Expected outcome = -$11.39 * 833 = -$9,494.87

You would expect to loose -$9,494.87.