Q1. 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 How many ways can 5 jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 2? Formula: \[ \begin{aligned} C(n, r) = \frac{n!}{(n - r)!*r!} \end{aligned} \] Number of permutation of 5 jellybeans where at most 1 are green: \[ \begin{aligned} C(5, 0)*C(7, 5) + C(5, 1)*C(7, 4) &= \frac{5!*7!}{5!*0!*2!*5!} + \frac{5!*7!}{4!*1!*3!*4!} \\ &= 196 \end{aligned} \]
Using R:
Q1 <- choose(5, 0)*choose(7, 5) + choose(5, 1)*choose(7, 4)
print(Q1)## [1] 196cat('The number of ways to withdraw 5 jellybeans with less than 2 green is', Q1)## The number of ways to withdraw 5 jellybeans with less than 2 green is 196Q2. 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 Formula: \[ \begin{aligned} C(n, r) = \frac{n!}{(n - r)!*r!} \end{aligned} \] Number of permutation of 5 members where at least 4 are representatives: \[ \begin{aligned} C(13, 4)*C(14, 1) + C(13, 5)*C(14, 0) &= \frac{13!*14!}{9!*4!*13!*1!} + \frac{13!*14!}{8!*5!*14!*0!} \\ &= 11297 \end{aligned} \]
Using R:
Q2 <- choose(13, 4)*choose(14, 1) + choose(13, 5)*choose(14, 0)
print(Q2)## [1] 11297cat('The number of ways to create a subcommitte of 5 where 4 are at least representative is', Q2)## The number of ways to create a subcommitte of 5 where 4 are at least representative is 11297Q3. 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? Formula: \[ \begin{aligned} \text{Choose without replacement} &= P(n, r) \\ &= \frac{n!}{(n - r)!} \end{aligned} \] Total permutations of 3 independent events: \[ \begin{aligned} 2^5 * 6^2 * P(52, 3) &= 2^5 * 6^2 * \frac{52!}{49!} \\ &= 152755200 \end{aligned} \]
Using R:
Q3 <- 2**5 * 6**2 * choose(52, 3)*factorial(3)
print(Q3)## [1] 152755200cat('The number of ways to create a subcommitte of 5 where 4 are at least representative is', Q3)## The number of ways to create a subcommitte of 5 where 4 are at least representative is 152755200Q4. 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. Formula: 1 - Probability of drawing 0 cards with a 3 on it \[ \begin{aligned} P(\text{at least one 3 picked}) &= 1 - \frac{\text{Total combination of not picking any 3}}{\text{Total Combination}} \\ \end{aligned} \] Probability of picking at least one 3: \[ \begin{aligned} \text{Total combination of not picking any 3} &= C(48, 3)\\ &= \frac{48!}{45!*3!} \\ &= 17296 \\ \text{Total combinaton} &= C(52, 3) \\ &= \frac{52!}{49!} \\ &= 22100 \\ P(\text{at least on 3 picked}) &= 1 - \frac{17296}{22100} \\ &= 0.2174 \end{aligned} \]
Using R:
Q4 <- 1 - round(choose(48, 3)/choose(52, 3), 4)
print(Q4)## [1] 0.2174cat('The probability of picking 3 cards where at least one card is a 3 is', Q4)## The probability of picking 3 cards where at least one card is a 3 is 0.2174Q5. 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.
Q5a. How many different combinations of 5 movies can he rent? Formula: \[ \begin{aligned} C(n, r) = \frac{n!}{(n - r)!*r!} \end{aligned} \] Total combinations of 5 movies: \[ \begin{aligned} n &= 17 \: documentaries + 14 \: mysteries \\ r &= 5 \: choises \\ C(31, 5) &= \frac{31!}{26!*5!} \\ &= 169911 \end{aligned} \]
Using R:
Q5a <- choose(31, 5)
print(Q5a)## [1] 169911cat('The number of ways to pick 5 movies is', Q5a)## The number of ways to pick 5 movies is 169911Q5b. How many different combinations of 5 movies can he rent if he wants at least one mystery? Formula: \[ \begin{aligned} \text{Total combination where at least one mystery picked} & = \text{Total combination} - \text{Total combination of picking no mystery} \end{aligned} \] Total combinations of 5 movies: \[ \begin{aligned} \text{Total combination} &= 169911 \\ \text{Total combinaiton of picking no mystery} &= C(14, 0)*C(17, 5) \\ &= \frac{14!*17!}{14!*0!*12!*5} \\ &= 6188 \\ \text{Total combination where at least one mystery picked} &= 169911 - 6188 \\ &= 163723 \end{aligned} \]
Using R:
Q5b <- choose(31, 5) - choose(14, 0)*choose(17, 5)
print(Q5b)## [1] 163723cat('The number of ways to pick 5 movies with at least one mystery', Q5b)## The number of ways to pick 5 movies with at least one mystery 163723Q6. 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. Formula: \[ \begin{aligned} C(n, r) = \frac{n!}{(n - r)!*r!} \end{aligned} \] Total number of schedule possibilities with equal number of symphonies: \[ \begin{aligned} C(4, 3)*C(104, 3)*C(17, 3) &= \frac{4!*104!*17!}{1!*3!*101!*3!*14!*3!} \\ &= 495322880 \end{aligned} \]
Using R:
Q6 <- format(choose(4, 3)*choose(104, 3)*choose(17, 3), scientific = TRUE, digits = 3)
print(Q6)## [1] "4.95e+08"cat('The number of ways to withdraw 5 jellybeans with less than 2 green is', Q6)## The number of ways to withdraw 5 jellybeans with less than 2 green is 4.95e+08Q7. 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.
