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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2?

There are 196 ways.
If we take 5 jellybeans so that the number of green ones withdrawn will be less than 2, we can have either all 5 red jellybeans or 4 red and 1 green. We will consider 2 cases:
1) We choose 1 red out of 5 reds and 4 green out of 7 greens. \[(_1^5)=\frac{5!}{1!(5-1)!}=1\] \[(_4^7)=\frac{7!}{4!(7-4)!}=35\] The amount of ways we can do this is \[(_1^5)*(_4^7)=5*35=175\]

2)We choose all red jellybeans, 5 out of 7 reds. \[(_5^7)=\frac{7!}{5!(7-5)!}=21\] The final result is \[(_1^5)*(_4^7) + (_5^7) = 175 + 21= 196\] Using R:

choose(5,1) * choose(7,4) + choose(7,5)
## [1] 196

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?

There are 11,297 ways.
If we take subcommittee of 5 be formed if at least 4 of the members must be representatives, we can have either all 5 representatives or 4 representatives and 1 senator. We will consider 2 cases:
1) We have 4 representatives out of 13 and 1 senator out of 14. \[(_4^{13})=\frac{13!}{4!(13-4)!}=715\] \[(_1^{14})=\frac{14!}{1!(14-1)!}=14\] The amount of ways we can do this is \[(_4^{13})*(_1^{14})=715*14=10,010\]

  1. We have all representatives, 5 out of 13 reds. \[(_5^{13})=\frac{13!}{5!(13-5)!}=1,287\] The final result is \[(_4^{13})*(_1^{14}) + (_5^{13}) = 10,010 + 1,287=11,297\] Using R:
choose(13,4)*choose(14,1) + choose(13,5)
## [1] 11297

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?

There are 152,755,200 different outcomes. We will consider each case separately:
1) A coin is tossed 5 times, a coin as 2 sides. After it is tossed 5 times, there are 32 different outcomes. For example, all tails, all heads, 1 head/9tails, etc: \[2^5=32\] 2) A six-sided die is rolled 2 times. A die has 6 sides. After it is rolled 2 times, there are 36 different outcomes. For example, 1 dot/1 dot, 1 dot/2 dots, 1 dot/3 dots, etc: \[6^2=36\] 3) A group of three cards are drawn from a standard deck of 52 cards without replacement. \[52*51*50=132,600\] The final result is \[2^5*6^2*(_3^{52})=32*36*132,600=152,755,200\] Using R:

2^5*6^2*choose(52,3)
## [1] 25459200

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.

The probability is 0.2174. There are 4 cards with 3 on it.
1) So we can take all 3 cards with 3 on it \[(_3^{4})=\frac{4!}{3!(4-3)!}=4\] 2) 1 card is 3 and 2 cards with something else (2 cards out of 48 cards since 52 cards minus 4 cards with 3 on them) \[(_1^{4})*(_2^{48})=\frac{4!}{1!(4-1)!}*\frac{48!}{2!(48-2)!}=4,512\] 3) 2 cards are 3 and 1 card with something else \[(_2^{4})*(_1^{48})=\frac{4!}{2!(4-2)!}*\frac{48!}{1!(48-1)!}=288\]

Any 3 cards from the deck: \[(_3^{52})=\frac{52!}{3!(52-3)!}=22,100\] The final result is the amount of options with 3 in it over the total ways to take any 3 cards: \[\frac{(_3^{4})+[(_1^{4})*(_2^{48})]+[(_2^{4})*(_1^{48})]}{(_3^{52})}=\frac{4+4,512+288}{22,100}=0.2174\] Using R:

round(((choose(4,1)*choose(48,2)+choose(4,3)+choose(4,2)*choose(48,1))/choose(52,3)),4)
## [1] 0.2174

We can also use the shorter way, just remove from 1 the probability that no card among 3 chosen is 3. \[1-\frac{(_3^{48})}{(_3^{52})}=1-\frac{17,296}{22,100}=0.2174\] Using R:

round((1-choose(48,3)/choose(52,3)),4)
## [1] 0.2174

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?

169,911 different combinations.
There are 31 movies to choose from: \[(_5^{31})=\frac{31!}{5!(31-5)!}=169,911\] Using R:

choose(31,5)
## [1] 169911

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

There are **163,723*8 different combinations There are several options:
1) He can take 1 mystery out of 14 and 4 documentaries out of 17: \[(_1^{14})*(_4^{17})=\frac{14!}{1!(14-1)!}*\frac{17!}{4!(17-4)!}=33,320\] 2) 2 mysteries out of 14 and 3 documentaries out of 17: \[(_2^{14})*(_3^{17})=\frac{14!}{2!(14-2)!}*\frac{17!}{3!(17-3)!}=61,880\] 3) 3 mysteries out of 14 and 2 documentaries out of 17: \[(_3^{14})*(_2^{17})=\frac{14!}{3!(14-3)!}*\frac{17!}{2!(17-2)!}=49,504\] 4) 4 mysteries out of 14 and 1 documentaries out of 17: \[(_4^{14})*(_1^{17})=\frac{14!}{4!(14-4)!}*\frac{17!}{1!(17-1)!}=17,017\] 5) 5 mysteries out of 14 and no documentaries: \[(_5^{14})=\frac{14!}{5!(14-5)!}=2,002\] The final result is \[(_1^{14})*(_4^{17}) + (_2^{14})*(_3^{17}) + (_3^{14})*(_2^{17}) + (_4^{14})*(_1^{17}) + (_5^{14}) = 33,320 + 61,880 + 49,504 + 17,017 + 2,002 = 163,723\]

choose(14,1)*choose(17,4) + choose(14,2)*choose(17,3) + choose(14,3)*choose(17,2) + choose(14,4)*choose(17,1) + choose(14,5)
## [1] 163723

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.

There are 4.95∙10^8 different schedules possible. Since he has 9 symphonies to be played, there will be 3 symphonies of each composer.
1) 3 symphonies of Brahms out of 4 total: \[(_3^{4})=\frac{4!}{3!(4-3)!}=4\] 2) 3 symphonies of Haydn out of 104 total: \[(_3^{104})=\frac{104!}{3!(104-3)!}=182,104\] 3) 3 symphonies of Mendelssohn out of 17 total: \[(_3^{17})=\frac{17!}{3!(17-3)!}=680\] Amount of different schedules:

\[(_3^{4})*(_3^{104})*(_3^{17})=4*182,104*680=4.95\cdot10^8\] Using R:

format((choose(4,3)*choose(104,3)*choose(17,3)), scientific = TRUE, digits=3)
## [1] "4.95e+08"

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.

There are 2.42∙10^6 different reading schedules. There are several options: 1) He can have 4 nonfiction books out of 5 and 9 other books out of 19 (6 novels, 6 plays, 7 poetry books): \[(_4^{5})*(_9^{19})=\frac{5!}{4!(5-4)!}*\frac{19!}{9!(19-9)!}=461,890\] 2) 3 nonfiction books out of 5 and 10 other books out of 19: \[(_3^{5})*(_{10}^{19})=\frac{5!}{3!(5-3)!}*\frac{19!}{10!(19-10)!}=923,780\] 3) 2 nonfiction books out of 5 and 11 other books out of 19: \[(_2^{5})*(_{11}^{19})=\frac{5!}{2!(5-2)!}*\frac{19!}{11!(19-11)!}=755,820\] 4) 1 nonfiction books out of 5 and 12 other books out of 19: \[(_1^{5})*(_{12}^{19})=\frac{5!}{1!(5-1)!}*\frac{19!}{12!(19-12)!}=251,940\] 5) 0 nonfiction books and 13 other books out of 19: \[(_{13}^{19})=\frac{19!}{13!(19-13)!}=27,132\] The total amount of ways: \[461,890+923,780+755,820+251,940+27,132=2.42\cdot10^6\]

Using R:

format((choose(5,4)*choose(19,9)+choose(5,3)*choose(19,10)+choose(5,2)*choose(19,11)+choose(5,1)*choose(19,12)+choose(19,13)),scientific = TRUE, digits=3)
## [1] "2.42e+06"

Shorter way is to take all possible ways of 13 books out of 24 and subtract when there are 5 nonfiction books with 8 other books: \[(_{13}^{24})-(_{5}^{5})*(_{8}^{19})=2496144-75582= 2420562=2.42\cdot10^6\]

choose(24,13)-choose(5,5)*choose(19,8)
## [1] 2420562

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.

There will be 6 plays out of 6 and 7 other books out of 18 other books (6 novels, 7 poetry books, 5 nonfiction books): \[(_6^{6})*(_{7}^{18})=\frac{6!}{6!(6-6)!}*\frac{18!}{7!(18-7)!}=31824=3.18\cdot10^4\] Using R:

format((choose(6,6)*choose(18,7)),scientific = TRUE, digits=3)
## [1] "3.18e+04"

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.

The probability is 0.0079. There are 2 possible ways: S S S S S C C C C C or C C C C C S S S S S.
The total amount of ways to plant 10 trees: \[(_5^{10})=\frac{10!}{5!(10-5)!}=252\] The probability is \[\frac{2}{252}=0.0079\] Using R:

round((2/choose(10,5)),4)
## [1] 0.0079

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.)

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

The expected value is $0.92.
There are 52 cards, 44 of them are a queen or lower, 8 of them are above queen. \[4*\frac{44}{52}-16*\frac{8}{52}=0.92\] #### 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. Win $768.92. \[833*0.92=768.92\]