Q7a. 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. Formula: \[ \begin{aligned} \text{Total combination where at most 4 nonfictions picked} & = \text{Total combination} - \text{Total combination of picking all 5 nonfiction books} \end{aligned} \] Total combinations of where at most 4 nonfictions picked: \[ \begin{aligned} \text{Total combination} &= C(24, 13) \\ &= \frac{24!}{11!*13!} \\ &= 2496144 \\ \text{Total Combination of picking all 5 nonfiction books} &= C(5, 5)*C(19, 8) \\ &= \frac{5!*19!}{0!*5!*11!*8!} \\ &= 75582 \\ \text{Total combinaiton where at most 4 nonfictions picked} &= 2496144 - 92378 \\ &= 2420562 \\ \end{aligned} \]
Using R:
Q7a <- format(choose(24, 13) - choose(55, 0)*choose(19, 8), scientific = TRUE, digits = 3)
print(Q7a)## [1] "2.42e+06"cat('The number of ways to pick novels with at most 4 nonfictions is', Q7a)## The number of ways to pick novels with at most 4 nonfictions is 2.42e+06Q7b. 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. Formula: \[ \begin{aligned} \text{Total combination where all 6 plays picked} & = C(6, 6)*C(18, 7) \end{aligned} \] Total combinations where all 6 plays picked: \[ \begin{aligned} \text{Total combination all 6 plays picked} &= \frac{6!*18!}{0!*6!*11!*7!} \\ &= 31824 \\ \end{aligned} \]
Using R:
Q7b <- format(choose(6, 6)*choose(18, 7), scientific = TRUE, digits = 3)
print(Q7b)## [1] "3.18e+04"cat('The number of ways to pick all 6 plays is', Q7b)## The number of ways to pick all 6 plays is 3.18e+04Q8. 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. Formula: \[ \begin{aligned} P(\text{Randomly, all 5 sycamores next to each other and all 5 cypress next to each other}) &= \frac{\text{Total combination where all 5 sycamores next to each other and all 5 cypriss next to each other}}{\text{Total combination of arranging all 10 trees}} \\ \text{Total combination where all 5 sycamores next to each other and all 5 cypriss next to each other} &= 2 \\ \text{Total combination of arranging all 10 trees} &= C(10, 5) \\ &= \frac{10!}{5!*5!} \\ &= 252 \end{aligned} \] Therefore; \[ \begin{aligned} P(\text{Randomly, all 5 sycamores next to each other and all 5 cypress next to each other}) &= \frac{2}{252} \\ &= 0.007936508 \end{aligned} \]
Using R:
Q8 <- round(2/choose(10, 5), 4)
print(Q8)## [1] 0.0079cat('Probability of randomly placing all 5 sycamores next to each other and all 5 cypress next to each other', Q8)## Probability of randomly placing all 5 sycamores next to each other and all 5 cypress next to each other 0.0079Q9. 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.)
Q9a. Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values. \[ \begin{aligned} E(X) &= Wins - Loss \\ Wins &= \text{Win Value} * P(\text{Picking a queen or lower}) \\ Loss &= \text{Loss Value} * P(\text{Picking higher than queen}) \\ \text{Win Value} &= 4 \\ P(\text{Picking a queen or lower}) &= \frac{44}{52} \\ \text{Loss Value} &= 16 \\ P(\text{Picking higher than queen}) &= \frac{8}{52} \end{aligned} \] Therefore; \[ \begin{aligned} E(X) &= 4 * \frac{44}{52} - 16 * \frac{8}{52} \\ &= 0.92 \end{aligned} \]
Q9b. 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. \[ \begin{aligned} \text{Total Expectation} &= 833 * 0.92 \\ &= 766.36 \end{aligned} \]